Differentiation (Part-38) | S N De

 

$~19(i)~~y=\sqrt{\frac{x^2+1}{x^2-1}}$

Sol. $~~y=\sqrt{\frac{x^2+1}{x^2-1}}\rightarrow(1)\\ \text{or,}~~ \log y=\log\left(\frac{x^2+1}{x^2-1}\right)^{1/2} \\ \text{or,}~~ \log y=\frac 12\log\left(\frac{x^2+1}{x^2-1}\right) \\ \text{or,}~~ \log y=\frac 12[~\log(x^2+1)-\log(x^2-1)~]\\ \therefore ~\text{Differentiating w.r.t}~~x,\\~~\frac 1y~\frac{dy}{dx}\\=\frac 12\left[\frac{d}{dx}~(\log(x^2+1))-\frac{d}{dx}(\log(x^2-1))\right]\\=\frac 12[\frac{1}{x^2+1}\cdot \frac{d}{dx}(x^2+1)-\frac{1}{x^2-1}\cdot \frac{d}{dx}(x^2-1)]\\=\frac 12[\frac{1}{x^2+1}\cdot (2x+0)-\frac{1}{x^2-1} \cdot (2x-0)]\\=\frac 12 \cdot 2x \left(\frac{1}{x^2+1}-\frac{1}{x^2-1}\right)\\=x \left[\frac{x^2-1-(x^2+1)}{(x^2+1)(x^2-1)}\right]\\=x~\cdot \frac{x^2-1-x^2-1}{(x^2+1)(x^2-1)}\\=\frac{x \cdot(-2)}{(x^2+1)(x^2-1)}\\=\frac{-2x}{(x^2+1)(x^2-1)} \\ \therefore \frac{dy}{dx}\\=y \cdot \frac{-2x}{(x^2+1)(x^2-1)}\\=\sqrt{\frac{x^2+1}{x^2-1}} \cdot \frac{-2x}{(x^2+1)(x^2-1)}~~[\text{By (1)}]\\=\frac{-2x}{\sqrt{x^2+1}\cdot \sqrt{x^2-1}~\cdot(x^2-1)} \\=-\frac{2x}{\sqrt{(x^2)^2-1}~ \cdot(x^2-1)}\\=-\frac{2x}{(x^2-1)~\sqrt{x^4-1}}~~\text{(ans.)}$

$~(ii)~~y=(1-x)(1-2x)(1-3x^2)$

Sol. $~~y=(1-x)(1-2x)(1-3x^2) \\ \text{or,}~~ \log y=\log[(1-x)(1-2x)(1-3x^2)]\\ \text{or,}~~ \log y=\log(1-x) +\log(1-2x) \\~~~~+\log(1-3x^2)\\ \therefore ~\text{Differentiating w.r.t}~~x,\\~~\frac 1y~\frac{dy}{dx}\\=\frac{d}{dx}~[\log(1-x)]+\frac{d}{dx}[\log(1-2x)]\\~~~~+\frac{d}{dx}[\log(1-3x^2)]\\=\frac{1}{1-x}~\cdot \frac{d}{dx}(1-x)+\frac{1}{1-2x}\cdot \frac{d}{dx}(1-2x)\\~~~~+\frac{1}{1-3x^2}~\cdot \frac{d}{dx}(1-3x^2)\\=\frac{1}{1-x}~\cdot(-1)+\frac{1}{1-2x}~\cdot(-2)\\~~~~+\frac{1}{1-3x^2}\cdot (-3\cdot 2x)\\=-\left(\frac{1}{1-x}+\frac{2}{1-2x}+\frac{6x}{1-3x^2}\right) \\ \therefore~\frac{dy}{dx}\\=-y\left(\frac{1}{1-x}+\frac{2}{1-2x}+\frac{6x}{1-3x^2}\right)\\=-(1-x)(1-2x)(1-3x^2)\\~~~~ \times\left(\frac{1}{1-x}+\frac{2}{1-2x}+\frac{6x}{1-3x^2}\right)~~\text{(ans.)}$

$~(iii)~~y=\sqrt{\frac{(x-3)(x^2+4)}{3x^2+4x+5}}$

Sol. $~~y=\sqrt{\frac{(x-3)(x^2+4)}{3x^2+4x+5}} \\ \text{or,}~~\log y=\log\left[\frac{(x-3)(x^2+4)}{3x^2+4x+5}\right]^{1/2} \\ \text{or,}~~\log y=\frac 12\log\left[\frac{(x-3)(x^2+4)}{3x^2+4x+5}\right]\\ \text{or,}~~ 2\log y=\log(x-3)+\log(x^2+4)\\~~~~-\log(3x^2+4x+5)\\ \therefore ~\text{Differentiating w.r.t}~~x, \\~~\frac 2y~\frac{dy}{dx}\\=\frac{1}{x-3}~\frac{d}{dx}(x-3)+\frac{1}{x^2+4}~\frac{d}{dx}(x^2+4)\\~~~~-\frac{1}{3x^2+4x+5}~\frac{d}{dx}(3x^2+4x+5)\\=\frac{1}{x-3}~\cdot 1+\frac{1}{x^2+4}~\cdot (2x)\\~~~~-\frac{1}{3x^2+4x+5}~\cdot (6x+4) \\ \therefore~\frac{dy}{dx}\\=\frac 12~y \left[\frac{1}{x-3}+\frac{2x}{x^2+4}-\frac{2(3x+2)}{3x^2+4x+5}\right]\\=\frac 12~\sqrt{\frac{(x-3)(x^2+4)}{3x^2+4x+5}} \left[\frac{1}{x-3}+\frac{2x}{x^2+4}\\~~~~-\frac{2(3x+2)}{3x^2+4x+5}\right]~\text{(ans.)}$

$~20.~$ Find the derivatives of the following :

$~(i)~~x^2~$ w.r.t. $~~\log x$

Sol. let $~~y=x^2,~~z=\log x~~$ so that we need to find the value of $~~\frac{dy}{dz}.$

Now, $~~y=x^2 \\ \therefore~\frac{dy}{dx}=\frac{d}{dx}(x^2)=2x\rightarrow(1)$

Again, $~~z=\log x \\ \therefore~\frac{dz}{dx}=\frac{d}{dx}(\log x)=\frac 1x \rightarrow(2)$

So, $~~\frac{dy}{dz}\\=\frac{dy}{dx} \cdot \frac{dx}{dz}\\=2x~\cdot x~~[\text{By (1),(2)}]\\=2x^2~~\text{(ans.)}$

$~(ii)~~\tan x~$ w.r.t. $~~\log x$

Sol. let $~~y=\tan x,~~z=\log x~~$ so that we need to find the value of $~~\frac{dy}{dz}.$

Now, $~~y=\tan x \\ \therefore~\frac{dy}{dx}=\frac{d}{dx}(\tan x)=\sec^2x\rightarrow(1)$

Again, $~~z=\log x \\ \therefore~\frac{dz}{dx}=\frac{d}{dx}(\log x)=\frac 1x \rightarrow(2)$

So, $~~\frac{dy}{dz}\\=\frac{dy}{dx} \cdot \frac{dx}{dz}\\=\sec^2x~\cdot x~~[\text{By (1),(2)}]\\=x~\sec^2x~~\text{(ans.)}$

$~(iii)~~\cot x~$ w.r.t. $~~\csc x$

Sol. let $~~y=\cot x,~~z=\csc x~~$ so that we need to find the value of $~~\frac{dy}{dz}.$

Now, $~~y=\cot x \\ \therefore~\frac{dy}{dx}=\frac{d}{dx}(\cot x)=-\csc^2x\rightarrow(1)$

Again, $~~z=\csc x \\ \therefore~\frac{dz}{dx}=\frac{d}{dx}(\csc x)=-\csc x \cot x\rightarrow(2)$

So, $~~\frac{dy}{dz}\\=\frac{dy}{dx} \cdot \frac{dx}{dz}\\=(-\csc^2x)~\cdot \frac{1}{-\csc x\cot x}~~[\text{By (1),(2)}]\\=\csc x~\cdot \tan x\\=\frac{1}{\sin x} \cdot \frac{\sin x}{\cos x}\\=\frac{1}{\cos x}\\=\sec x~~\text{(ans.)}$

$~(iv)~~\log_{10}x~$ w.r.t. $~~\tan^{-1}x$

Sol. let $~~y=\log_{10}x,~~z=\tan^{-1}x~$ so that we need to find the value of $~~\frac{dy}{dz}.$

Now, $~~y=\log_{10}x=\log_ex~ \cdot~\log_{10}e  \\ \therefore~\frac{dy}{dx}\\=\log_{10}e~\cdot\frac{d}{dx}(\log_ex)\\=\log_{10}e~\cdot \frac 1x\rightarrow(1)$

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

So, $~~\frac{dy}{dz}\\=\frac{dy}{dx} \cdot \frac{dx}{dz}\\=\frac{\log_{10}e}{x}~\cdot (1+x^2)~~[\text{By (1),(2)}]\\=\frac{1+x^2}{x}~\cdot \log_{10}e~~\text{(ans.)}$

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