Question

If $y=y(x)$ and it follows the relation $e^{x{y}^2}+y\cos \left(x^2\right)=5$ then $y^{\prime }(0)$ is equal to

• A. 4
• B. -16
• C. -4
• D. 16

Answer 1

Answer: (B)

We have $e^{xy}{}^2+y\cos \left(x^2\right)=5$ .....(1)
Put $x=0$, we get
$1+y=5\Rightarrow y=4\Rightarrow (0,4)$ lies on the given curve.
Now, differentiating (1) with respect to $x$, we get
$\Rightarrow e^{xy}\left[x\cdot 2y\cdot \frac{dy}{dx}+y^2\right]-y\cdot 2x\cdot \sin \left(x^2\right)+\cos \left(x^2\right)\frac{dy}{dx}=0$
As $(0,4)$ satisfy it, we get
${{16+\frac{dy}{dx}]}_{(0,4)}=0\Rightarrow \frac{dy}{dx}]}_{(0,4)}=-16$