Question

The slope of the tangent of the curve $y^2{e}^{xy}=9e^{-3}{x}^2$ at $\left(-1,3\right)$ is

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

Answer 1

The correct option is C

$15$

Explanation for the correct answer:

Compute the slope:

$y^2{e}^{xy}=9e^{-3}{x}^2$

Differentiate with respect to $x$

$2y{e}^{xy}\frac{dy}{dx}+y^2{e}^{xy}(y+x\frac{dy}{dx})=18e^{-3}x$

Substitute $x=-1,y=3$ in this equation we get,

$2\times 3e^{-3}\frac{dy}{dx}+9e^{-3}\left(3-\frac{dy}{dx}\right)=-18e^{-3}$

$\Rightarrow$ $6\frac{dy}{dx}+27-9\frac{dy}{dx}=-18$

$\Rightarrow$ $-3\frac{dy}{dx}=-45$

$\Rightarrow$ $\frac{dy}{dx}=15$

Hence, the slope of the tangent is $15$.

Hence, option $\left(C\right)$ is the correct answer.