Question

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

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

Hint:

For terms like \(e^{xy}\), first keep the exponential unchanged and differentiate the exponent separately using product rule.

Answer 1

The Correct Option is D

Concept: The slope of the tangent to a curve is given by \[ \frac{dy}{dx} \] Since the equation contains both \(x\) and \(y\), we use implicit differentiation.

Step 1: Differentiate the equation implicitly.
Given, \[ y^2e^{xy}=9e^{-3}x^2 \] Differentiate both sides with respect to \(x\): \[ \frac{d}{dx}(y^2e^{xy}) = \frac{d}{dx}(9e^{-3}x^2) \] Using product rule on the left side: \[ 2y\frac{dy}{dx}e^{xy} + y^2e^{xy}\frac{d}{dx}(xy) = 18e^{-3}x \] Now, \[ \frac{d}{dx}(xy)=y+x\frac{dy}{dx} \] Therefore, \[ 2y\frac{dy}{dx}e^{xy} + y^2e^{xy}\left(y+x\frac{dy}{dx}\right) = 18e^{-3}x \]

Step 2: Substitute the point \((-1,3)\).
At the point, \[ x=-1,\qquad y=3 \] Also, \[ xy=(-1)(3)=-3 \] Substituting: \[ 2(3)\frac{dy}{dx}e^{-3} + (3)^2e^{-3}\left(3-\frac{dy}{dx}\right) = 18e^{-3}(-1) \] \[ 6e^{-3}\frac{dy}{dx} + 9e^{-3}\left(3-\frac{dy}{dx}\right) = -18e^{-3} \] Divide throughout by \(e^{-3}\): \[ 6\frac{dy}{dx}+9\left(3-\frac{dy}{dx}\right)=-18 \]

Step 3: Solve for \( \frac{dy}{dx} \).
\[ 6\frac{dy}{dx}+27-9\frac{dy}{dx}=-18 \] \[ -3\frac{dy}{dx}=-45 \] \[ \frac{dy}{dx}=15 \] Hence, the slope of the tangent is \[ \boxed{15} \]