Question

If $x{e}^{xy}=y+{\sin }^{2}x$, then at $x=0$, $\frac{dy}{dx}=$

• A. $-1$
• B. $-2$
• C. $1$
• D. $2$

Answer 1

The correct option is C

$1$

Explanation for the correct option.

Step- 1: Find the value of $y$ at $x=0$.

In the given equation $x{e}^{xy}=y+{\sin }^{2}x$, substitute $0$ for $x$ to find the value of $y$.

$$\begin{gathered} 0e^{0·y}=y+{\sin }^{2}0 \\ \Rightarrow 0=y+0 \\ \Rightarrow 0=y \end{gathered}$$

So at $x=0$, the value of $y$ is $0$.

Step 2. Find the value of $\frac{dy}{dx}$ at $x=0$.

Differentiate the equation $x{e}^{xy}=y+{\sin }^{2}x$ with respect to $x$.

$$\begin{gathered} \frac{d}{dx}\left(x{e}^{xy}\right)=\frac{d}{dx}\left(y+{\sin }^{2}x\right) \\ \Rightarrow e^{xy}+x\left[e^{xy}\left(x\frac{dy}{dx}+y.1\right)\right]=\frac{dy}{dx}+2\sin x\cos x \end{gathered}$$

Now, substitute $x=0$ and $y=0$ to get the value of $\frac{dy}{dx}$ at $x=0$.

$$\begin{gathered} e^{0·0}+0·\left[e^{0·0}\left(0\frac{dy}{dx}+0\frac{dy}{dx}\right)\right]=\frac{dy}{dx}+2\sin 0\cos 0 \\ \Rightarrow e^0+0=\frac{dy}{dx}+2\times 0\times 1 \\ \Rightarrow \frac{dy}{dx}=1 \end{gathered}$$

Hence, the correct option is C.