If x = exp left{ {{{tan }^{ - 1}}left( {{{y - | vedclass

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(A) Given $x = \exp \left\{ {{{	an }^{ - 1}}\left( {\frac{{y - {x^2}}}{{{x^2}}}} ight)} ight\}$.Taking natural logarithm on both sides,we get $\l

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given $x = \exp \left\{ {{{	an }^{ - 1}}\left( {\frac{{y - {x^2}}}{{{x^2}}}} ight)} ight\}$.Taking natural logarithm on both sides,we get $\log x = {	an ^{ - 1}}\left( {\frac{{y - {x^2}}}{{{x^2}}}} ight)$.This implies $\frac{{y - {x^2}}}{{{x^2}}} = 	an (\log x)$.Rearranging for $y$,we get $y = {x^2}	an (\log x) + {x^2}$.Differentiating both sides with respect to $x$ using the product rule and chain rule:$\frac{{dy}}{{dx}} = \frac{d}{{dx}}({x^2}	an (\log x)) + \frac{d}{{dx}}({x^2})$.$\frac{{dy}}{{dx}} = [2x \cdot 	an (\log x) + {x^2} \cdot {\sec ^2}(\log x) \cdot \frac{1}{x}] + 2x$.$\frac{{dy}}{{dx}} = 2x	an (\log x) + x{\sec ^2}(\log x) + 2x$.Factoring out $2x$ from the first and last terms,we get $\frac{{dy}}{{dx}} = 2x[1 + 	an (\log x)] + x{\sec ^2}(\log x)$.

If $x = \exp \left\{ {{{\tan }^{ - 1}}\left( {{{y - {x^2}} \over {{x^2}}}} \right)} \right\}$,then $\frac{dy}{dx}$ equals

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