Maths Solve the Following Question.(3 Marks)

Consider $x^7 \cdot y^5=(x+y)^{12}$
Taking log of both sides,
$\log \left(x^2 y^5\right)=\log (x+y)^{12}$
$\log x^7+\log y^5=12 \log (x+y)$
$7 \log x+5 \log y=12 \log (x+y)$
Differentiating w.r.t. $x$
$\frac{7}{x}+\frac{5}{y} \cdot \frac{d y}{d x}=\frac{12}{x+y}\left(1+\frac{d y}{d x}\right)$
$\therefore \frac{d y}{d x}\left(\frac{5}{y}-\frac{12}{x+y}\right)=\frac{12}{x+y}-\frac{7}{x}$
$\therefore \frac{d y}{d x}\left(\frac{5 x+5 y-12 y}{y(x+y)}\right)=\frac{12 x-7 x-7 y}{x(x+y)}$
$\therefore \quad \frac{d y}{d x}\left(\frac{5 x-7 y}{y(x+y)}\right)=\frac{(5 x-7 y)}{x(x+y)}$
$\therefore \quad \frac{d y}{d x}=\frac{y}{x}$

Let $y=\tan ^{-1}\left(\frac{5 x+1}{3-x-6 x^2}\right)$
$=\tan ^{-1}\left(\frac{5 x+1}{1+2-x-6 x^2}\right)$
$=\tan ^{-1}\left(\frac{5 x+1}{1-(3 x+2)(2 x-11)}\right)$
$=\tan ^{-1}\left(\frac{(3 x+2)+(2 x-1)}{(1-(3 x+2))(2 x-1)}\right)$
$y=\tan ^{-1}(3 x+2)+\tan ^{-1}(2 x-1)$
Differentiate w.r.t. x
$\therefore \frac{d y}{d x}=\frac{3}{1+(3 x+2)^2}+\frac{2}{1+(2 x-1)^2}$
$=\frac{3}{1+9 x^2+12 x+4}+\frac{2}{1+4 x^2-4 x+1}$
$=\frac{3}{9 x^2+12 x+5}+\frac{1}{2 x^2-2 x+1}$

We have $\frac{d y}{d x}=\frac{ dy / dx }{ dx / dt }, dx / dt \neq 0$...(1)
Now $y=a \sin ^3 t=a(\sin t)^3 \Rightarrow \sin t=\left(\frac{y}{a}\right)^{\frac{1}{3}}$
$\therefore \frac{d y}{d x}=a \frac{d}{d t}(\sin t)^3=a .3(\sin t)^2 \frac{d}{d t}(\sin t)$
$=3 a \sin ^2 t \cos t \ldots$ (2)
Also, $x=a \cos ^3 t=a(\cos t)^3 \Rightarrow \cos t=\left(\frac{x}{a}\right)^{\frac{1}{3}}$
$\therefore \frac{d x}{d t}=a .3 \cos ^2 t \frac{d}{d t}(\cos t)$
$=3 a \cos ^2 t(-\sin t)$
$=-3 a \cos ^2 t \sin t \quad \ldots 3$
From (1), (2) and (3),
$\frac{d y}{d x}=\frac{3 a \sin ^2 t \cos t}{-3 a \cos ^2 t \sin t}=-\frac{\sin t}{\cos t}=-\left(\frac{y}{x}\right)^{\frac{1}{3}}$

$y=\cos ^{-1}\left(2 x \sqrt{1-x^2}\right)$
put $x=\sin \theta$
$\theta=\sin ^{-1} x$
$=\cos ^{-1}\left(2 \sin \theta \sqrt{1-\sin ^2 \theta}\right)$
$=\cos ^{-1}(\sin 2 \theta)$
$=\cos ^{-1}\left(\cos \left(\frac{\pi}{2}-2 \theta\right)\right)$
$y=\frac{\pi}{2}-2 \theta=\frac{\pi}{2}-2 \sin ^{-1} x$
Differentiating with respect to 'x', we get
$\frac{d y}{d x}=0-\frac{2}{\sqrt{1-x^2}}=\frac{-2}{\sqrt{1-x^2}}$

Let δx be a small increment in x.
Let δy and δu be the corresponding increments in y and u respectively
As δx → 0, δy → 0, δu → 0.
As u is differentiable function, it is continuous.
Consider the incrementary ratio $\frac{\delta y}{\delta x}$
We have,$\frac{\delta y}{\delta x}=\frac{\delta y}{\delta u} \times \frac{\delta u}{\delta x}$
Taking limit as δx → 0, on both sides,
$\lim _{\delta x \rightarrow 0} \frac{\delta y}{\delta x}=\lim _{\delta x \rightarrow 0}\left(\frac{\partial t y}{\delta u} \times \frac{\delta u}{\delta x}\right)$
$\lim _{\delta x \rightarrow 0} \frac{\delta y}{\delta x}=\lim _{\delta u \rightarrow 0} \frac{\delta y}{\delta u} \times \lim _{\delta x \rightarrow 0} \frac{\delta u}{\delta x} \ldots(1)$
Since $y$ is a differentiable function of $u , \lim _{\delta u \rightarrow 0} \frac{\delta y}{\delta u}$ exists
and $\lim _{\delta u \rightarrow 0} \frac{\delta y}{\delta x}$ exists as $u$ is a differentiable function of $x$.
Hence, R.H.S. of (1) exists
now $\lim _{\delta u \rightarrow 0} \frac{\delta y}{\delta u}=\frac{d y}{d u}$ and $\lim _{\delta u \rightarrow 0} \frac{\delta u}{\delta x}=\frac{d u}{d x}$
$\lim _{\delta x \rightarrow 0} \frac{\delta y}{\delta x}=\frac{d y}{d u} \times \frac{d u}{d x}$
Since R.H.S. exists, L.H.S. of (1) also exists and 
$\lim _{\delta x \rightarrow 0} \frac{\delta y}{\delta x}=\frac{d y}{d x}$
$\frac{d y}{d x}=\frac{d y}{d u} \times \frac{d u}{d x}$

Let $\delta y$ be the increment in $y$ corresponding to an increment $\delta x$ in $x$.
as $\delta x \rightarrow 0, \delta y \rightarrow 0$
Now y is a differentiable function of x.
$\therefore \lim _{\delta x \rightarrow 0} \frac{\delta y}{\delta x}=\frac{d y}{d x}$
Now $\frac{\delta y}{\delta x} \times \frac{\delta x}{\delta y}=1$
$\therefore \frac{\delta x}{\delta y}=\frac{1}{\frac{\delta y}{\delta x}}$
Taking limits on both sides as $\delta x \rightarrow 0, w e \geq t$
$\lim _{\delta x \rightarrow 0} \frac{\delta x}{\delta y}=\lim _{\delta x \rightarrow 0}\left[\frac{1}{\frac{\delta y}{\delta x}}\right]=\frac{1}{\lim _{d x \rightarrow 0} \frac{\delta y}{\delta x}}$
$\lim _{\delta x \rightarrow 0} \frac{\delta x}{\delta y}=\frac{1}{\lim _{d x \rightarrow 0} \frac{\delta y}{\delta x}} \quad \ldots .[$ as $\delta x \rightarrow 0, \delta y \rightarrow 0]$
Since limit in R.H.S. exists 
limit in L.H.S. also exists and we have,
$\lim _{\delta y \rightarrow 0} \frac{\delta x}{\delta y}=\frac{d x}{d y}$
$\frac{d x}{d y}=\frac{1}{\frac{d y}{d x}}$, where $\frac{d y}{d x} \neq 0$
Let $y=\tan ^{-1} x$
$x=\tan y \Rightarrow \cos y=\frac{1}{\sqrt{1+\tan ^2 y}}=\frac{1}{\sqrt{1+x^2}}$
$\therefore \sec ^y \cdot \frac{d y}{d x}=1 \Rightarrow \frac{d x}{d y}=\sec ^2 y$
$\frac{d y}{d x}=\frac{1}{\frac{d x}{d y}}=\frac{1}{\sec ^2 y}=\cos ^2 y \Rightarrow \frac{d y}{d x}=\cos ^y$
$\frac{d\left(\tan ^{-1} x\right)}{d x}=\cos ^2 y=(\cos y)^2=\left(\frac{1}{\sqrt{1+x^2}}\right)^2$
$\therefore \frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{1+x^2}$