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# If ${{\text{e}}^{\sin \left( {{x^2} + {y^2}} \right)}} = {\text{tan }}\dfrac{{{y^2}}}{4} + {\sin ^{ - 1}}x$, then ${\text{y'}}\left( 0 \right)$ can be A) $\dfrac{1}{{3\sqrt \pi }}$B) $- \dfrac{1}{{3\sqrt \pi }}$C) $- \dfrac{1}{{5\sqrt \pi }}$D) $- \dfrac{1}{{3\sqrt {5\pi } }}$

Last updated date: 14th Sep 2024
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Hint: This problem comes under implicit function on differentiation. We need to solve separately and want to differentiate on the function and want to find y and then there will be re arranging and substitute to the equation to compare the coordinates for the solving $y'(0) = \dfrac{{dy}}{{dx}}$ which first order differentiation and there will be multiple solvable equation and then complete step by step explanation.

$\dfrac{{dy}}{{dx}} = \dfrac{1}{{3\sqrt {5\pi } }}$ ${{\text{e}}^{\operatorname{Sin} ({x^2} + {y^2})}} = {\text{ tan }}\dfrac{{{y^2}}}{4} + {\sin ^{ - 1}}x - - - - - - - - (1)$
When $x = 0$, then the inverse function will be zero and the values of $x$ will be so, then we get
$\Rightarrow {{\text{e}}^{\sin {y^2}}} = {\text{ tan }}\dfrac{{{y^2}}}{4}$
Taking $\log$ on both sides,
$\Rightarrow \sin {y^2} = {\text{ log tan }}\dfrac{{{y^2}}}{4}$
Hence we get,
$y = \pm \sqrt x , \pm \sqrt {5x} ,..............$
Now differentiating equation (1) with respect to x, we get
Let us solve separately Left hand side and Right hand side,
Now differentiate Left hand side
$\Rightarrow {{\text{e}}^{\sin \left( {{x^2} + {y^2}} \right)}}$
$\Rightarrow {{\text{e}}^{\sin \left( {{x^2} + {y^2}} \right)}}\cos \left( {{x^2} + {y^2}} \right)\left( {2x + 2y\dfrac{{dy}}{{dx}}} \right)$
Now differentiate Right hand side
$\Rightarrow {\text{tan }}\dfrac{{{y^2}}}{4} + {\sin ^{ - 1}}x$
Differentiating we get,
$\Rightarrow \dfrac{{2y}}{4}{\sec ^2}\dfrac{{{y^2}}}{4}\dfrac{{dy}}{{dx}} + \dfrac{1}{{\sqrt {1 - {x^2}} }}$
Now again substitute $x = 0$ and compare,
$\Rightarrow {{\text{e}}^{\sin {y^2}}}\left[ {\cos {y^2}\left( {2y\dfrac{{dy}}{{dx}}} \right)} \right] = \dfrac{y}{2}{\sec ^2}\dfrac{{{y^2}}}{4}\dfrac{{dy}}{{dx}} - - - - (2)$
Now substitute $x$ value and $y$ value in equation (2), we get $y'(0) = \dfrac{{dy}}{{dx}}$
Now, at $x = 0$, $y = \sqrt \pi$ in equation in (2),
$\Rightarrow {{\text{e}}^{\sin {{(\sqrt \pi )}^2}}}[\cos {(\sqrt \pi )^2}2\sqrt \pi \dfrac{{dy}}{{dx}} = \dfrac{{\sqrt \pi }}{2}{\sec ^2}\dfrac{{{{\sqrt \pi }^2}}}{4}\dfrac{{dy}}{{dx}}$
Now separate $\dfrac{{dy}}{{dx}}$,
$\Rightarrow {{\text{e}}^{\sin \pi }}[\cos \pi ]2\sqrt \pi \dfrac{{dy}}{{dx}} = \dfrac{{\sqrt \pi }}{2}{\sec ^2}\dfrac{\pi }{4}\dfrac{{dy}}{{dx}}$
We know that $\sin \pi = 0$ and $\cos \pi = - 1$,
$\Rightarrow {{\text{e}}^0}[ - 1]2\sqrt \pi \dfrac{{dy}}{{dx}} = \dfrac{{\sqrt \pi }}{2}{(\sec \dfrac{\pi }{4})^2}\dfrac{{dy}}{{dx}}$
We know that ${\sec ^2}\dfrac{\pi }{4}\dfrac{{dy}}{{dx}} = 2$ and ${e^0} = 1$,
$\Rightarrow - 2\sqrt \pi \dfrac{{dy}}{{dx}} = \dfrac{{\sqrt \pi }}{2} \times 2\dfrac{{dy}}{{dx}}$
Cancelling the term $2$ in denominator and numerator,
$\Rightarrow$$- 2\sqrt \pi \dfrac{{dy}}{{dx}} = \sqrt \pi \dfrac{{dy}}{{dx}}$
Rearranging the terms we get,
$\Rightarrow$$- 2\sqrt \pi \dfrac{{dy}}{{dx}} - \sqrt \pi \dfrac{{dy}}{{dx}} = 0$
Taking common term same as in both terms,
$\Rightarrow$$( - 2\sqrt \pi - \sqrt \pi )\dfrac{{dy}}{{dx}} = 0$
Subtracting the terms we get,
$\Rightarrow$$- 3\sqrt \pi \dfrac{{dy}}{{dx}} = 0$
Hence we get,
$\Rightarrow$$\dfrac{{dy}}{{dx}} = \dfrac{{ - 1}}{{3\sqrt \pi }}$
Similarly we can find for when $x = 0,y = - \sqrt \pi$ we get
$\Rightarrow$$\dfrac{{dy}}{{dx}} = \dfrac{1}{{3\sqrt \pi }}$
Similarly we can find for when$x = 0,y = \sqrt {5\pi }$, we get
$\Rightarrow$$\dfrac{{dy}}{{dx}} = - \dfrac{1}{{3\sqrt {5\pi } }}$
Similarly we can find for when $x = 0,y = - \sqrt {5\pi }$, we get
$\Rightarrow$$\dfrac{{dy}}{{dx}} = \dfrac{1}{{3\sqrt {5\pi } }}$
There will be multiple answers and for this question, these are,
$\Rightarrow$$\dfrac{1}{{3\sqrt \pi }}$, $- \dfrac{1}{{3\sqrt \pi }}$ and $- \dfrac{1}{{3\sqrt {5\pi } }}$

$\therefore$ The correct answers are option A) $\dfrac{1}{{3\sqrt \pi }}$, B) $- \dfrac{1}{{3\sqrt \pi }}$ and D) $- \dfrac{1}{{3\sqrt {5\pi } }}$

Note: This problem needs attention on differentiation and some trigonometric identities, this kind of problem will be able to solve when the when before differentiation and after differentiation for when find x values and y values for finding the first order differential and then simple basic calculation for that arrange and substitute the value in order to find solution.