Differentiation (Part-38) | S N De

 

$~4(xvii)~~x\cos x+2^x\sec x$

Sol. let $~~y=x\cos x+2^x\sec x \\ \therefore ~\frac{dy}{dx}=x\frac{d}{dx}(\cos x)+ \cos x ~\frac{d}{dx}(x)\\~~~~~~~~~~+2^x ~\frac{d}{dx}(\sec x)+\sec x~\frac{d}{dx}(2^x) \\ \text{or,}~~ \frac{dy}{dx}=-x\sin x+\cos x \cdot 1\\~~~~~~~~~~+2^x \sec x\tan x+\sec x\cdot 2^x \log_e2  \\ \text{or,}~~ \frac{dy}{dx}= \cos x-x \sin x\\~~~~~~~~~~+2^x\sec x(\tan x+\log_e2)~~\text{(ans.)}$

$~(xviii)~~\frac{\tan x}{x}~\log\left(\frac{e^x}{x^x}\right)$

Sol. let $~~y=\frac{\tan x}{x}~\log\left(\frac{e^x}{x^x}\right) \\ \text{or,}~~  y=\frac{\tan x}{x} \cdot \log\left(\frac ex\right)^x \\ \text{or,}~~  y=\frac{\tan x}{x} \cdot x \log(e/x) \\  \text{or,}~~  y=\tan x\cdot (\log_ee-\log_e x) \\ \text{or,}~~ y= \tan x(1-\log_e x) \\ \therefore ~\frac{dy}{dx}\\=\frac{d}{dx} \left[\tan x(1-\log_ex)\right]\\=\frac{d}{dx}(\tan x) \cdot (1- \log_ex) \\~~~~+\tan x \cdot \frac{d}{dx}(1- \log_ex)\\=\sec^2x(1-\log_ex)\\~~~~+\tan x\left(0-\frac 1x\right)\\=(1- \log_ex)\sec^2x-\frac{\tan x}{x}~~\text{(ans.)}$

$~(xix)~~\frac{\log x}{\cos x}$

Sol. let $~~y=\frac{\log_ex}{\cos x} \\ \therefore \frac{dy}{dx}\\=\frac{\cos x \cdot \frac{d}{dx}(\log_ex)-\log_ex \cdot \frac{d}{dx}(\cos x)}{\cos^2x}\\=\frac{\cos x \cdot \frac 1x-\log_ex \cdot (-\sin x)}{\cos^2x}\\=\frac{\cos x+x\sin x \log_ex}{x\cos^2x}~~\text{(ans.)}$

$~(xx)~~\frac{e^x}{\sin x}$

Sol. let $~~y=\frac{e^x}{\sin x} \\ \therefore~ \frac{dy}{dx}\\=\frac{d}{dx}\left(\frac{e^x}{\sin x}\right)\\=\frac{\sin x \cdot \frac{d}{dx}(e^x)-e^x~ \cdot \frac{d}{dx}(\sin x)}{\sin^2x}\\=\frac{\sin x~\cdot e^x-e^x \cdot \cos x}{\sin^2x}\\=\frac{e^x (\sin x-\cos x)}{\sin^2x}~~\text{(ans.)}$

Advanced Problems In Mathematics For JEE Main & Advanced For Examination 2023-2024 Paperback 

Find $\,\,\frac{dy}{dx} :$

$~5(i)~~\log(xy)=x^2-y^2~~\text{when}~~x=1,y=1.$

Sol. $~~\log_e(xy)=x^2-y^2 \\ \text{or,}~~ \log_ex+\log_ey=x^2-y^2 \\ \therefore~\frac{d}{dx}(\log_ex+\log_ey)=\frac{d}{dx}(x^2-y^2) \\ \text{or,}~~ \frac 1x+\frac 1y~\frac{dy}{dx}=2x-2y~\frac{dy}{dx} \\ \text{or,}~~\left(2y+\frac 1y\right)~\frac{dy}{dx}=2x-\frac 1x \\ \text{or,}~~ \frac{2y^2+1}{y}~\frac{dy}{dx}=\frac{2x^2-1}{x} \\ \text{or,}~~ \frac{dy}{dx}=\frac{2x^2-1}{x} \cdot \frac{y}{2y^2+1} \\ \therefore~\left[\frac{dy}{dx}\right]_{(x=1,~y=1)}\\=\frac{2 \times 1^2-1}{1} \times \frac{1}{2 \times 1^2+1}\\=\frac 13~~\text{(ans.)}$

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$~(ii)~~x^y=y^x$

Sol. let $~~x^y=y^x \\ \text{or,}~~ \log_e(x^y)=\log_e(y^x) \\ \text{or,}~~ y \log_e x=x \log_e y \\ \therefore \frac{d}{dx} (y \log_e x)=\frac{d}{dx}(x \log_e y)\\ \text{or,}~~ \frac{dy}{dx} \cdot \log_ex+ y~\frac{d}{dx}(\log_ex) \\~~~~=\frac{d}{dx}(x)~ \log_ey+x \frac{d}{dx}(\log_ey)\\ \text{or,}~~ \frac{dy}{dx}~ \log_ex+y \cdot \frac 1x\\~~~~=1 \cdot \log_ey+x \cdot \frac 1y~\frac{dy}{dx} \\ \text{or,}~~ \left(\log_ex-\frac xy\right)~\frac{dy}{dx}=\log_ey-\frac yx \\ \text{or,}~~ \frac{dy}{dx}\\=\frac{\log_ey-y/x}{\log_ex-x/y}\\=\frac yx \left(\frac{x \log_ey-y}{y\log_ex-x}\right)~~\text{(ans.)}$

$~(iii)~~x^y \cdot y^x=1$

Sol. let $~~x^y \cdot y^x=1 \\ \text{or,}~~ \log(x^y \cdot y^x)=\log 1 \\ \text{or,}~~ \log(x^y) + \log(y^x)=0 \\ \text{or,}~~ y \log x+x\log y=0 \rightarrow(1)\\ \therefore \frac{d}{dx}(y \log x)+\frac{d}{dx}(x \log y)=0 \\ \text{or,}~~ \frac{dy}{dx}~(\log x)+y \frac{d}{dx}(\log x)\\~~~+\frac{d}{dx}(x)\cdot \log y+x \cdot \frac{d}{dx}(\log y)=0 \\ \text{or,}~~ (\log x)~\frac{dy}{dx}+y \cdot \frac 1x+1 \cdot \log y\\~~~+x \cdot \frac 1y~\frac{dy}{dx}=0 \\ \text{or,}~~ \left(\log x+\frac xy\right)~\frac{dy}{dx}=-\left(\frac yx+\log y\right) \\ \text{or,}~~ \frac{dy}{dx}=-\frac{\left(\frac yx+\log y\right)}{\left(\log x+\frac xy\right)} \\ \therefore ~\frac{dy}{dx}\\=-\frac yx \left(\frac{y+x \log y}{x+y\log x}\right) \\=-\frac yx \left(\frac{y-y\log x}{x-x\log y}\right)~~[~\text{By (1)}]\\=-\frac{y^2}{x^2} \left(\frac{1-\log x}{1-\log y}\right)~~\text{(ans.)}$