The equation $y^{2} \mathrm{e}^{xy} = 9 \mathrm{e}^{-3} \cdot x^{2}$ defines $y$ as a differentiable function of $x$. The value of $\frac{dv}{dx}$ for $x = -1$ and $y = 3$ is
A$-\frac{15}{2}$
B$\frac{15}{2}$
C3
D15
Answer & Solution
Correct answer: D. 15
Differentiate implicitly using $y=y(x)$. Starting from $$y^{2}\mathrm{e}^{xy}=9\mathrm{e}^{-3}x^{2}$$ we get $$\frac{d}{dx}\left(y^{2}\mathrm{e}^{xy}\right)=\frac{d}{dx}\left(9\mathrm{e}^{-3}x^{2}\right)$$ Using the product rule on the left, $$2yy'\mathrm{e}^{xy}+y^{2}\mathrm{e}^{xy}\frac{d}{dx}(xy)=18\mathrm{e}^{-3}x$$ and $$\frac{d}{dx}(xy)=xy'+y$$ So $$2yy'\mathrm{e}^{xy}+y^{2}\mathrm{e}^{xy}(xy'+y)=18\mathrm{e}^{-3}x$$ Factor $\mathrm{e}^{xy}$ and collect $y'$ terms: $$\mathrm{e}^{xy}\left(2yy'+xy^{2}y'+y^{3}\right)=18\mathrm{e}^{-3}x$$ $$y'\mathrm{e}^{xy}(2y+xy^{2})=18\mathrm{e}^{-3}x-y^{3}\mathrm{e}^{xy}$$ $$y'=\frac{18\mathrm{e}^{-3}x-y^{3}\mathrm{e}^{xy}}{\mathrm{e}^{xy}(2y+xy^{2})}$$ At $x=-1$ and $y=3$, $$xy=-3$$ so $$\mathrm{e}^{xy}=\mathrm{e}^{-3}$$ Then $$y'=\frac{18\mathrm{e}^{-3}(-1)-27\mathrm{e}^{-3}}{\mathrm{e}^{-3}(6-9)}$$ $$y'=\frac{-45\mathrm{e}^{-3}}{-3\mathrm{e}^{-3}}=15$$ This matches option $D$.
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