Differentiation (Part-38) | S N De

 

In each of the following cases find $~\frac{dy}{dx}~$ using the rule $~~\frac{dy}{dx}=\frac{dy}{du} \cdot \frac{du}{dx}~[1-4]$

$~(1)~~y=\csc^3x~$ assuming $~~\csc x=u.$

Sol. $~~~~~y=\csc^3x \\ \text{or,}~~y=u^3~~[\text{where}~~u=\csc x~]\\ \therefore \frac{dy}{du}=3u^2\rightarrow(1) $

Again, $~~~~~~u=\csc x \\ \therefore~~\frac{du}{dx}=-\csc x \cot x\rightarrow(2)$

So, $~~\frac{dy}{dx}\\=\frac{dy}{du} \cdot \frac{du}{dx}\\=3u^2 \cdot (-\csc x \cot x)~~~~[\text{By (1),(2)}]\\=3\csc^2x \cdot (-\csc x \cot x)\\=-3\csc^3x \cot x~~\text{(ans.)}$

$~~[*]:~$ By $~\csc x~$ we mean cosec$~x.$

$~2.~y=m^{ax^2+bx+c}~~$ assuming $~~ax^2+bx+c=u.$

Sol. $~~~y=m^{ax^2+bx+c} \\ \text{or,}~~y=m^u \\~~[~\text{where}~~~u=ax^2+bx+c ] \\ \therefore ~\frac{dy}{du}=m^u \cdot \log_em \rightarrow(1)$

Also, $~~u=ax^2+bx+c \\ \therefore~\frac{du}{dx}=2ax+b\rightarrow(2)$

Now, $~\frac{dy}{dx}\\=\frac{dy}{du} \cdot \frac{du}{dx}\\=m^u~ \log_em \cdot(2ax+b)~~~~[\text{By (1),(2)}]\\=m^{ax^2+bx+c}~\log_em \cdot(2ax+b)~\text{(ans.)}$

$~3.~~~y=\log(ax+b)^3~$ assuming $~ax+b=u.$

Sol. $~~~~~~y=\log(ax+b)^3 \\ \text{or,}~~ y=\log u^3 \\ \text{or,}~~ y=3\log u \\ \therefore \frac{dy}{du}=\frac{3}{u}\rightarrow(1)$

Also, $~~u=ax+b \\ \therefore~\frac{du}{dx}=a\cdot 1=a \rightarrow(2)$

So, $~~~~\frac{dy}{dx}\\=\frac{dy}{du} \cdot \frac{du}{dx}\\=\frac 3u \cdot a~~~~[\text{By (1),(2)}]\\=\frac{3a}{ax+b}~~\text{(ans.)}$

$~~4.~~y=\sqrt{\tan^{-1}~x}~~$assuming $~~\tan^{-1}~x=u.$

Sol. $~~y=\sqrt{\tan^{-1}~x}=u^{1/2} \\ \therefore~\frac{dy}{du}=\frac 12~u^{\frac 12-1}=\frac 12~u^{-\frac 12}\rightarrow(1)$

Again, $~~u=\tan^{-1}~x \\ \therefore~\frac{du}{dx}\\=\frac{d}{dx}(\tan^{-1}x)\\=\frac{1}{1+x^2}\rightarrow(2)$

Now, $~~~\frac{dy}{dx}\\=\frac{dy}{du} \cdot \frac{du}{dx}\\=\frac 12u^{-\frac 12}\cdot \frac{1}{1+x^2}~~[\text{By (1),(2)}]\\=\frac{1}{2\sqrt{u}~(1+x^2)}\\=\frac{1}{2(1+x^2)\sqrt{\tan^{-1}~x}}~~\text{(ans.)}$

$~5.~~$ If $~u~$ and $~v~$ are two differentiable functions of $~x~$ show that, $~~\frac{d}{dx}\left(\tan^{-1}\frac uv\right)=\frac{v~\frac{du}{dx}-u\frac{dv}{dx}}{u^2+v^2}.$

Sol. $~~\frac{d}{dx}\left(\tan^{-1}\frac uv\right)\\=\frac{1}{1+(u/v)^2}~\cdot\frac{d}{dx}\left(\frac uv\right)\\=\frac{v^2}{v^2+u^2} \cdot \frac{v~\frac{du}{dx}-u\frac{dv}{dx}}{v^2}\\=\frac{v~\frac{du}{dx}-u\frac{dv}{dx}}{u^2+v^2}~~\text{(showed)}$

$~~6.~~$Using the formula $~~\frac{d}{dx}(\tan^{-1}x)=\frac{1}{1+x^2},~~$ deduce that $~~\frac{d}{dx}(\cot^{-1}x)=-\frac{1}{1+x^2}.$

Sol. We know, $~~\tan^{-1}x+\cot^{-1}x=\pi/2 \\ \text{or,}~~ \cot^{-1}x=\pi/2-\tan^{-1}x \\ \therefore~\frac{d}{dx}(\cot^{-1}x)\\=\frac{d}{dx}(\pi/2)-\frac{d}{dx}(\tan^{-1}x)\\=0-\frac{1}{1+x^2}\\=-\frac{1}{1+x^2}.$