Question

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 } }}$

Answer 1

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.

Complete step-by-step answer:
\[\dfrac{{dy}}{{dx}} = \dfrac{1}{{3\sqrt {5\pi } }}\] \[{{\text{e}}^{\operatorname{Sin} ({x^2} + {y^2})}} = {\text{ tan }}\dfrac{{{y^2}}}{4} + {\sin ^{ - 1}}x - - - - - - - - (1)\]
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.