Calculus 1 - Derivative

Contents: Limit of a Function Derivative of a Function Definition Laws of Derivation Derivation of Trigonometric Function and Its Inverse L'Hôpital's Rule Sample Problems Applications of Derivative Function of Two or More Variables

Definition

The derivative of the function f with respect to x is the function f':
$f'(x)=\lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}$

Laws of Derivation

Given that c are constants, u and v are functions:

$\begin{align*} \text{1. }&f(x)=c \\ &f'(x)=0\\ \text{2. }&f(x)= cu\\ &f'(x)=c.u' \\ \text{3. }&f(x)=u\pm v \\ &f'(x)=u'\pm v'\\ \text{4. }&f(x)=u.v \\ &f'(x)=u'v+uv'\\ \text{5. }&f(x)=\frac{u}{v}\\ &f'(x)=\frac{u'v-uv'}{v^{2}}\\ \text{6. }&f(x)=g(h(x))=(g \circ h)(x)\\ &f'(x)=g'(h(x)).h'(x)\\ \text{7. }&f(x)=\log_{a}{u}\\ &f'(x)=\frac{u'}{u \ln{a}}\\ \text{8. }&f(x)=a^{u}\\ &f'(x)=(\ln{a}) a^{u} u' \\ \text{9. }&f(x)=u^{v}\\ &f'(x)=u^v (v\ln{u})' \\ \end{align*}$

Derivation of Trigonometric Functions and Its Inverse

Given that u as a function:

$\DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \begin{align*} \text{10. }&f(x)=\sin(u) \\ &f'(x)=u'.\cos(u)\\ \text{11. }&f(x)=\cos(u)\\ &f'(x)=-u'.\sin(u) \\ \text{12. }&f(x)=\tan(u) \\ &f'(x)=u'.\sec^{2}(u)\\ \text{13. }&f(x)=\csc(u)\\ &f'(x)=-u'.\csc(u).\cot(u)\\ \text{14. }&f(x)=\sec(u)\\ &f'(x)=u'.\sec(u).\tan(u)\\ \text{15. }&f(x)=\cot(u)\\ &f'(x)=-u'.\csc^{2}(u) \\ \text{16. }&f(x)=\sin^{-1}(u) \\ &f'(x)=\frac{u'}{\sqrt{1-u^2}}\\ \text{17. }&f(x)=\cos^{-1}(u)\\ &f'(x)=-\frac{u'}{\sqrt{1-u^2}}\\ \text{18. }&f(x)=\tan^{-1}(u) \\ &f'(x)=\frac{u'}{1+u^2}\\ \text{19. }&f(x)=\csc^{-1}(u)\\ &f'(x)=-\frac{u'}{|u| \sqrt{u^2-1}}\\ \text{20. }&f(x)=\sec^{-1}(u)\\ &f'(x)=\frac{u'}{|u| \sqrt{u^2-1}}\\ \text{21. }&f(x)=\cot^{-1}(u)\\ &f'(x)=\frac{-u'}{1+u^2}\\ \text{22. }&f(x)=\sinh(u) \\ &f'(x)=u'\cosh{u}\\ \text{23. }&f(x)=\cosh(u)\\ &f'(x)=u'\sinh{u}\\ \text{24. }&f(x)=\tanh(u) \\ &f'(x)=u'\sech^2{u}\\ \text{25. }&f(x)=\csch(u)\\ &f'(x)=-u'\csch{u} \coth{u}\\ \text{26. }&f(x)=\sech(u)\\ &f'(x)=-u'\sech{u} \tanh{u}\\ \text{27. }&f(x)=\coth(u)\\ &f'(x)=-u'\csc^2{u}\\ \end{align*}$

Rather than memorizing all the inverse or hyperbolic trigonometric function, we can also derive them manually if we want to. For example, we want to know the the derivative of the inverse tangent but we don't remember them. So, we can make use the implicit differentiation, and also basic trigonometric identity to find it. Recall that 1 + tan²(a) = sec²(a):
$\begin{align*} y&=\tan^{-1}(u)\\ \tan(y)&=u\\ \frac{d}{dx} \tan(y) &= \frac{d}{dx}(u)\\ \frac{dy}{dx}\sec^2{x} &= \frac{du}{dx}\\ \frac{dy}{dx} &= \frac{du}{dx} \frac{1}{\sec^2{x}}\\ &= \frac{du}{dx} \frac{1}{1+\tan^2(x)}\\ &= \frac{u'}{1+u^2}\\ \end{align*}$

sinh, cosh, tanh, csch, sech, and coth are hyperbolic trigonometric function. They are defined as such:
$\begin{align*} \sinh(x)&=\frac{e^{x}-e^{-x}}{2}\\ \cosh(x)&=\frac{e^{x}+e^{-x}}{2}\\ \tanh(x)&=\frac{\sinh(x)}{\cosh(x)}\\ \csch(x)&=\frac{1}{\sinh(x)}\\ \sech(x)&=\frac{1}{\cosh(x)}\\ \tanh(x)&=\frac{1}{\coth(x)} \end{align*}$

Though, some of their derivation isn't the same with the original trigonometric function. For example, the derivative of sec(x) is sec(x)tan(x). The derivative of sech(x) is -sec(x)tan(x).
$\begin{align*} y&=\sech(x)\\ &=\frac{1}{\cosh(x)}\\ \frac{dy}{dx}&=\frac{0(\cosh(x)) - \sinh(x)}{\cosh^2(x)}\\ &=\frac{-\sinh(x)}{\cosh^2(x)}\\ &=-\tanh(x) \frac{1}{\cosh(x)}\\ &=-\tanh(x) \sech(x)\\ &=-\sech(x) \tanh(x) \end{align*}$

L'Hôpital's Rule

Let f and g are differentable on an open interval L that conains a, with the possible exception of a itself, and g(x) = 0 for all x in L. If $\lim_{x \rightarrow a}\frac{f(x)}{g(x)}$ is an indeterminate form, then: $\lim_{x \rightarrow a}\frac{f(x)}{g(x)} = \lim_{x \rightarrow a}\frac{f'(x)}{g'(x)}$.

Sample Problems

Problem 1: Derivative of a Function
Given f(x), find the first derivative of the given function using the definition of derivative.
m and n are functions.

1. f(x)=log(x)
2. f(x)=cot(x)
3. f(x)=m(x).n(x)
4. f(x)=cos(4x)
5. f(x)=m(n(x))

Sub-Problem A
The following laws of logarithm are used:
$\begin{align*} \log_{a}b^n&=n\log_{a}b\\ \log_{a}b\log_{b}c&=\log_{a}c\\ \log_{a}b - \log_{a}c &=\log_{a}\left ( \frac{b}{c}\right ) \end{align*}$

Let us begin:
$\begin{align*} f(x)&=\log{x}\\ f'(x)&=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}\\ &=\lim_{h\rightarrow 0}\frac{\log{(x+h)-\log{x}}}{h}\\ &=\lim_{h\rightarrow 0}\frac{1}{h}\log{\left ( 1+\frac{h}{x} \right )}\\ \end{align*}$

Let us define a new variable, such that $u=\frac{x}{h}$, and the limit will be converted accordingly. If h approaches 0, then u would approach infinity:
$\begin{align*} f'(x)&=\lim_{h\rightarrow 0}\frac{1}{h}\log{\left ( 1+\frac{h}{x} \right )}\\ &=\lim_{u\rightarrow \infty}\frac{x}{u}\log{\left ( 1+\frac{h}{x} \right )}\\ &=\lim_{u\rightarrow \infty}\frac{1}{u}\log{\left ( 1+\frac{h}{x} \right )^x}\\ \end{align*}$

The logarithm form above looks familiar, and similiar to the limit definition of euler's number (Formula 12: Derivative). So, the log form can be changed to ln:
$\begin{align*} f'(x)&=\lim_{u\rightarrow \infty}\frac{1}{x}\log{\left ( 1+\frac{1}{u} \right )^u}\\ &=\lim_{u\rightarrow \infty}\frac{1}{x}\times\log_{10}e\times\ln{\left ( 1+\frac{1}{u} \right )^u}\\ &=\lim_{u\rightarrow \infty}\frac{1}{x}\times\frac{1}{\ln{10}}\times\ln{\left ( 1+\frac{1}{u} \right )^u}\\ &=\frac{1}{x}\times\frac{1}{\ln{10}}\times{\ln{e}}\\ &=\frac{1}{x}\times\frac{1}{\ln{10}}\times{1}\\ &=\frac{1}{x \ln{10}} \end{align*}$

Sub-Problem B
The following identity is used:
$\begin{align*} \tan{(A+B)}&=\frac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}\\ 1+\cot^2{x}&=\csc^2{x}\\ \end{align*}$

So:
$\begin{align*} f(x)&=\cot{x}\\ f'(x)&=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}\\ &=\lim_{h\rightarrow 0}\frac{\cot{(x+h)-\cot{(x)}}}{h}\\ &=\lim_{h\rightarrow 0}\frac{1}{h}\left ( \frac{1}{\tan{(x+h)}}-\frac{1}{\tan{x}} \right )\\ &=\lim_{h\rightarrow 0}\frac{1}{h}\left ( \frac{1-\tan{x}\tan{h}}{\tan{x}+\tan{h}}-\frac{1}{\tan{x}} \right )\\ &=\lim_{h\rightarrow 0}\frac{1}{h}\left ( \frac{1-\tan{x}\tan{h} - \left (1 +\frac{\tan{h}}{\tan{x}} \right )}{\tan{x}+\tan{h}} \right )\\ &=\lim_{h\rightarrow 0}\frac{1}{h}\left ( \frac{-\tan{x}\tan{h} - \frac{\tan{h}}{\tan{x}}}{\tan{x}+\tan{h}} \right )\\ &=\lim_{h\rightarrow 0}\left ( \frac{-\frac{\tan{x}\tan{h}}{h} - \frac{\tan{h}}{h\tan{x}}}{\tan{x}+\tan{h}} \right )\\ &=\left (\frac{-\tan{x}-\frac{1}{\tan{x}}}{\tan{x}} \right )\\ &=-1-\cot^2{x}=-(1+\cot^2{x})\\ &=-\csc^2{x}\\ \end{align*}$

Sub-Problem C
$\begin{align*} f(x)&=m(x)n(x)\\ f'(x)&=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}\\ &=\lim_{h\rightarrow 0}\frac{m(x+h)n(x+h)-m(x)n(x)}{h}\\ &=\lim_{h\rightarrow 0}\frac{m(x+h)n(x+h)-m(x)n(x+h)+m(x)n(x+h)-m(x)n(x)}{h}\\ &=\lim_{h\rightarrow 0}\frac{n(x+h) [m(x+h)-m(x)]+m(x)[n(x+h)-n(x)]}{h}\\ &=\lim_{h\rightarrow 0}\frac{n(x+h) [m(x+h)-m(x)]}{h}+\lim_{h\rightarrow 0}\frac{m(x)[n(x+h)-n(x)]}{h}\\ &=\left ( \lim_{h\rightarrow 0}\frac{m(x+h)-m(x)}{h}\right)\left ( \lim_{h\rightarrow 0}n(x+h)\right)+\left ( \lim_{h\rightarrow 0}m(x)\right)\left ( \lim_{h\rightarrow 0}\frac{n(x+h)-n(x)}{h}\right)\\ &=m'(x)n(x)+m(x)n'(x)\\ \end{align*}$

Sub-Problem D
$\begin{align*} f(x)&=\cos{4x}\\ f'(x)&=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}\\ &=\lim_{h\rightarrow 0}\frac{\cos{(4x+4h)}-\cos{(4x)}}{h}\\ &=\lim_{h\rightarrow 0}\frac{\cos{(4x)\cos{(4h)}-\sin{(4x)\sin{(4h)}}}-\cos{(4x)}}{h}\\ &=\lim_{h\rightarrow 0}\frac{\cos{(4x)\left [\cos{(4h)} -1\right ]-\sin{(4x)\sin{(4h)}}}}{h}\\ &=\cos{(4x)}\lim_{h\rightarrow 0}\frac{\left [\cos{(4h)} -1\right ]}{h}-\lim_{h\rightarrow 0}\frac{\left [\sin{(4x)}\sin{(4h)}\right ]}{h}\\ &=-4\sin{(4x)} \end{align*}$

Sub-Problem E
$\begin{align*} f(x)&=m(n(x))\\ f'(x)&=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}\\ &=\lim_{h\rightarrow 0}\frac{m(n(x+h))-m(n(x))}{h}\\ &=\lim_{h\rightarrow 0}\frac{m(n(x+h))-m(n(x))}{h}\times\frac{n(x+h)-n(x)}{n(x+h)-n(x)}\\ \end{align*}$

Let us define the variable k, such that: $k=n(x+h)-n(x)$. So, if h approaches 0, then k also approaches 0. Then:

$\begin{align*} f'(x)&=\lim_{h\rightarrow 0}\frac{m(n(x+h))-m(n(x))}{h}\times\frac{n(x+h)-n(x)}{n(x+h)-n(x)}\\ &=\lim_{h\rightarrow 0}\frac{m(n(x+h))-m(n(x))}{n(x+h)-n(x)}\times\frac{n(x+h)-n(x)}{h}\\ &=\lim_{h\rightarrow 0}\left (\frac{m(n(x+h))-m(n(x))}{n(x+h)-n(x)} \right ).\lim_{h\rightarrow 0}\left ( \frac{n(x+h)-n(x)}{h} \right )\\ &=\lim_{k\rightarrow 0}\left (\frac{m(n(x)+k)-m(n(x))}{k} \right ).\,n'(x)\\ &=m'(n(x)).n'(x) \end{align*}$

Problem 2: Derivative of a Function
Given f(x), find the first and second derivative of the function:
$\begin{align*} \text{A. } &f(x)=x^3-x^2-x+1\\ \text{B. } &f(x)=\frac{2x^3-5x^2-3x+1}{5x^2}\\ \text{C. } &f(x)=2x^2(4x+1)^5\\ \text{D. } &f(x)=\frac{2x+1}{\sqrt{x^2-1}}\\ \text{E. } &f(x)=\sqrt{\frac{x+1}{x-1}}\\ \text{F. } &f(x)=e^x(x^2+3)(x^3+2)\\ \text{G. } &f(x)=\log{x}+\ln{x}\\ \text{H. } &f(x)=\frac{3^x}{5^x+1}\\ \text{I. } &f(x)=(4x-2)^{\ln{x}}\\ \text{J. } &f(x)=e^{\sqrt{x}}\log{x}\\ \end{align*}$

Sub-Problem A
- First Derivative:
$\begin{align*} f(x)&=x^3-x^2-x+1\\ f'(x)&=3x^2-2x-1 \\ \end{align*}$

- Second Derivative:
$\begin{align*} f'(x)&=3x^2-2x-1\\ f''(x)&=6x-2\\ \end{align*}$

- Second Derivative:
$\begin{align*} f'(x)&=\frac{2}{5}+\frac{3}{5}x^{-2}-\frac{2}{5}x^{-3}\\ f''(x)&=-\frac{6}{5}x^{-3}+\frac{6}{5}x^{-4}\\ \end{align*}$

Sub-Problem C
- First Derivative:
Let us define a variable u and v, so that:
$\begin{align*} u&=2x^2\\ u'&=4x\\ v&=(4x+1)^5\\ v'&=20(4x+1)^4 \end{align*}$

We can proceed using the product rule:
$\begin{align*} f(x)&=2x^2(4x+1)^5\\ &uv\\ f'(x)&=u'v+uv'\\ &=4x(4x+1)^5+40x^2(4x+1)^4 \end{align*}$

- Second Derivative:
Let us define a variable u and v, so that:
$\begin{align*} u&=(4x+1)^4\\ u'&=16(4x+1)^3\\ v&=56x^2+4x\\ v'&=112x+4 \end{align*}$

We can proceed using the product rule:
$\begin{align*} f'(x)&=4x(4x+1)^5+40x^2(4x+1)^4\\ &=(4x+1)^4(4x(4x+1)+40x^2)\\ &=(4x+1)^4(56x^2+4x)\\ &=uv\\ f''(x)&=u'v+uv'\\ &=16(4x+1)^3(56x^2+4x)+(4x+1)^4(112x+4) \end{align*}$

Sub-Problem D
- First Derivative:
Let us define a variable u and v, so that:
$\begin{align*} u&=2x+1\\ u'&=2\\ v&=\sqrt{x^2-1}\\ v'&=\frac{x}{\sqrt{x^2-1}}\\ \end{align*}$

We can proceed using the $f'(x) = \frac{u'v-uv'}{v^2}$ form:
$\begin{align*} f(x)&=\frac{2x+1}{\sqrt{x^2-1}}\\ &=\frac{u}{v}\\ f'(x)&=\frac{u'v-uv'}{v^2}\\ &=\frac{2\sqrt{x^2-1}-\frac{x(2x+1)}{\sqrt{x^2-1}}}{x^2-1}\\ &=\frac{2\sqrt{x^2-1}}{x^2-1}-\frac{2x^2+x}{\sqrt{(x^2-1)^3}}\\ &=\frac{2(x^2-1)-2x^2-x}{\sqrt{(x^2-1)^3}}\\ &=\frac{-x-2}{\sqrt{(x^2-1)^3}} \end{align*}$

- Second Derivative:
Let us define another u and v:
$\begin{align*} u&=-x-2\\ u'&=-1\\ v&=(x^2-1)^{\frac{3}{2}}\\ v'&=3x(x^2-1)^{\frac{1}{2}} \end{align*}$

We can proceed using the $f'(x) = \frac{u'v-uv'}{v^2}$ form:
$\begin{align*} f'(x)&=\frac{-x-2}{\sqrt{(x^2-1)^3}}\\ &=\frac{u}{v}\\ f''(x)&=\frac{u'v-uv'}{v^2}\\ &=\frac{-(x^2-1)^{\frac{3}{2}}-3x(-x-2)(x^2-1)^{\frac{1}{2}}}{(x^2-1)^3}\\ &=\frac{(\sqrt{x^2-1})( -(x^2-1) -3x(-x-2))}{(x^2-1)^3}\\ &=\frac{(\sqrt{x^2-1})( -x^2+1+3x^2+6x)}{(x^2-1)^3}\\ &=\frac{(\sqrt{x^2-1})( 2x^2+6x+1)}{(x^2-1)^3}\\ \end{align*}$

Sub-Problem E
- First Derivative:
Let us define a variable u and v, so that:
$\begin{align*} u&=\sqrt{x+1}\\ u'&=\frac{1}{2\sqrt{x+1}}\\ v&=\sqrt{x-1}\\ v'&=\frac{1}{2\sqrt{x-1}}\\ \end{align*}$

We can proceed using the quotient rule:
$\begin{align*} f(x)&=\sqrt{\frac{x+1}{x-1}}\\ &=\frac{\sqrt{x+1}}{\sqrt{x-1}}\\ &=\frac{u}{v}\\ f'(x)&=\frac{u'v-uv'}{v^2}\\ &=\frac{\frac{\sqrt{x-1}}{2\sqrt{x+1}}-\frac{\sqrt{x+1}}{2\sqrt{x-1}}}{x-1}\\ &=\frac{\sqrt{x-1}}{2\sqrt{x+1}(x-1)}-\frac{\sqrt{x+1}}{2\sqrt{x-1}(x-1)}\\ &=\frac{-2}{2\sqrt{(x+1)(x-1)}(x-1)}\\ &=-\frac{1}{(x-1)^2\sqrt{\frac{x+1}{x-1}}}\\ \end{align*}$

- Second Derivative:
Let us define a variable u, v, m and n, so that:
$\begin{align*} m&=(x-1)^2\\ m'&=2(x-1)\\ n&=\sqrt{\frac{x+1}{x-1}}\\ n'&=-\frac{1}{(x-1)^2\sqrt{\frac{x+1}{x-1}}}\\ u&=1\\ u'&=0\\ v&=mn\\ v'&=m'n+mn'\\ &=2(x-1)\sqrt{\frac{x+1}{x-1}}-\frac{(x-1)^2}{(x-1)^2\sqrt{\frac{x+1}{x-1}}}\\ &=2(x-1)\sqrt{\frac{x+1}{x-1}}-\sqrt{\frac{x-1}{x+1}} \end{align*}$

Then, using the quotient rule:
$\begin{align*} f'(x)&=-\frac{1}{(x-1)^2\sqrt{\frac{x+1}{x-1}}}\\ &=-\frac{u}{v}\\ f''(x)&=-\frac{u'v-uv'}{v^2}\\ &=\frac{2(x-1)\sqrt{\frac{x+1}{x-1}}-\sqrt{\frac{x-1}{x+1}}}{(x-1)^4 \left (\frac{x+1}{x-1} \right )}\\ \end{align*}$

Sub-Problem F
- First Derivative:
Let us define a variable u and v, so that:
$\begin{align*} u&=e^x\\ u'&=e^x\\ v&=(x^5+3x^3+2x^2+6)\\ v'&=5x^4+9x^2+4x \end{align*}$

We can proceed using the product rule:
$\begin{align*} f(x)&=e^x(x^2+3)(x^3+2)\\ &=e^x(x^5+3x^3+2x^2+6)\\ &=uv\\ f'(x)&=u'v+uv'\\ &=e^x(x^5+3x^3+2x^2+6)+e^x(5x^4+9x^2+4x)\\ &=e^x(x^5+5x^4+3x^3+11x^2+4x+6) \end{align*}$

- Second Derivative:
Let us define another variable u and v, so that:
$\begin{align*} u&=e^x\\ u'&=e^x\\ v&=(x^5+5x^4+3x^3+11x^2+4x+6)\\ v'&=5x^4+20x^3+9x^2+22x+4 \end{align*}$

We can proceed using the product rule:
$\begin{align*} f'(x)&=e^x(x^5+5x^4+3x^3+11x^2+4x+6)\\ &=uv\\ f''(x)&=u'v+uv'\\ &=e^x(v+v')\\ &=e^x(x^5+5x^4+3x^3+11x^2+4x+6+5x^4+20x^3+9x^2+22x+4)\\ &=e^x(x^5+10x^4+23x^3+20x^2+26x+10)\\ \end{align*}$

Sub-Problem G
- First Derivative:
$\begin{align*} f(x)&=\log{x}+\ln{x}\\ f'(x)&=\frac{1}{x \ln{10}}+\frac{1}{x} \end{align*}$

- Second Derivative:
$\begin{align*} f'(x)&=\frac{1}{x \ln{10}}+\frac{1}{x}\\ &=(x\ln{10})^{-1}+x^{-1}\\ f''(x)&=-\ln{10}(x \ln{10})^{-2} - x^{-2} \end{align*}$

Sub-Problem H
- First Derivative:
Let us define the variable u and v, so that:
$\begin{align*} u&=3^x\\ u'&=3^x \ln{3} \\ v&=5^x+1\\ v'&=5^x \ln{5} \end{align*}$

We can proceed using the quotient rule:
$\begin{align*} f(x)&=\frac{3^x}{5^x+1}\\ &=\frac{u}{v}\\ f'(x)&=\frac{u'v-uv'}{v^2}\\ &=\frac{3^x (5^x+1) \ln{3} - 5^x (3^x) \ln{5}}{(5^x+1)^2}\\ &=\frac{3^x \left [ (5^x+1)\ln{3} - 5^x \ln{5}\right ]}{(5^x+1)^2}\\ \end{align*}$

- Second Derivative:
Let us define the variables u, and v, so that:
$\begin{align*} u&=\frac{3^x}{5^x+1} \\ u'&=\frac{3^x \left [ (5^x+1)\ln{3} - 5^x \ln{5}\right ]}{(5^x+1)^2}\\ &=\frac{3^x\ln{3}}{(5^x+1)}-\frac{15^x\ln{5}}{(5^x+1)^2}\\ v&=\frac{(5^x+1)\ln{3}-5^x\ln{5}}{5^x+1}\\ &=\ln{3}-\frac{5^x\ln{5}}{5^x+1}\\ v'&=-\frac{\left ( 5^x(\ln{5})^2(5^x+1)\right ) + 5^{2x}(\ln{5})^2}{(5^x+1)^2}\\ &=-\frac{5^x(\ln{5})^2}{(5^x+1)^2}\\ \end{align*}$

Product rule:

$\begin{align*} f'(x)&=\frac{3^x \left [ (5^x+1)\ln{3} - 5^x \ln{5}\right ]}{(5^x+1)^2}\\ &=\frac{3^x}{5^x+1} \times \frac{(5^x+1)\ln{3}-5^x\ln{5}}{5^x+1}\\ &=uv\\ f''(x)&=u'v+uv'\\ &=\left(\frac{3^x\ln{3}}{(5^x+1)}-\frac{15^x\ln{5}}{(5^x+1)^2} \right )\left (\ln{3}-\frac{5^x\ln{5}}{5^x+1} \right)+\left(\frac{3^x}{5^x+1} \right )\left(-\frac{5^x(\ln{5})^2}{(5^x+1)^2} \right )\\ \end{align*}$

Sub-Problem I
- First Derivative:
Through Formula 10: Derivative, or the Generalized Power Rule:
$\begin{align*} f(x)&=(4x-2)^{\ln{x}}\\ f'(x)&=(4x-2)^{\ln{x}}\left (\ln{(4x-2)} \ln{(x)} \right )'\\ &=(4x-2)^{\ln{x}}\left (\frac{4\ln{x}}{4x-2}+\frac{\ln{(4x-2)}}{x} \right )\\ &=(4x-2)^{\ln{x}}\left (\frac{2\ln{x}}{2x-1}+\frac{\ln{(4x-2)}}{x} \right )\\ \end{align*}$

- Second Derivative:
Let us define the variables u, v, m, n, p and q, so that:
$\begin{align*} u&=(4x-2)^{\ln{x}}\\ u'&=(4x-2)^{\ln{x}}\left (\frac{4\ln{x}}{4x-2}+\frac{\ln{(4x-2)}}{x} \right )\\ v&=\left (\frac{2\ln{x}}{2x-1}+\frac{\ln{(4x-2)}}{x} \right )\\ &=\left (\frac{m}{n} + \frac{p}{q} \right ) \\ m&=2\ln{x}\\ m'&=\frac{2}{x}\\ n&=2x-1\\ n'&=2\\ p&=\ln{(4x-2)}\\ p'&=\frac{4}{4x-2}\\ q&=x\\ q'&=1\\ v'&=\left (\frac{m'n-mn'}{n^2} + \frac{p'q-pq'}{q^2} \right ) \\ &=\left (\frac{\frac{4x-2}{x}-4\ln{x}}{(2x-1)^2}+\frac{\frac{4x}{4x-2}-\ln{(4x-2)}}{x^2} \right )\\ \end{align*}$

We can proceed using the product rule:

$\begin{align*} f'(x)&=(4x-2)^{\ln{x}}\left (\frac{2\ln{x}}{2x-1}+\frac{\ln{(4x-2)}}{x} \right )\\ &=uv\\ f''(x)&=u'v+uv'\\ &=(4x-2)^{\ln{x}}\left ( \left (\frac{2\ln{x}}{2x-1}+\frac{\ln{(4x-2)}}{x} \right )^2+\left (\frac{\frac{4x-2}{x}-4\ln{x}}{(2x-1)^2}+\frac{\frac{2x}{2x-1}-\ln{(4x-2)}}{x^2} \right ) \right ) \end{align*}$

Sub-Problem J
- First Derivative:
Let us define the variables u, and v so that:
$\begin{align*} u&=e^{\sqrt{x}}\\ u'&=\frac{e^{\sqrt{x}}}{2\sqrt{x}}\\ v&=\log{x}\\ v'&=\frac{1}{x\ln{10}}\\ \end{align*}$

We can proceed using the product rule:
$\begin{align*} f(x)&=e^{\sqrt{x}}\log{x}\\ &=uv\\ f'(x)&=u'v+uv'\\ &=\frac{e^{\sqrt{x}}\log{x}}{2\sqrt{x}}+\frac{e^{\sqrt{x}}}{x\ln{10}} \end{align*}$

- Second Derivative:
Let us define the variables m, n, p and q so that:
$\begin{align*} m&=e^{\sqrt{x}}\log{x}\\ m'&=\frac{e^{\sqrt{x}}\log{x}}{2\sqrt{x}}+\frac{e^{\sqrt{x}}}{x\ln{10}}\\ n&=2\sqrt{x}\\ n'&=\frac{1}{\sqrt{x}}\\ p&=e^{\sqrt{x}}\\ p'&=\frac{e^{\sqrt{x}}}{2\sqrt{x}}\\ q&=x\ln{10}\\ q'&=\ln{10}\\ \end{align*}$

We can proceed using the quotient rule:

$\begin{align*} f'(x)&=\frac{e^{\sqrt{x}}\log{x}}{2\sqrt{x}}+\frac{e^{\sqrt{x}}}{x\ln{10}}\\ &=\frac{m}{n}+\frac{p}{q}\\ f''(x)&=\frac{m'n-mn'}{n^2}+\frac{p'q-pq'}{q^2}\\ &=\frac{1}{4x}\left({e^{\sqrt{x}}\log{x}}+\frac{e^{\sqrt{x}}2\sqrt{x}}{x\ln{10}} -\frac{e^{\sqrt{x}}\log{x}}{\sqrt{x}} \right )+\frac{1}{(x\ln{10})^2}\left(\frac{e^{\sqrt{x}}x\ln{10}}{2\sqrt{x}}-e^{\sqrt{x}}\ln{10} \right )\\ \end{align*}$

Problem 3: Derivative of a Function
Find $\frac{dy}{dx}$ through implicit differentation:
$\begin{align*} \text{A. } &x^4+y^3=25\\ \text{B. } &y=\sin^{-1}(x)\\ \text{C. } &y=x^{x^{x^{...}}}\\ \text{D. } &\ln{\sqrt{xy}}+\sin{xy}+5x^2y^3=0\\ \text{E. } &x^3y^2+\tan{y}=x^2-4y\\ \end{align*}$

Sub-Problem A
$\begin{align*} x^4+y^3&=25\\ \frac{d}{dx}(x^4+y^3)&=\frac{d}{dx}25\\ \frac{d}{dx}x^4 + \frac{d}{dx}y^3 &= 0\\ 4x^3 + \frac{dy}{dx} 3y^2 &= 0\\ \frac{dy}{dx} &= \frac{-4x^3}{3y^2}\\ \end{align*}$

Sub-Problem B
Recall that sin²+cos²=1:
$\begin{align*} y&=\sin^{-1}(x)\\ \sin(y)&=x\\ \frac{d}{dx}\sin(y)&=\frac{d}{dx}x\\ \frac{dy}{dx}\cos(y)&=1\\ \frac{dy}{dx}&=\frac{1}{\cos(y)}\\ \frac{dy}{dx}&=\frac{1}{\sqrt{1-\sin^2(y)}}\\ &=\frac{1}{\sqrt{1-x^2}}\\ \end{align*}$

Sub-Problem C
The given equation, can be changed to the following:
$\begin{align*} y&=x^{x^{x^{...}}}\\ y&=x^y\\ \ln{(y)}&=\ln{(x^y)}\\ \ln{(y)}&=y\ln{(x)}\\ \frac{d}{dx} \left ( \ln{(y)} \right ) &=\frac{d}{dx} \left ( y\ln{(x)} \right )\\ \frac{1}{y} \frac{dy}{dx} &= \frac{dy}{dx}\ln{(x)} + \frac{y}{x}\\ \frac{1}{y} \frac{dy}{dx} - \frac{dy}{dx}\ln{(x)} &= \frac{y}{x}\\ \frac{dy}{dx} \left ( \frac{1-y\ln{(x)}}{y}\right ) &= \frac{y}{x}\\ \frac{dy}{dx} &= \frac{y^2}{x-xy\ln{(x)}}\\ \end{align*}$

Sub-Problem D

$\begin{align*} \ln{\sqrt{xy}}+\sin{xy}+5x^2y^3&=0\\ \frac{d}{dx} (\ln{\sqrt{xy}}+\sin{xy}+5x^2y^3) &= 0\\ \frac{\frac{1}{2}(y+x\frac{dy}{dx})(\frac{1}{\sqrt{xy}})}{\sqrt{xy}}+(y+x\frac{dy}{dx})\cos{xy}+10xy^3+15x^2y^2\frac{dy}{dx}&=0\\ \frac{1}{2x} + \frac{1}{2y}\frac{dy}{dx} + y \cos{xy} + x \cos{xy} \frac{dy}{dx} + 10xy^3 + 15x^2y^2 \frac{dy}{dx} &= 0 \\ \frac{1}{2y}\frac{dy}{dx} + x \cos{xy} \frac{dy}{dx} + 15x^2y^2 \frac{dy}{dx} &= -\frac{1}{2x} - y \cos{xy}-10xy^3 \\ \frac{dy}{dx} &= \frac{ -\frac{1}{2x} - y \cos{xy}-10xy^3 }{\frac{1}{2y}+x\cos{xy} + 15x^2y^2}\\ &= \frac{ -y - 2xy^2 \cos{xy}-20x^2y^4 }{x+2x^2y\cos{xy} + 30x^3y^3} \end{align*}$

Sub-Problem E
$\begin{align*} x^3y^2+\tan{y}&=x^2-4y\\ \frac{d}{dx} (x^3y^2 + \tan{y}) &= \frac{d}{dx} (x^2-4y)\\ 3x^2y^2 + 2x^3y\frac{dy}{dx} + \frac{dy}{dx} \sec^2{y} &= 2x-4\frac{dy}{dx}\\ 2x^3y\frac{dy}{dx} + \frac{dy}{dx} \sec^2{y} + 4\frac{dy}{dx} &= -3x^2y^2 + 2x\\ \frac{dy}{dx} &= -\frac{-3x^2y^2 + 2x}{2x^3y+\sec^2{y}+4}\\ \end{align*}$

Problem 4: Derivative of a Function
Given f(x), find the first and second derivative of the function:

$\begin{align*} \text{A. } &f(x)=(\sin{(x)})^{\sin{(x)}}\\ \text{B. } &f(x)=4\sin^2{(x)}-3\tan{(x)}\\ \text{C. } &f(x)=3\cos^2{(x)}\sin^2{(3x)}\\ \text{D. } &f(x)=\frac{2\tan{(4x)}}{2-\sec^2{(4x)}}\\ \text{E. } &f(x)=\ln\left(\sin^{2}\left(\frac{1}{\tan^{2}\left(x\right)-1}\right)\right)\\ \text{F. } &f(x)=\tan^{-1}\left ( \frac{x-1}{x+1} \right )\\ \text{G. } &f(x)=\ln\left(\sin^{-1}\left(\frac{x}{2}\right)\right)-\sin^{-1}\left(x\right)\\ \text{H. } &f(x)=\frac{\sin^{-1}\left(x\right)}{\cos^{-1}\left(x\right)}\\ \text{I. } &f(x)=x\csc^{-1}\left(\frac{\pi}{x}\right) \\ \text{J. } &f(x)=e^{\cot^{-1}\left(x\right)} \\ \end{align*}$

Sub-Problem A
- First Derivative:
Through Formula 10: Derivative, or the Generalized Power Rule, and also the Product Rule:
$\begin{align*} u&=\sin{(x)}\\ v&=\ln{\sin{(x)}}\\ u'&=\cos{(x)}\\ v'&=\frac{\cos{(x)}}{\sin{(x)}}\\ u'v+uv'&=\cos{(x)}\ln{\sin{(x)}}+\sin{(x)}\left ( \frac{\cos{(x)}}{\sin{(x)}} \right ) \end{align*}$

$\begin{align*} f(x)&=(\sin{(x)})^{\sin{(x)}}\\ f'(x)&=(\sin{(x)})^{\sin{(x)}}(\sin{(x)}\ln(\sin{(x)}))'\\ &=(\sin{(x)})^{\sin{(x)}}(\cos{(x)}\ln(\sin{(x)})+\cos{(x)})\\ \end{align*}$

- Second Derivative:
Using product rule:

$\begin{align*} f'(x)&=(\sin{(x)})^{\sin{(x)}}(\cos{(x)}\ln(\sin{(x)})+\cos{(x)})\\ \\ u&=(\sin{(x)})^{\sin{(x)}}\\ v&=(\cos{(x)}\ln(\sin{(x)})+\cos{(x)})\\ u'&=(\sin{(x)})^{\sin{(x)}}(\cos{(x)}\ln(\sin{(x)}))+\cos{(x)})\\ v'&=\left (-\sin{(x)}\ln{(\sin{(x)})} + \frac{\cos^2{(x)}}{\sin{(x)}} -\sin{(x)}\right )\\ \\ f''(x)&=(\sin{(x)})^{\sin{(x)}} \left ( (\cos{(x)}\ln(\sin{(x)})+\cos{(x)})^2-\sin{(x)}\ln{(\sin{(x)})} + \frac{\cos^2{(x)}}{\sin{(x)}} -\sin{(x)} \right )\\ \end{align*}$

Sub-Problem B
- First Derivative:
$\begin{align*} f(x)&=4\sin^2{(x)}-3\tan{(x)}\\ &=4\sin{(x)}\sin{(x)}-3\tan{(x)}\\ f'(x)&=4(\cos{(x)}\sin{(x)}+\sin{(x)}\cos{(x)})-3\sec^2{(x)}\\ &=4(2\sin{(x)}\cos{(x)})-3\sec^2{(x)}\\ &=4\sin{2x}-3\sec^2{(x)}\\ \end{align*}$

- Second Derivative:
Let us define a variable u and v, so that:
$\begin{align*} u=v&=\sec{(x)}\\ u'=v'&=\sec{(x)}\tan{(x)}\\ u'v+uv'&=2\sec^2{(x)}\tan{(x)}\\ \\ f'(x)&=4\sin{(2x)}-3\sec^2{(x)}\\ &=4\sin{(2x)}-3uv\\ f''(x)&=8\cos{(2x)}-6\sec^2{(x)}\tan{(x)}\\ \end{align*}$

Sub-Problem C
The following identity is used:
$\begin{align*} &2\cos{A}\cos{B}=\cos{(A+B)}+\cos{(A-B)}\\ &2\sin{A}\sin{B}=-\left \{ \cos{(A+B)}-\cos{(A-B)} \right \} \\ &2\sin{A}\cos{B}=\sin{(A+B)}+\sin{(A-B)}\\ &2\cos{A}\sin{B}=\sin{(A+B)}-\sin{(A-B)}\\ \end{align*}$

- First Derivative:

$\begin{align*} f(x)&=3\cos^2{(x)}\sin^2{(3x)}\\ &=3(\cos{(x)}\cos{(x)})(\sin{(3x)}\sin{(3x)})\\ &=3\left (\frac{\cos{(2x)}+\cos{(0)}}{2}\right)\left (\frac{-\cos{(6x)}+\cos{(0)}}{2} \right)\\ &=3\left (\frac{\cos{(2x)}+1}{2}\right)\left (\frac{-\cos{(6x)}+1}{2} \right)\\ &=\frac{3}{4}(\cos{(2x)+1})(-\cos{(6x)+1})\\ &=\frac{3}{4}uv\\ \\ u&=\cos{(2x)}+1\\ u'&=-2\sin{(2x)}\\ v&=-\cos{(6x)}+1\\ v'&=6\sin{(6x)}\\ \\ f(x)&=\frac{3}{4}uv\\ f'(x)&=\frac{3}{4}(uv)'\\ &=\frac{3}{4}(u'v+uv')\\ &=\frac{3}{4}((-2\sin{(2x)}.(-\cos{(6x)}+1)+(\cos{(2x)}+1)).(6\sin{(6x)}))\\ &=\frac{3}{4}(((2\sin{(2x)}\cos{(6x)}-2\sin{(2x)})+(6\sin{(6x)}\cos{(2x)}+6\sin{(6x)}))\\ &=\frac{3}{4}((\sin{(8x)}+\sin{(-4x)}-2\sin{(2x)})+(3(\sin{(8x)+\sin{(4x)}})+6\sin{(6x)}))\\ &=\frac{3}{4}((\sin{(8x)}-\sin{(4x)}-2\sin{(2x)})+(3\sin{(8x)+3\sin{(4x)}}+6\sin{(6x)}))\\ &=\frac{3}{4}(4\sin{(8x)}+6\sin{(6x)}+2\sin{(4x)}-2\sin{(2x)}) \end{align*}$

- Second Derivative:
$\begin{align*} f'(x)&=\frac{3}{4}(4\sin{(8x)}+6\sin{(6x)}+2\sin{(4x)}-2\sin{(2x)})\\ f''(x)&=\frac{3}{4}(32\cos{(8x)}+36\cos{(6x)}+8\cos{(4x)}-4\cos{(2x)})\\ \end{align*}$

Sub-Problem D
The following identity is used:
$\begin{align*} &2\cos{A}\cos{B}=\cos{(A+B)}+\cos{(A-B)}\\ &2\sin{A}\sin{B}=-\left \{ \cos{(A+B)}-\cos{(A-B)} \right \} \\ &2\sin{A}\cos{B}=\sin{(A+B)}+\sin{(A-B)}\\ &2\cos{A}\sin{B}=\sin{(A+B)}-\sin{(A-B)}\\ \end{align*}$

- First Derivative:
Using the following identity:
$\begin{align*} \tan{2x}&=\frac{2\tan{x}}{1-\tan^2{x}}\\ \tan^2{x}+1&=\sec^2{x} \\ f(x)&=\frac{2\tan{(4x)}}{2-\sec^2{(4x)}}\\ &=\frac{2\tan{(4x)}}{2-(1+\tan^2{(4x)})}\\ &=\frac{2\tan{(4x)}}{1-\tan^2{(4x)})}\\ &=\tan{(8x)}\\ f'(x)&=8\sec^2{(8x)}\\ \end{align*}$

- Second Derivative:
Let us define a variable u and v, such that:
$\begin{align*} u=v&=\sec{(8x)}\\ u'=v'&=8\sec{(8x)}\tan{(8x)} \\ f'(x)&=8\sec^2{(8x)}\\ &=8uv\\ f''(x)&=8(u'v+uv')\\ &=8(16\sec^2{(8x)}\tan{(8x)}) \end{align*}$

Sub-Problem E
- First Derivative:
Let's start:

$\begin{align*} f(x)&=\ln\left(\sin^{2}\left(\frac{1}{\tan^{2}\left(x\right)-1}\right)\right)\\ &=g(h(i(x)))\\ g(x)&= \ln{x}\\ g'(x)&=\frac{1}{x}\\ h(x)&= \sin^2{x}\\ h'(x)&=\sin{2x}\\ i(x)&= \frac{1}{\tan^2{(x)-1}}\\ i'(x)&= -\frac{2\tan{(x)}\sec^2{(x)}}{(\tan^2{(x)-1})^2}\\ f'(x)&=i'(x).h'(i(x)).g'(h(i(x)))\\ &=-\frac{2\tan{(x)}\sec^2{(x)}}{(\tan^2{(x)-1})^2} \times \sin{\left(\frac{2}{\tan^2{(x)-1}}\right)} \times \frac{1}{\sin^{2}\left(\frac{1}{\tan^{2}\left(x\right)-1}\right)}\\ &=-\frac{2\tan{(x)}\sec^2{(x)}}{(\tan^2{(x)-1})^2} \times 2\sin{\left(\frac{1}{\tan^2{(x)-1}}\right)}\cos{\left(\frac{1}{\tan^2{(x)-1}}\right)} \times \frac{1}{\sin^{2}\left(\frac{1}{\tan^{2}\left(x\right)-1}\right)}\\ &=-\frac{4\tan{(x)}\sec^2{(x)}}{(\tan^2{(x)-1})^2} \times \cot{\left(\frac{1}{\tan^2{(x)-1}}\right)}\\ \end{align*}$

- Second Derivative:
Let us define several variables:

$\begin{align*} j&=\tan{(x)}\\ j'&=\sec^2{(x)}\\ k&=\sec^2{(x)}\\ k'&=2\sec^2{(x)}\tan{(x)}\\ b&=(\tan^2{(x)-1})^2\\ b'&=(4\sec^2{(x)}\tan{(x)})(\tan^2{(x)-1})\\ v&=\cot{\left(\frac{1}{\tan^2{(x)-1}}\right)}\\ v'&=-\frac{2\tan{(x)}\sec^2{(x)}}{(\tan^2{(x)-1})^2} \csc^2{\left(\frac{1}{\tan^2{(x)-1}}\right)}\\ \\ f'(x)&=-\frac{4\tan{(x)}\sec^2{(x)}}{(\tan^2{(x)-1})^2} \times \cot{\left(\frac{1}{\tan^2{(x)-1}}\right)}\\ &=-4uv\\ &=-4\left ( \frac{a}{b} \right )v\\ &=-4\left ( \frac{jk}{b}\right )v\\ \frac{f''(x)}{-4}&=\left ( \frac{jk}{b}\right )^{'}v+\left ( \frac{jk}{b}\right )v'\\ &= \left ( \frac{(jk)'b-b'(jk)}{b^2}\right )v + \left ( \frac{jk}{b}\right )v'\\ &= \left ( \frac{(j'k+k'j)b-b'(jk)}{b^2}\right )v + \left ( \frac{jk}{b}\right )v'\\ &= \left ( \frac{j'k+k'j}{b}-\frac{b'(jk)}{b^2}\right )v + \left ( \frac{jk}{b}\right )v'\\ f''(x)&= -4\left ( \left ( \frac{\sec^4{(x)}+2\sec^2{(x)}\tan^2{(x)}}{(\tan^2{(x)-1})^2}-\frac{(4\sec^4{(x)}\tan^2{(x)})}{(\tan^2{(x)-1})^3} \right ) \cot{\left(\frac{1}{\tan^2{(x)-1}}\right)}+ \left (\frac{2\tan^2{(x)}\sec^4{(x)}\csc^2{\left(\frac{1}{\tan^2{(x)-1}}\right)}}{(\tan^2{(x)-1})^4} \right ) \right ) \end{align*}$

Sub-Problem F
- First Derivative:
$\begin{align*} f(x)&=\tan^{-1}\left ( \sqrt{\frac{x-1}{x+1}} \right )\\ &=\tan^{-1}\left ( u\right )\\ f'(x)&=\frac{u'}{u^2+1}\\ &=\frac{1}{u^2+1}u'\\ &= \frac{1}{u^2+1} \left ( \frac{\frac{\sqrt{x+1}}{2\sqrt{x-1}}-\frac{\sqrt{x-1}}{2\sqrt{x+1}}}{x+1} \right )\\ &= \frac{1}{u^2+1} \left ( \frac{\frac{1}{\sqrt{(x-1)(x+1)}} }{x+1}\right )\\ &= \frac{1}{\frac{x-1}{x+1}+1} \left ( \frac{1}{\sqrt{(x-1)(x+1)^3}} \right )\\ &= \frac{1}{\frac{2x}{x+1}} \left ( \frac{1}{\sqrt{(x-1)(x+1)^3}} \right )\\ &= \frac{1}{2x\sqrt{(x-1)(x+1)}}\\ f'(x)&= \frac{1}{\sqrt{4(x^{4}-x^{2})}} \end{align*}$

- Second Derivative:
$\begin{align*} f'(x)&=\frac{1}{\sqrt{4(x^{4}-x^{2})}}\\ f''(x)&= \frac{- \frac{16x^3-8x}{2 \sqrt{4(x^{4}-x^{2})}} }{4(x^{4}-x^{2})}\\ &= -\frac{2x^3-x}{\sqrt{4(x^4-x^2)^3}}\\ \end{align*}$

Sub-Problem G
- First Derivative:
$\begin{align*} f(x)&=\ln\left(\sin^{-1}\left(\frac{x}{2}\right)\right)-\sin^{-1}\left(2x\right)\\ f'(x)&=\frac{\frac{1}{2 \sqrt{1-\frac{x^2}{4}}}}{\sin^{-1}\left(\frac{x}{2}\right)} - \frac{2}{ \sqrt{1-4x^2}}\\ &=\frac{1}{2 \sin^{-1}\left(\frac{x}{2}\right) \sqrt{1-\frac{x^2}{4}}} - \frac{2}{ \sqrt{1-4x^2}}\\ \end{align*}$

- Second Derivative:
Let us declare some variables first:
$\begin{align*} u&=2\sin^{-1}\left ( \frac{x}{2} \right )\\ u'&=\frac{1}{2\sqrt{1-\frac{x^2}{4}}}\\ v&=\sqrt{1-\frac{x^2}{4}}\\ v'&=\frac{-x}{4\sqrt{1-\frac{x^2}{4}}}\\ \\ f'(x)&=\frac{1}{2 \sin^{-1}\left(\frac{x}{2}\right) \sqrt{1-\frac{x^2}{4}}} - \frac{2}{ \sqrt{1-4x^2}}\\ &=(uv)^{-1} - \frac{2}{ \sqrt{1-4x^2}}\\ f''(x)&=-(uv)'(uv)^{-2}-\frac{8x}{\sqrt{(1-4x^2)^3}}\\ &=-\frac{u'v+uv'}{\left ( 2 \sin^{-1} \left ( \frac{x}{2} \right ) \sqrt{1-\frac{x^2}{4}} \right )^2}-\frac{8x}{\sqrt{(1-4x^2)^3}}\\ &=-\frac{1+ \frac{-x \sin^{-1}\left ( \frac{x}{2} \right )}{2\sqrt{1-\frac{x^2}{4}}}}{\left ( 2 \sin^{-1} \left ( \frac{x}{2} \right ) \sqrt{1-\frac{x^2}{4}} \right )^2}-\frac{8x}{\sqrt{(1-4x^2)^3}}\\ \end{align*}$

Sub-Problem H
- First Derivative:
$\begin{align*} f(x)&=\frac{\sin^{-1}\left(x\right)}{\cos^{-1}\left(x\right)}\\ &=\frac{u}{v}\\ f'(x)&=\frac{\frac{\cos^{-1}(x)}{\sqrt{1-x^2}}+\frac{\sin^{-1}(x)}{\sqrt{1-x^2}}}{(\cos^{-1}{(x)})^2}\\ &=\frac{\left ( \frac{\cos^{-1}(x)+\sin^{-1}(x)}{\sqrt{1-x^2}} \right )}{(\cos^{-1}{(x)})^2}\\ \end{align*}$

- Second Derivative:
Let us declare some variables first:

$\begin{align*} f'(x)&=\frac{\left ( \frac{\cos^{-1}(x)+\sin^{-1}(x)}{\sqrt{1-x^2}} \right )}{(\cos^{-1}{(x)})^2}\\ &=\frac{a}{b}\\ \\ a&=\frac{\cos^{-1}(x)+\sin^{-1}(x)}{\sqrt{1-x^2}}\\ a'&=\frac{\frac{x}{\sqrt{1-x^2}}(\cos^{-1}(x)+\sin^{-1}(x))}{1-x^2}\\ b&=(\cos^{-1}(x))^2\\ b'&=-\frac{2\cos^{-1}(x)}{\sqrt{1-x^2}}\\ \\ f''(x)&=\frac{\left ( \frac{\frac{x}{\sqrt{1-x^2}}(\cos^{-1}(x)+\sin^{-1}(x))}{1-x^2} \right ) (\cos^{-1}(x))^2 - \left ( \frac{\cos^{-1}(x)+\sin^{-1}(x)}{\sqrt{1-x^2}} \right )\left ( -\frac{2\cos^{-1}(x)}{\sqrt{1-x^2}}\right ) }{(\cos^{-1}(x))^4} \end{align*}$

Sub-Problem I
The following formulas will be used: (a is a constant)
$\begin{align*} \sin^{-1}\left ( \frac{a}{x} \right )&= \csc^{-1}\left ( \frac{x}{a} \right ) \end{align*}$
- First Derivative:
$\begin{align*} f(x)&=x\csc^{-1}\left(\frac{\pi}{x}\right)\\ &=x \sin^{-1}\left (\frac{x}{\pi}\right )\\ f'(x)&=\sin^{-1}\left (\frac{x}{\pi}\right )+ x\frac{\frac{1}{\pi}}{\sqrt{1-\frac{x^2}{pi^2}}}\\ &=\sin^{-1}\left (\frac{x}{\pi}\right )+ \frac{x}{\sqrt{\pi^2-x^2}}\\ \end{align*}$

- Second Derivative:
Let us declare some variables first:
$\begin{align*} f'(x)&=\sin^{-1}\left (\frac{x}{\pi}\right )+ \frac{x}{\sqrt{\pi^2-x^2}}\\ &=\sin^{-1}\left (\frac{x}{\pi}\right )+ \frac{a}{b}\\ \\ a&=x\\ a'&=1\\ b&=\sqrt{\pi^2-x^2}\\ b'&=\frac{-x}{\sqrt{\pi^2-x^2}}\\ \\ f''(x)&=\frac{\frac{1}{\pi}}{\sqrt{1-\frac{x^2}{\pi^2}}}+ \frac{a'b-ab'}{b^2} \\ &=\frac{1}{\sqrt{\pi^2-x^2}}+ \frac{\sqrt{\pi^2-x^2} + \frac{x^2}{\sqrt{\pi^2-x^2}}}{\pi^2-x^2} \\ &=\frac{\sqrt{\pi^{2}-x^{2}}}{\pi^{2}-x^{2}}+\frac{\sqrt{\pi^{2}-x^{2}}}{\pi^{2}-x^{2}}+\frac{x^{2}}{\sqrt{\left(\pi^{2}-x^{2}\right)^{3}}}\\ f''(x)&=\frac{2\sqrt{\pi^{2}-x^{2}}}{\pi^{2}-x^{2}}+\frac{x^{2}}{\sqrt{\left(\pi^{2}-x^{2}\right)^{3}}}\\ \end{align*}$

Sub-Problem J
- First Derivative:
$\begin{align*} f(x)&=e^{\cot^{-1}\left(e^x\right)}\\ f'(x)&=e^{\cot^{-1}\left(e^x\right)} \frac{-e^x}{1+e^{2x}}\\ &=-\frac{e^{x+\cot^{-1}(e^x)}}{1+e^{2x}}\\ \end{align*}$

- Second Derivative:
Let us declare some variables first:
$\begin{align*} f'(x)&=-\frac{e^{x+\cot^{-1}(e^x)}}{1+e^{2x}}\\ &=-\frac{a}{b}\\ \\ a&=e^{x+\cot^{-1}(e^x)}\\ a'&=e^{x+\cot^{-1}(e^x)} \left ( 1 + \frac{-e^x}{1+e^{2x}} \right ) \\ b&=1+e^{2x}\\ b'&=2e^{2x}\\ \\ f''(x)&= - \frac{a'b-ab'}{b^2}\\ &= - \frac{e^{x+\cot^{-1}(e^x)} \left ( 1 + \frac{-e^x}{1+e^{2x}} \right ) (1+e^{2x})-(e^{x+\cot^{-1}(e^x)})(2e^{2x})}{(1+e^{2x})^2}\\ \end{align*}$

Problem 5: Derivative of a Function
Find the limit of the given function with the L'Hôpital's rule:

$\begin{align*} \text{A. } &\lim_{x\rightarrow 0}\frac{\sin{2x}}{\ln{(\sin{x}+1)}}\\ \text{B. } &\lim_{x\rightarrow \infty}\left(\frac{x^{2}+7x+12}{x^{2}+7x+11}\right)^{\frac{x^{2}+4x+8}{\log_{2}\left(x^{2}+12x+36\right)}}\\ \text{C. } &\lim_{x\rightarrow 0^{+}}{\ln{x}\sin{x}}\\ \text{D. } &\lim_{x\rightarrow 0^{+}}{(\csc{x}+\cot{x})^{\sin{x}}}\\ \text{E. } &\lim_{x\rightarrow 0}{\frac{e^{x^2}-1}{\cos{x}-1}}\\ \end{align*}$

Sub-Problem A
If we directly substitute x = 0, then will get an indeterminate form 0/0. So, we can use L'Hôpital's rule:
$\begin{align*} \lim_{x\rightarrow 0}\frac{\sin{2x}}{\ln{(\sin{x}+1)}}&=\lim_{x\rightarrow 0}\frac{2\cos{2x}}{\left ( \frac{\cos{x}}{\sin{x}+1} \right )}\\ &=\frac{2(1)}{\frac{1}{(0)+1}}\\ &=2 \end{align*}$

Sub-Problem B
If we directly substitute x = 0, then will get an indeterminate form 0/0. But, we must find a way to convert the function into fractions f(x)/g(x) so that we can use L'Hôpital's rule.

$\begin{align*} &\lim_{x\rightarrow \infty}\left(\frac{x^{2}+7x+12}{x^{2}+7x+11}\right)^{\frac{x^{2}+4x+8}{\log_{x}\left(x^{2}+12x+36\right)}}\\&= \lim_{x\rightarrow \infty}\left(\frac{x^{2}+7x+12}{x^{2}+7x+11}-1+1\right)^{\frac{x^{2}+4x+8}{\log_{x}\left(x^{2}+12x+36\right)}}\\ &= \lim_{x\rightarrow \infty}\left(\frac{x^{2}+7x+12}{x^{2}+7x+11}-\frac{x^{2}+7x+11}{x^{2}+7x+11}+1\right)^{\frac{x^{2}+4x+8}{\log_{x}\left(x^{2}+12x+36\right)}}\\ &= \lim_{x\rightarrow \infty}\left ( \left(\frac{1}{x^{2}+7x+11}+1\right)^{(x^2+7x+11)} \right )^{\frac{x^{2}+4x+8}{(x^2+7x+11) \log_{x}\left(x^{2}+12x+36\right)}}\\ &= \lim_{x\rightarrow \infty} e^{\frac{x^{2}+4x+8}{(x^2+7x+11) \log_{x}\left(x^{2}+12x+36\right)}}\\ &= e^{\lim_{x\rightarrow \infty}\frac{x^{2}+4x+8}{(x^2+7x+11) \log_{x}\left(x^{2}+12x+36\right)}}\\ &= e^{\lim_{x\rightarrow \infty}\frac{x^{2}+4x+8}{x^2+7x+11} \times \lim_{x\rightarrow \infty} \frac{1}{\log_{x}\left(x^{2}+12x+36\right)}}\\ &= e^{\lim_{x\rightarrow \infty}\frac{1+\frac{4}{x}+\frac{8}{x^2}}{1+\frac{7}{x}+\frac{11}{x^2}} \times \lim_{x\rightarrow \infty} \frac{\ln{x}}{\ln{x}\log_{x}\left(x^{2}+12x+36\right)}}\\ &= e^{\lim_{x\rightarrow \infty} \frac{\ln{x}}{\ln{x^{2}+12x+36}}}\\ \end{align*}$

We can use L'Hôpital's rule:
$\begin{align*} &\lim_{x\rightarrow \infty}\left(\frac{x^{2}+7x+12}{x^{2}+7x+11}\right)^{\frac{x^{2}+4x+8}{\log_{x}\left(x^{2}+12x+36\right)}}\\ &= e^{\lim_{x\rightarrow \infty} \frac{\ln{x}}{\ln{x^{2}+12x+36}}}\\ &= e^{\lim_{x\rightarrow \infty} \frac{ \left( \frac{1}{x} \right )}{ \left ( \frac{2x+12}{x^2+12x+36} \right ) }}\\ &= e^{\lim_{x\rightarrow \infty} \frac{ \left( \frac{1}{x} \right )}{ \left ( \frac{2x+12}{x^2+12x+36} \right ) }}\\ &= e^{\lim_{x\rightarrow \infty} \frac{x^2+12x+36}{2x^2+12x}} \\ &=\sqrt{e}\\ \end{align*}$

Sub-Problem C
If we directly substitute x = 0, then will get an indeterminate form -∞.0. But, we must find a way to convert the function into fractions f(x)/g(x) so that we can use L'Hôpital's rule.
$\begin{align*} \lim_{x\rightarrow 0^+}\ln{x}\sin{x}&= \lim_{x\rightarrow 0^+}\frac{\ln{x}}{\csc{x}}\\ &= \lim_{x\rightarrow 0^+}\frac{\frac{1}{x}}{-\cot{x}\csc{x}}\\ &= \lim_{x\rightarrow 0^+}\frac{1}{-x\cot{x}\csc{x}}\\ &= \lim_{x\rightarrow 0^+}\frac{\tan{x}\sin{x}}{-x}\\ &= \lim_{x\rightarrow 0^+}-\frac{\sin{x}}{x}\\ &= 0 \end{align*}$

Sub-Problem D
If we directly substitute x = 0, then will get an indeterminate form ∞⁰. But, we must find a way to convert the function into fractions f(x)/g(x) so that we can use L'Hôpital's rule. Also, the following laws of logarithm is used in the following solution:
$\begin{align*} e^{\ln{x}}&=x\\ \ln{x^a}&=a\ln{x} \end{align*}$

So:
$\begin{align*} &\lim_{x\rightarrow 0^+}(\csc{x}+\cot{x})^{\sin{x}}\\ &=\lim_{x\rightarrow 0^+}e^{\ln{(\csc{x}+\cot{x})^{\sin{x}}}}\\ &=e^{\lim_{x\rightarrow 0^+}\ln{(\csc{x}+\cot{x})^{\sin{x}}}}\\ &=e^{\lim_{x\rightarrow 0^+}\sin{x}\ln{(\csc{x}+\cot{x})}}\\ &=e^{\lim_{x\rightarrow 0^+}\frac{\ln{(\csc{x}+\cot{x})}}{\csc{x}}}\\ \end{align*}$

Now, we can use the rule:
$\begin{align*} &\lim_{x\rightarrow 0^+}(\csc{x}+\cot{x})^{\sin{x}}\\ &=e^{\lim_{x\rightarrow 0^+}\frac{\ln{(\csc{x}+\cot{x})}}{\csc{x}}}\\ &=e^{\lim_{x\rightarrow 0^+}\frac{\frac{-\csc{x}\cot{x}-\csc^2{x}}{(\csc{x}+\cot{x})}}{-\csc{x}\cot{x}}}\\ &=e^{\lim_{x\rightarrow 0^+}\frac{\csc{x}\cot{x}+\csc^2{x}}{(\csc{x}+\cot{x})(\csc{x}\cot{x})}}\\ &=e^{\lim_{x\rightarrow 0^+}\frac{\cot{x}+\csc{x}}{(\csc{x}+\cot{x})(\cot{x})}}\\ &=e^{\lim_{x\rightarrow 0^+}\frac{\cot{x}+\csc{x}}{(\csc{x}+\cot{x})(\cot{x})}\times \frac{\sin{x}}{\sin{x}}} \\ &=e^{\lim_{x\rightarrow 0^+}\frac{\cos{x}+1}{(1+\cos{x})(\cot{x})}}\\ &=e^{\lim_{x\rightarrow 0^+}\tan{x}}\\ &=1 \end{align*}$

Sub-Problem E
If we directly substitute x = 0, then will get an indeterminate form 0/0. So, we can use L'Hôpital's rule:
$\begin{align*} &\lim_{x\rightarrow 0}{\frac{e^{x^2}-1}{\cos{x}-1}}\\ &=\lim_{x\rightarrow 0}-{\frac{e^{x^2}2x}{\sin{x}}}\\ &=\lim_{x\rightarrow 0}-e^{x^2}2 \times \lim_{x\rightarrow 0}\frac{x}{\sin{x}}\\ &=-2\\ \end{align*}$