Question:
If $x=exp\left\{tan^{-1}\left(\frac{y-x^{2}}{x^{2}}\right)\right\}$, then $\frac{dy}{dx}$ equals
If $x=exp\left\{tan^{-1}\left(\frac{y-x^{2}}{x^{2}}\right)\right\}$, then $\frac{dy}{dx}$ equals
Updated On: May 16, 2024
- $2x[1+ tan(logx)] + x sec^2(log\,x)$
- $x[1 + tan(logx)] + sec^2(log \,x)$
- $2x[1 + tan(logx)] + x^2sec^2(log\,x)$
- $2x[1 + tan(logx)] + sec^2(log\,x)$
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The Correct Option is A
Solution and Explanation
Given that, $x=exp\left\{tan^{-1}\left(\frac{y-x^{2}}{x^{2}}\right)\right\}$
Taking log on both sides, we get
$log\,x=tan^{-1}\left(\frac{y-x^{2}}{x^{2}}\right)$
$\Rightarrow \frac{y-x^{2}}{x^{2}}=tan\left(log\,x\right)$
$\Rightarrow y=x^{2}\,tan\left(logx\right)+x^{2}$
Differentiating $w$.$r$.$t$. $x$, we get
$\frac{dy}{dx}=2x\,tan\left(log\,x\right)+x^{2} \frac{sec^{2}\left(log\,x\right)}{x}+2x$
$\Rightarrow \frac{dy}{dx}=2x\left[1+tan\left(log\,x\right)\right]+x\,sec^{2}\left(log\,x\right)$
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Concepts Used:
Continuity & Differentiability
Definition of Differentiability
f(x) is said to be differentiable at the point x = a, if the derivative f ‘(a) be at every point in its domain. It is given by
Definition of Continuity
Mathematically, a function is said to be continuous at a point x = a, if
It is implicit that if the left-hand limit (L.H.L), right-hand limit (R.H.L), and the value of the function at x=a exist and these parameters are equal to each other, then the function f is said to be continuous at x=a.
If the function is unspecified or does not exist, then we say that the function is discontinuous.