Question

The equation $y^2 e^{x y}=9 e^{-3} \cdot x^2$ defines $y$ as a differentiable function of $x$. The value of $\frac{d y}{d x}$ for $x=-1$ and $y=3$ is

• A. $-\frac{15}{2}$
• B. $-\frac{9}{5}$
• C. 3
• D. 15

Answer 1

Answer: (D)

Using product rule
$y^2\left(e^{x y}\left(x \frac{d y}{d x}+y\right)\right)+e^{x y} \cdot 2 y \frac{d y}{d x}=9 e^{-3} \cdot 2 x $
$\text { put } x=-1 \text { and } y=3$
$9\left(e^{-3}\left(-1 \frac{d y}{d x}+3\right)\right)+e^{-3} \cdot 6 \frac{d y}{d x}=-9 e^{-3} \cdot 2$
$-9\left(\frac{d y}{d x}-3\right)+6 \frac{d y}{d x}=-18 $
$3 \frac{d y}{d x}=45 \Rightarrow \frac{d y}{d x}=15 $