Implicit Differentiation — Question 5

Question 5

Consider the equation: $$x^2y + e^{xy} = y^3 + 4x$$

• Use implicit differentiation to find $\dfrac{dy}{dx}$.

• Find the slope of the tangent line to the curve at the point $(0,1)$.

Question 5 - Solution

We are given: $$x^2y + e^{xy} = y^3 + 4x$$

Differentiate both sides implicitly with respect to $x$: $$\frac{d}{dx}[x^2y] + \frac{d}{dx}[e^{xy}] = \frac{d}{dx}[y^3] + \frac{d}{dx}[4x]$$

Using the product rule and chain rule: $$x^2\frac{dy}{dx} + 2xy + e^{xy}\left(x\frac{dy}{dx}+y\right) = 3y^2\frac{dy}{dx}+4$$

Expand: $$x^2\frac{dy}{dx}+2xy+xe^{xy}\frac{dy}{dx}+ye^{xy} = 3y^2\frac{dy}{dx}+4$$

Move the $\dfrac{dy}{dx}$ terms to one side: $$x^2\frac{dy}{dx}+xe^{xy}\frac{dy}{dx}-3y^2\frac{dy}{dx} = 4-2xy-ye^{xy}$$

Factor: $$\left(x^2+xe^{xy}-3y^2\right)\frac{dy}{dx} = 4-2xy-ye^{xy}$$

Therefore, $$\boxed{ \frac{dy}{dx} = \frac{4-2xy-ye^{xy}}{x^2+xe^{xy}-3y^2} }$$

At $(0,1)$, we have $xy=0$, so $e^{xy}=e^0=1$. Thus, $$\frac{dy}{dx} = \frac{4-2(0)(1)-(1)(1)} {0^2+0(1)-3(1)^2} = \frac{3}{-3} = \boxed{-1}$$

Therefore, the slope of the tangent line at $(0,1)$ is: $$\boxed{-1}$$