# Calculus Exercise (8) : Derivative of Trigonometric Functions

## Trigonometric Limits

1. $\displaystyle \lim_{x \to 0} \left( {\sin x} \right)=0$
2. $\displaystyle \lim_{x \to 0} \left(\cos x\right)=1$
3. $\displaystyle \lim_{x \to 0} \left(\dfrac{\sin x}{x}\right)=1$
4. ဖော်ပြပါပုံသေနည်း များဖြစ်ပေါ်လာပုံကို ဒီနေရာမှာ ဖတ်ပါ။

## Derivative of $\sin x$ and $\cos x$ from first principles

အထက်ဖော်ပြပါ limit များကို အသုံးချ၍ $\sin x$ နှင့် $\cos x$ ၏ derivatives များကို အောက်ပါအတိုင်း ရှာယူနိုင်ပါသည်။

 \begin{aligned} y &=\sin x \\\\ y+\delta y &=\sin (x+\delta x) \\\\ \delta y &=(y+\delta y)-y \\\\ &=\sin (x+\delta x)-\sin x \\\\ &=2 \cos \frac{x+\delta x+x}{2} \sin \frac{x+\delta x-x}{2} \\\\ \frac{\delta y}{\delta x} &=\frac{2}{\delta x} \cos \left(x+\frac{\delta x}{2}\right) \sin \frac{\delta x}{2} \\\\ &=\cos \left(x+\frac{\delta x}{2}\right) \frac{\sin \frac{\delta x}{2}}{\frac{\delta x}{2}} \\\\ \frac{d y}{d x} &=\lim _{\delta x \rightarrow 0} \frac{\delta y}{\delta x} \\\\ &=\lim _{\delta x \rightarrow 0}\left[\cos \left(x+\frac{\delta x}{2}\right) \cdot \frac{\sin \frac{\delta x}{2}}{\frac{\delta x}{2}}\right] \\\\ &=\cos (x+0)(1) \\\\ &=\cos x \end{aligned} $$\begin{array}{|c|} \hline \dfrac{d}{d x}(\sin x)=\cos x\\ \hline \end{array}$$ \begin{aligned} y &=\cos x \\\\ y+\delta y &=\cos (x+\delta x) \\\\ \delta y &=(y+\delta y)-y \\\\ &=\cos (x+\delta x)-\cos x \\\\ &=-2 \sin \left(\frac{x+\delta x+x}{2}\right) \sin \left(\frac{x+\delta x-x}{2}\right) \\\\ \frac{\delta y}{\delta x} &=-\frac{2}{\delta x} \sin \left(x+\frac{\delta x}{2}\right) \sin \frac{\delta x}{2} \\\\ &=-\sin \left(x+\frac{\delta x}{2}\right) \frac{\sin \frac{\delta x}{2}}{\frac{\delta x}{2}} \\\\ \frac{d y}{d x} &=\lim _{\delta x \rightarrow 0} \frac{\delta y}{\delta x} \\\\ &=\lim _{\delta x \rightarrow 0}\left[-\sin \left(x+\frac{\delta x}{2}\right) \cdot \frac{\sin \frac{\delta x}{2}}{\frac{\delta x}{2}}\right] \\\\ &=-\sin x \end{aligned} $$\begin{array}{|c|} \hline \dfrac{d}{d x}(\cos x)=-\sin x\\ \hline \end{array}$$

## Drivative of Other Trigonometric Functions

$\sin x$ နှင့် $\cos x$ ၏ derivatives များကို သိလျှင် အခြား trigonometric fuctions များ ဖြစ်ကြသော $\tan x, \cot x, \sec x$ နှင့် $\csc x$ တို့၏ derivatives များကို chain rule, product rule, quotient rule များကို သုံး၍ အလွယ်တကူရှာနိုင်ပါသည်။

### Derivative of $\tan x$

\begin{aligned} \frac{d}{d x}(\tan x) &=\frac{d}{d x}\left(\frac{\sin x}{\cos x}\right) \\\\ &=\frac{\cos x \frac{d}{d x}(\sin x)-\sin x \frac{d}{d x}(\cos x)}{\cos ^{2} x} \\\\ &=\frac{\cos ^{2} x+\sin ^{2} x}{\cos ^{2} x} \\\\ &=\frac{1}{\cos ^{2} x} \\\\ &=\sec ^{2} x \end{aligned}

### Derivative of $\cot x$

\begin{aligned} \dfrac{d}{d x}(\cot x) &=\dfrac{d}{d x}\left(\dfrac{\cos x}{\sin x}\right) \\\\ &=\dfrac{\sin x \dfrac{d}{d x}(\cos x)-\cos x \dfrac{d}{d x}(\sin x)}{\sin ^{2} x} \\\\ &=\dfrac{-\sin ^{2} x-\cos ^{2} x}{\sin ^{2} x} \\\\ &=\dfrac{-\left(\sin ^{2} x+\cos ^{2} x\right)}{\sin ^{2} x} \\\\ &=-\dfrac{1}{\sin ^{2} x} \\\\ &=-\csc^{2} x \end{aligned}

### Derivative of $\sec x$

\begin{aligned} \frac{d}{d x}(\sec x) &=\frac{d}{d x}\left(\frac{1}{\cos x}\right) \\\\ &=-\frac{1}{\cos ^{2} x} \frac{d}{d x}(\cos x) \\\\ &=-\frac{1}{\cos ^{2} x}(-\sin x) \\\\ &=\frac{1}{\cos x} \cdot \frac{\sin x}{\cos x} \\\\ &=\sec x \cdot \tan x \end{aligned}

### Derivative of $\csc x$

\begin{aligned} \frac{d}{d x}(\csc x) &=\frac{d}{d x}\left(\frac{1}{\sin x}\right) \\\\ &=-\frac{1}{\sin ^{2} x} \frac{d}{d x}(\sin x) \\\\ &=-\frac{1}{\sin x} \frac{\cos x}{\sin x} \\\\ &=-\csc x \cot x \end{aligned}

## Formulas for Derivatives of Trigonometric Functions

 $$\begin{array}{|l|l|l|} \hline 1 & \dfrac{d}{d x} \sin x=\cos x & \dfrac{d}{d x} \sin u=\cos u \dfrac{d u}{d x} \\ \hline 2 & \dfrac{d}{d x} \cos x=-\sin x & \dfrac{d}{d x} \cos u=-\sin u \dfrac{d u}{d x} \\ \hline 3 & \dfrac{d}{d x} \tan x=\sec ^{2} x & \dfrac{d}{d x} \tan u=\sec ^{2} u \dfrac{d u}{d x} \\ \hline 4 & \dfrac{d}{d x} \cot x=-\csc^{2} x & \dfrac{d}{d x} \cot u=-\csc^{2} u \dfrac{d u}{d x} \\ \hline 5 & \dfrac{d}{d x} \sec x=\sec x \tan x & \dfrac{d}{d x} \sec u=\sec u \tan u \dfrac{d u}{d x} \\ \hline 6 & \dfrac{d}{d x} \csc x=-\csc x \cot x & \dfrac{d}{d x} \csc u=-\csc u \cot u \dfrac{d u}{d x} \\ \hline \end{array}$$

## Problems

1. Differentiate the following with respect to $x$.
\begin{aligned} &\\ \text {(a) }\quad & \sin 5 x\\\\ \text {(b) }\quad & \cos \left(7 x^{2}-2\right)\\\\ \text {(c) }\quad & \tan (6 x+7)\\\\ \text {(d) }\quad & 5 \sec (3 x+1)\\\\ \text {(e) }\quad & \sin (2 x+3) \\\\ \text {(f) }\quad & \cos \left(\dfrac{3}{x}\right)\\\\ \text {(g) }\quad & x^{3} \cos 2 x\\\\ \text {(h) }\quad & \cos 7 x+\sin 3 x\\\\ \text {(i) }\quad & \dfrac{\cot (1-2 x)}{3}\\\\ \text {(j) }\quad & -2 \csc 3 x\\\\ \text {(k) }\quad & \sin x \cos 2 x\\\\ \text {(l) }\quad & \cos ^{2}(5 x)\\\\ \text {(m) }\quad & \tan ^{3} \sqrt{x}\\\\ \text {(n) }\quad & \sin (\cos x)\\\\ \text {(o) }\quad & \dfrac{\sin x}{1+\tan x}\\\\ \text {(p) }\quad & \sqrt{\sin x+\cos x}\\\\ \text {(q) }\quad & (x+\tan x)^{3}\\\\ \text {(r) }\quad & \dfrac{\tan 2 x}{1+\cot x} \end{aligned}

2. \begin{aligned} \text{(a)}\quad & \dfrac{d}{d x}(\sin 5 x)\\\\ =&\ \cos 5 x \dfrac{d}{d x}(5 x) \\\\ =&\ 5 \cos 5 x\\\\\\ \text{(b)}\quad & \dfrac{d}{d x}\left[\cos \left(7 x^{2}-2\right)\right]\\\\ =&\ -\sin \left(7 x^{2}-2\right) \dfrac{d}{d x}\left(7 x^{2}-2\right) \\\\ =&\ -14 x \sin \left(7 x^{2}-2\right)\\\\\\ \text{(c)}\quad & \dfrac{d}{d x}[\tan (6 x+7)] \\\\ =&\ \sec ^{2}(6 x+7) \dfrac{d}{d x}(6 x+7) \\\\ =&\ 6 \sec ^{2}(6 x+7)\\\\\\ \text{(d)}\quad & \dfrac{d}{d x}[5 \sec (3 x+1)] \\\\ =&\ 5 \sec (3 x+1) \tan (3 x+1) \dfrac{d}{d x}(3 x+1) \\\\ =&\ 15 \sec (3 x+1) \tan (3 x+1)\\\\\\ \text{(e)}\quad & \dfrac{d}{d x}[\sin (2 x+3)] \\\\ =&\ \cos (2 x+3) \dfrac{d}{d x}(2 x+3) \\\\ =&\ 2 \cos (2 x+3)\\\\\\ \text{(f)}\quad & \dfrac{d}{d x}\left[\cos \left(\dfrac{3}{x}\right)\right] \\\\ =&\ -\sin \left(\dfrac{3}{x}\right) \dfrac{d}{d x}\left(\dfrac{3}{x}\right) \\\\ =&\ \dfrac{3}{x^{2}} \sin \left(\dfrac{3}{x}\right)\\\\\\ \text{(g)}\quad &\dfrac{d}{d x}\left(x^{3} \cos 2 x\right) \\\\ =&\ x^{3} \dfrac{d}{d x}(\cos 2 x)+\cos 2 x \dfrac{d}{d x}\left(x^{3}\right) \\\\ =&\ -2 x^{3} \sin 2 x+3 x^{2} \cos 2 x\\\\\\ \text{(h)}\quad & \dfrac{d}{d x}(\cos 7 x+\sin 3 x) \\\\ =&\ -\sin 7 x \dfrac{d}{d x}(7 x)+\cos 3 x \dfrac{d}{d x}(3 x) \\\\ =&\ -7 \sin 7 x+3 \cos 3 x\\\\\\ \text{(i)}\quad & \dfrac{d}{d x}\left[\dfrac{\cot (1-2 x)}{3}\right] \\\\ =&\ -\dfrac{1}{3} \csc^{2}(1-2 x) \dfrac{d}{d x}(1-2 x) \\\\ =&\ \dfrac{2}{3} \csc^{2}(1-2 x)\\\\\\ \text{(j)}\quad & \dfrac{d}{d x}(-2 \csc 3 x) \\\\ =&\ -2(-\csc 3 x \cot 3 x) \dfrac{d}{d x}(3 x) \\\\ =&\ 6 \csc 3 x \cot 3 x\\\\\\ \text{(k)}\quad & \dfrac{d}{d x}(\sin x \cos 2 x)\\\\ =&\ \sin x \dfrac{d}{d x}(\cos 2 x)+\cos 2 x \dfrac{d}{d x}(\sin x)\\\\ =&\ -\sin x \sin 2 x \dfrac{d}{d x}(2 x)+\cos x \cos 2 x\\\\ =&\ -2 \sin x \sin 2 x+\cos x \cos 2 x\\\\\\ \text{(l)}\quad & \dfrac{d}{d x}\left[\cos ^{2}(5 x)\right]\\\\ =&\ 2 \cos 5 x \dfrac{d}{d x}(\cos 5 x)\\\\ =&\ -2 \sin 5 x \cos 5 x \dfrac{d}{d x}(5 x)\\\\ =&\ -10 \sin 5 x \cos 5 x\\\\\\ \text{(m)}\quad & \dfrac{d}{d x}\left(\tan ^{3} \sqrt{x}\right)\\\\ =&\ 3 \tan ^{2} \sqrt{x} \dfrac{d}{d x}(\tan \sqrt{x})\\\\ =&\ 3 \tan ^{2} \sqrt{x} \sec ^{2} \sqrt{x} \dfrac{d}{d x}(\sqrt{x})\\\\\\ \text{(n)}\quad & \dfrac{d}{d x}[\sin (\cos x)] \\\\ =& \cos (\cos x) \dfrac{d}{d x}(\cos x) \\\\ =&-\sin x[\cos (\cos x)] \\\\\\ \text{(o)}\quad & \dfrac{d}{d x}\left(\dfrac{\sin x}{1+\tan x}\right) \\\\ =& \dfrac{(1+\tan x) \dfrac{d}{d x}(\sin x)-\sin x \dfrac{d}{d x}(1+\tan x)}{(1+\tan x)^{2}} \\\\ =& \dfrac{\cos x(1+\tan x)-\sin x \cdot \sec ^{2} x}{(1+\tan x)^{2}}\\\\\\ \text{(p)}\quad & \dfrac{d}{d x} \sqrt{\sin x+\cos x} \\\\ =& \dfrac{1}{2 \sqrt{\sin x+\cos x}} \dfrac{d}{d x}(\sin x+\cos x) \\\\ =& \dfrac{\cos x-\sin x}{2 \sqrt{\sin x+\cos x}} \\\\\\ \text{(q)}\quad & \dfrac{d}{d x}(x+\tan x)^{3} \\\\ =& 3(x+\tan x)^{2} \dfrac{d}{d x}(x+\tan x) \\\\ =& 3(x+\tan x)^{2}\left(1+\sec ^{2} x\right) \\\\\\ \text{(r)}\quad & \dfrac{d}{d x}\left(\dfrac{\tan 2 x}{1+\cot x}\right) \\\\ =& \dfrac{(1+\cot x) \dfrac{d}{d x}(\tan 2 x)-(\tan 2 x) \dfrac{d}{d x}(1+\cot x)}{(1+\cot x)^{2}} \\\\ =& \dfrac{2 \tan 2 x(1+\cot x)+\tan 2 x \csc ^{2} x}{(1+\cot x)^{2}} \end{aligned}

3. Find $\dfrac{d y}{d x}$.
\begin{aligned} &\\ \text {(a) }\quad & y=\sin^2x\\\\ \text {(b) }\quad & y=\cos \sqrt{x} \\\\ \text {(c) }\quad & y=\tan ^{2}\left(x^{2}\right) \\\\ \text {(d) }\quad & y=\sin 2x – x \cos x\\\\ \text {(e) }\quad & y=\dfrac{x}{\tan x}\\\\ \text {(f) }\quad & y=\sin x \cos ^{2} x\\\\ \text {(g) }\quad & y=\sin \left(1-x^{2}\right)\\\\ \text {(h) }\quad & y=2 \pi x+2 \cos \pi x\\\\ \text {(i) }\quad & y=\sin ^{2} x \cos {3 x}\\\\ \text {(j) }\quad & y=x^{2} \sin \left(\dfrac{1}{x}\right)\\\\ \text {(k) }\quad & 3 x^{2}+2 \sin y=y^{2}\\\\ \text {(l) }\quad & \sin x \cos y=2 y\\\\ \text {(m) }\quad & 2 x y+\sin x=3\\\\ \text {(n) }\quad & y=\dfrac{\cos (1-2 x)}{\sqrt{x}}\\\\ \text {(o) }\quad & y=\sin ^{4} x \cos (3 x)\\\\ \text {(p) }\quad & x+\sin y=\cos (x y)\\\\ \text {(q) }\quad & x+y^{2}=\sin (x+y)\\\\ \text {(r) }\quad & 2 \sin x=x+\cos y \end{aligned}

4. \begin{aligned} \text{(a)}\quad y &=\sin ^{2} x \\\\ \dfrac{d y}{d x} &=2 \sin x \dfrac{d}{d x}(\sin x) \\\\ &=2 \sin x \cos x \\\\ &=\sin 2 x \\\\\\ \text{(b)}\quad y &=\cos \sqrt{x} \\\\ \dfrac{d y}{d x} &=2(-\sin \sqrt{x}) \dfrac{d}{d x}(\sqrt{x}) \\\\ &=-\dfrac{\sin \sqrt{x}}{\sqrt{x}} \\\\\\ \text{(c)}\quad y &=\tan ^{2}\left(x^{2}\right) \\\\ \dfrac{d y}{d x} &=2 \tan \left(x^{2}\right) \dfrac{d}{d x}\left[\tan \left(x^{2}\right)\right] \\\\ &=2 \tan \left(x^{2}\right) \sec ^{2}\left(x^{2}\right) \dfrac{d}{d x}\left(x^{2}\right) \\\\ &=4 x \tan \left(x^{2}\right) \sec ^{2}\left(x^{2}\right)\\\\\\ \text{(d)}\quad y &=\sin 2 x-x \cos x \\\\ \dfrac{d y}{d x} &=\cos 2 x \dfrac{d}{d x}(2 x)-(-x \sin x+\cos x) \\\\ &=2 \cos 2 x+x \sin x-\cos x \\\\\\ \text{(e)}\quad y &=\dfrac{x}{\tan x} \\\\ \dfrac{d y}{d x} &=\dfrac{\tan x-x \sec ^{2} x}{\tan ^{2} x} \\\\ &=\dfrac{1}{\tan x}-x \dfrac{\cos ^{2} x}{\sin ^{2} x} \dfrac{1}{\cos ^{2} x} \\\\ &=\cot x-x \operatorname{cosec}^{2} x \\\\\\ \text{(f)}\quad y &=\sin x \cos ^{2} x \\\\ \dfrac{d y}{d x} &=\sin x(2 \cos x) \dfrac{d}{d x}(\cos x)+\cos ^{2} x \cdot \cos x \\\\ &=-2 \sin ^{2} x \cos x+\cos ^{3} x\\\\\\ \text{(g)}\quad y &=\sin \left(1-x^{2}\right) \\\\ \dfrac{d y}{d x} &=\cos \left(1-x^{2}\right) \dfrac{d}{d x}\left(1-x^{2}\right) \\\\ &=-2 x \cos \left(1-x^{2}\right) \\\\\\ \text{(h)}\quad y &=2 \pi x+2 \cos \pi x \\\\ \dfrac{d y}{d x} &=2 \pi-2 \sin \pi x \dfrac{d}{d x}(\pi x) \\\\ &=2 \pi(1-\sin \pi x) \\\\\\ \text{(i)}\quad y &=\sin ^{2} x \cos 3 x \\\\ \dfrac{d y}{d x} &=-\sin ^{2} x \sin 3 x \dfrac{d}{d x}(3 x)+\cos 3 x(2 \sin x) \dfrac{d}{d x}(\sin x) \\\\ &=-3 \sin ^{2} x \sin 3 x+2 \sin x \cos x \cos 3 x \\\\ &=-3 \sin ^{2} x \sin 3 x+\sin 2 x \cos 3 x \\\\\\ \text{(j)}\quad y &=x^{2} \sin \left(\dfrac{1}{x}\right) \\\\ \dfrac{d y}{d x} &=x^{2} \cos \left(\dfrac{1}{x}\right) \dfrac{d}{d x}\left(\dfrac{1}{x}\right)+2 x \sin \left(\dfrac{1}{x}\right) \\\\ &=x^{2}\left(-\dfrac{1}{x^{2}}\right) \cos \left(\dfrac{1}{x}\right)+2 x \sin \left(\dfrac{1}{x}\right) \\\\ &=-\cos \left(\dfrac{1}{x}\right)+2 x \sin \left(\dfrac{1}{x}\right)\\\\\\ \text{(k)}\quad & 3 x^{2}+2 \sin y=y^{2}\\\\ &\text{Differentiate with respect to } x,\\\\ &6 x+2 \cos y \dfrac{d y}{d x}=2 y \dfrac{d y}{d x} \\\\ &\dfrac{d y}{d x}=\dfrac{3 x}{y-\cos y} \\\\\\ \text{(l)}\quad & \sin x \cos y=2 y\\\\ &\text{Differentiate with respect to } x,\\\\ &-\sin x \sin y \dfrac{d y}{d x}+\cos x \cos y=2 \dfrac{d y}{d x} \\\\ &\dfrac{d y}{d x}=\dfrac{\cos x \cos y}{2+\sin x \sin y} \\\\\\ \text{(m)}\quad &2 x y+\sin x=3\\\\ &\text{Differentiate with respect to } x,\\\\ &2 x \dfrac{d y}{d x}+2 y+\cos x=0 \\\\ &\dfrac{d y}{d x}=-\dfrac{\cos x+2 y}{2 x}\\\\\\ \text{(n)}\quad y &=\dfrac{\cos (1-2 x)}{\sqrt{x}} \\\\ \dfrac{d y}{d x} &=\dfrac{-\sqrt{x} \sin (1-2 x) \dfrac{d}{d x}(1-2 x)-\dfrac{\cos (1-2 x)}{2 \sqrt{x}}}{x} \\\\ &=\dfrac{1}{x}\left[2 \sqrt{x} \sin (1-2 x)-\dfrac{\cos (1-2 x)}{2 \sqrt{x}}\right] \\\\ &=\dfrac{4 x \sin (1-2 x)-\cos (1-2 x)}{2 x \sqrt{x}} \\\\\\ \text{(o)}\quad y &=\sin ^{4} x \cos (3 x) \\\\ \dfrac{d y}{d x} &=-\sin ^{4} x \sin 3 x \dfrac{d}{d x}(3 x)+4 \sin ^{3} x \cos (3 x) \dfrac{d}{d x}(\sin x) \\\\ &=-3 \sin ^{4} x \sin 3 x+4 \sin ^{3} x \cos (3 x) \cos x \\\\\\ \text{(p)}\quad & x+ \sin y=\cos (x y) \\\\ &\text{Differentiate with respect to } x,\\\\ &1+\cos y \dfrac{d y}{d x}=-\sin (x y) \dfrac{d}{d x}(x y) \\\\ &1+\cos y \dfrac{d y}{d x}=-\sin (x y)\left(x \dfrac{d y}{d x}+y\right) \\\\ &{[x \sin (x y)+\cos y] \dfrac{d y}{d x}=-[1+y \sin (x y)]} \\\\ &\dfrac{d y}{d x}=-\dfrac{1+y \sin (x y)}{x \sin (x y)+\cos y} \\\\\\ \text{(q)}\quad & x+\sin y=\cos (x y)\\\\ &\text{Differentiate with respect to } x,\\\\ &1+\cos y \dfrac{d y}{d x}=-\sin (x y) \dfrac{d}{d x}(x y) \\\\ &1+\cos y \dfrac{d y}{d x}=-\sin (x y)\left(x \dfrac{d y}{d x}+y\right) \\\\ &{[x \sin (x y)+\cos y] \dfrac{d y}{d x}=-[1+y \sin (x y)]} \\\\ &\dfrac{d y}{d x}=-\dfrac{1+y \sin (x y)}{x \sin (x y)+\cos y} \\\\\\ \text{(r)}\quad & x+y^{2}=\sin (x+y)\\\\ &1+2 y \dfrac{d y}{d x}=\cos (x+y) \dfrac{d}{d x}(x+y) \\\\ &1+2 y \dfrac{d y}{d x}=\cos (x+y)\left(1+\dfrac{d y}{d x}\right) \\\\ &{[2 y-\cos (x+y)] \dfrac{d y}{d x}=\cos (x+y)-1} \\\\ &\dfrac{d y}{d x}=\dfrac{\cos (x+y)-1}{2 y-\cos (x+y)} \end{aligned}

5. Given that $y=x \sin x$, find $\dfrac{d^{2} y}{d x^{2}}$.

6. \begin{aligned} y &=x \sin x \\\\ \frac{d y}{d x} &=x \cos x+\sin x \\\\ \frac{d^{2} y}{d x^{2}} &=-x \sin x+\cos x+\cos x \\\\ &=2 \cos x-x \sin x \end{aligned}

7. Given that $y=\cos ^{2} x$, prove that $\dfrac{d^{2} y}{d x^{2}}+4 y=2$.

8. \begin{aligned} y &=\cos ^{2} x \\\\ \frac{d y}{d x} &=2 \cos x \frac{d}{d x}(\cos x) \\\\ &=-2 \sin x \cos x \\\\ &=-\sin 2 x \\\\ \frac{d^{2} y}{d x^{2}} &=-\cos 2 x \frac{d}{d x}(2 x) \\\\ &=-2 \cos 2 x \\\\ \frac{d^{2} y}{d x^{2}}+4 y &=-2 \cos 2 x+4 \cos ^{2} x \\\\ &=-2\left(\cos ^{2} x-\sin ^{2} x\right)+4 \cos ^{2} x \\\\ &=-2 \cos ^{2} x+2 \sin ^{2} x+4 \cos ^{2} x \\\\ &=2\left(\sin ^{2} x+\cos ^{2} x\right) \\\\ &=2 \end{aligned}

9. Given that $y=\dfrac{1}{3} \cos ^{3} x-\cos x$, prove that $\dfrac{d y}{d x}=\sin ^{3} x$.

10. \begin{aligned} y&= \frac{1}{3} \cos ^{3} x-\cos x \\\\ \frac{d y}{d x} &=\cos ^{2} x \frac{d}{d x}(\cos x)-(-\sin x) \\\\ &=-\sin x \cos ^{2} x+\sin x \\\\ &=\sin x\left(1-\cos ^{2} x\right) \\\\ &=\sin x\left(\sin ^{2} x\right) \\\\ &=\sin ^{3} x \end{aligned}

11. If $x \cos y=\sin x$, prove that $\dfrac{d y}{d x}=\dfrac{\cos y(\cos y-\cos x)}{\sin y \sin x}$.

12. \begin{aligned} &x \cos y=\sin x\\\\ &\text{Differentiate with respect to } x,\\\\ &-x \sin y \frac{d y}{d x}+\cos y=\cos x \\\\ \therefore \quad & \frac{d y}{d x}=\frac{1}{x} \cdot \frac{\cos y-\cos x}{\sin y} \\\\ & \text { Since } x \cos y=\sin x, \frac{1}{x}=\frac{\cos y}{\sin x} \\\\ \therefore \quad & \frac{d y}{d x}=\frac{\cos y(\cos y-\cos x)}{\sin y \sin x} \end{aligned}

13. Find the value $a$ and $b$ for which $\dfrac{d}{d x}\left(\dfrac{\sin x}{2+\cos x}\right)=\dfrac{a+b \cos x}{(2+\cos x)^{2}}$.

14. \begin{aligned} \frac{d}{d x}\left(\frac{\sin x}{2+\cos x}\right) &= \frac{a+b \cos x}{(2+\cos x)^{2}} \\\\ (2+\cos x) \frac{d}{d x}(\sin x)-\sin x \frac{d}{d x}(2+\cos x) &= \frac{a+b \cos x}{(2+\cos x)^{2}} \\\\ \frac{(2+\cos x)(\cos x)^{2}-\sin x(-\cos x)}{(2+\cos x)^{2}} &= \frac{a+b \cos x}{(2+\cos x)^{2}} \\\\ \frac{\sin ^{2} x+\cos ^{2} x+2 \cos x}{(2+\cos x)^{2}} &= \frac{a+b \cos x}{(2+\cos x)^{2}} \\\\ \frac{1+2 \cos x}{(2+\cos x)^{2}} &= \frac{a+b \cos x}{(2+\cos x)^{2}} \\\\ \therefore a= 1,\ \ b&= 2 \end{aligned}

15. Find the equation of the tangent line to the curve $y=2 \sin x-3$ at $x=\dfrac{\pi}{6}$.

16. \begin{aligned} \text {Curve }: y &=2 \sin x-3\\\\ \text {When } x &=\frac{\pi}{6}\\\\ y &=2 \sin \frac{\pi}{6}-3 \\\\ &=1-3 \\\\ &=-2\left[\because \sin \frac{\pi}{6}=\frac{1}{2}\right]\\\\ \text { Let } \left(x_{1}, y_{1}\right) &=\left(\frac{\pi}{6},-2\right) \\\\ \frac{d y}{d x} &=2 \cos x \\\\ m &=\left.\frac{d y}{d x}\right|_{\left(\frac{\pi}{6},-2\right)} \\\\ &=2 \cos \frac{\pi}{6} \\\\ &=2\left(\frac{\sqrt{3}}{2}\right) \\\\ &=\sqrt{3}\\\\ \text{Equation } & \text{ of tangent at } \left(x_{1}, y_{1}\right) \text { is}\\\\ y-y_{1} &=m\left(x-x_{1}\right) \\\\ \therefore \quad y-(-2) &=\sqrt{3}\left(x-\frac{\pi}{6}\right) \\\\ y &=\sqrt{3}\left(x-\frac{\pi}{6}\right)-2 \end{aligned}

17. If $y=\sin 6 x+\cos ^{2} x$, find the value of $\dfrac{d y}{d x}$ when $x=\dfrac{\pi}{3}$.

18. \begin{aligned} y &=\sin 6 x+\cos ^{2} x \\\\ \frac{d y}{d x} &=\cos 6 x \frac{d}{d x}(6 x)+2 \cos x \frac{d}{d x}(\cos x) \\\\ \frac{d y}{d x} &=6 \cos 6 x-2 \sin x \cos x=6 \cos 6 x-\sin 2 x \\\\ \text { When } x &=\frac{\pi}{3} \\\\ \frac{d y}{d x} &=6 \cos 6\left(\frac{\pi}{3}\right)-\sin 2\left(\frac{\pi}{3}\right) \\\\ &=6 \cos 2 \pi-\sin \frac{2 \pi}{3} \\\\ &=6-\frac{\sqrt{3}}{2} \end{aligned}

19. Given that $y=\sin 3 x+\cos ^{3} x$, find the value of $\dfrac{d y}{d x}$ when $x=\dfrac{\pi}{4}$.

20. \begin{aligned} y &=\sin 3 x+\cos ^{3} x \\\\ \frac{d y}{d x} &=\cos 3 x \frac{d}{d x}(3 x)+3 \cos ^{2} x \frac{d}{d x}(\cos x) \\\\ \frac{d y}{d x} &=3 \cos 3 x-3 \sin x \cos ^{2} x=3\left(\cos 3 x-\sin x \cos ^{2} x\right)\\\\ \text{When } x&=\frac{\pi}{4},\\\\ \frac{d y}{d x} &=3\left(\cos \frac{3 \pi}{4}-\sin \frac{\pi}{4} \cos ^{2} \frac{\pi}{4}\right) \\\\ &=3\left(-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2} \cdot \frac{1}{2}\right) \\\\ &=-\frac{9 \sqrt{2}}{4} \end{aligned}

21. Find the equation of tangent to the curve $y=\sin \left(\dfrac{x}{3}\right)$ at the point where $x=\pi$.

22. \begin{aligned} \text{ Curve }: y&=\sin \left(\frac{x}{3}\right)\\\\ \text{ When } x&=\pi,\\\\ y &=\sin \frac{\pi}{3} \\\\ &=\frac{\sqrt{3}}{2} \\\\ \text { Let }\left(x_{1}, y_{1}\right) &=\left(\pi, \frac{\sqrt{3}}{2}\right) . \\\\ \frac{d y}{d x} &=\frac{1}{3} \cos \left(\frac{x}{3}\right) \\\\ m &=\frac{d y}{d x}\left(_{\pi, \frac{\sqrt{3}}{2}}\right) \\\\ &=\frac{1}{3} \cos \frac{\pi}{3} \\\\ &=\frac{1}{3}\left(\frac{1}{2}\right) \\\\ &=\frac{1}{6}\\\\ \text{ Equation } & \text{ of tangent at } \left(x_{1}, y_{1}\right) \text{ is}\\\\ y-y_{1} &=m\left(x-x_{1}\right) \\\\ \therefore y-\frac{\sqrt{3}}{2} &=\frac{1}{6}(x-\pi) \\\\ x-6 y &=\pi-3 \sqrt{3} \end{aligned}

23. If $y=\sin ^{2} x$, prove that $4 y+y^{\prime \prime}=2$.

24. \begin{aligned} y &=\sin ^{2} x \\\\ y^{\prime} &=2 \sin x \cos x \\\\ &=\sin 2 x \\\\ y^{\prime \prime} &=2 \cos 2 x \\\\ 4 y+y^{\prime \prime} &=4 \sin ^{2} x+2 \cos 2 x \\\\ &=4 \sin ^{2} x+2\left(\cos ^{2} x-\sin ^{2} x\right) \\\\ &=2\left(\sin ^{2} x+\cos ^{2} x\right) \\\\ &=2 \end{aligned}

25. If $y=3 \sin 2 x+2 \cos 2 x$, prove that $4 y+y^{\prime \prime}=0$.

26. \begin{aligned} y &=3 \sin 2 x+2 \cos 2 x \\\\ y^{\prime} &=6 \cos 2 x-4 \sin 2 x \\\\ y^{\prime \prime} &=-12 \sin 2 x-8 \cos 2 x \\\\ 4 y+y^{\prime \prime} &=12 \sin 2 x+8 \cos 2 x-12 \sin 2 x-8 \cos 2 x \\\\ \therefore \quad 4 y+y^{\prime \prime} &=0 \end{aligned}

27. If $y=\sin 3 x-2 \cos 3 x$, prove that $\dfrac{d^{2} y}{d x^{2}}=-9 y$.

28. \begin{aligned} y &=\sin 3 x-2 \cos 3 x \\\\ \frac{d y}{d x} &=3 \cos 3 x+6 \sin 3 x \\\\ \frac{d^{2} y}{d x^{2}} &=-9 \sin 3 x+18 \cos 3 x \\\\ &=-9(\sin 3 x-2 \cos 3 x) \end{aligned}

29. Given that $y=\sin ^{2} x$ for $0 \leqslant x \leqslant \pi$, find the exact values of the $x$-coordinates of the points on the curve where the gradient is $\dfrac{\sqrt{3}}{2}$.

30. \begin{aligned} \text { Curve: } y &=\sin ^{2} x, 0 \leq x \leq \pi \\\\ \frac{d y}{d x} &=2 \sin x \frac{d}{d x}(\sin x) \\\\ &=2 \sin x \cos x \\\\ &=\sin 2 x \\\\ \frac{d y}{d x} &=\frac{\sqrt{3}}{2} \\\\ \sin 2 x &=\frac{\sqrt{3}}{2} \\\\ \therefore 2 x &=\frac{\pi}{3} \text { or } 2 x=\frac{2 \pi}{3} \\\\ \therefore x &=\frac{\pi}{6} \text { or } x=\frac{\pi}{3} \end{aligned}

31. Prove that the normal to the curve $y=x \sin x$ at the point $P\left(\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$ intersects the $x$-axis at the point $(\pi, 0)$.

32. \begin{aligned} \text{Curve }: y&=x \sin x\\\\ \frac{d y}{d x} &=x \frac{d}{d x}(\sin x)+\sin x \frac{d}{d x}(x) \\\\ &=x \cos x+\sin x\\\\ \text{ At the point } & P\left(\frac{\pi}{2}, \frac{\pi}{2}\right)\\\\ \frac{d y}{d x} &=\frac{\pi}{2} \cos \frac{\pi}{2}+\sin \frac{\pi}{2} \\\\ &=\frac{\pi}{2}(0)+1 \\\\ &=1\\\\ \therefore \quad \text{ The gradient } & \text{ of normal is } -1.\\\\ \text{ Equation } & \text{ of normal at } P \text{ is }\\\\ y-\frac{\pi}{2}&=-1\left(x-\frac{\pi}{2}\right) \\\\ y&=\pi-x\\\\ \text{ When} y&=0 ,\\\\ \pi-x &=0 \\\\ x &=\pi\\\\ \therefore \quad \text{ The normal at } P & \text{ cuts x-axis at } (\pi, 0). \end{aligned}

စာဖတ်သူ၏ အမြင်ကို လေးစားစွာစောင့်မျှော်လျက်!