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If x=\exp \left\{\tan ^{-1}\left(\frac{y-x^{2}}{x^{2}}\right)\right\}, then \frac{d y}{d x} equals ?

Option: 1

{2 x[1+\tan (\log x)]+x \sec ^{2}(\log x)}


Option: 2

\\ {x[1+\tan (\log x)]+\sec ^{2}(\log x)}


Option: 3

\\ {2 x[1+\tan (\log x)]+x^{2} \sec ^{2}(\log x)}


Option: 4

\\ {2 x[1+\tan (\log x)]+\sec ^{2}(\log x)}


Answers (1)

best_answer

 

 

Differentiation of Function and Relation -

Differentiation of Function and Relation

To solve the question of the type where functional relation in two independent variables with some conditions are given and asked to find the derivative of the function at some value of x or it is asked to find the function. 

Since, there are two independent variables in the functional relation we can differentiate the relation w.r.t. any one variable considering another variable constant, 

Let’s go through some illustration to understand how to deal with such questions.

Illustration 1

\\\mathrm{Let\;\;f\left(\frac{x+y}{2}\right)=\frac{f(x)+f(y)}{2}\;for\;all\;real\;.\;x \;and\;y.\;If\;f'(0)\;exists\;}\\\mathrm{and\;equal\;to\;-1\;and\;f(0)=1.\;Then\;f'(x)\;is}

\\\text { Given equation is } f\left(\frac{x+y}{2}\right)=\frac{f(x)+f(y)}{2}\;\;\;\;\;\;\;\;\;\;\;\;\;\;...(i)\\\text { Putting } y=0 \text { and } f(0)=1 \text { in (i), we have, }\\f\left(\frac{x}{2}\right)=\frac{1}{2}[f(x)+1] \quad \Rightarrow f(x)=2 f_{i}\left(\frac{x}{2}\right)-1\;\;\;\;\;\;\;\;\;\;\;\;\;...(ii)\\\text{Now,}\;\;\\\begin{aligned} f^{\prime}(x) &=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\ &=\lim _{h \rightarrow 0} \frac{f\left(\frac{2 x+2 h}{2}\right)-f(x)}{h} \\ &=\lim _{h \rightarrow 0} \frac{\frac{f(2 x)+f(2 h)}{2}-f(x)}{h} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;[\text { using }(\mathrm{i})]\\ &=\lim _{h \rightarrow 0} \frac{f(2 x)+f(2 h)-2 f(x)}{2 h} \\ &=\lim _{h \rightarrow 0} \frac{2 f(x)-1+f(2 h)-2 f(x)}{2 h} \;\;\;\;\;\;\;\;\;\;\;[\text { using }(\mathrm{ii})]\end{aligned}\\\mathrm{\Rightarrow \lim_{h\rightarrow 0}\frac{f(2h)-1}{2h}=f'(0)}\\\mathrm{\therefore \;\;f'(x)=-1}  

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x=\exp \left\{\tan ^{-1}\left(\frac{y-x^{2}}{x^{2}}\right)\right\}\\ \log x=\tan ^{-1}\left(\frac{y-x^{2}}{x^{2}}\right)\\ \tan (\log x)=\frac{y-x^{2}}{x^{2}}\\ y=x^{2} \tan (\log x)+x^{2}\\ \frac{d y}{d x}=x^{2} \cdot \frac{\sec ^{2}(\log x)}{x}+2 x \cdot \tan (\log x)+2 x\\ \frac{d y}{d x}=x \sec ^{2}(\log x)+2 x[1+\tan (\log x)]

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seema garhwal

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