Question

use implicit differentiation to find $\frac{dy}{dx}$ given the equation $e^{xy}=y^{4}$. $\frac{dy}{dx}=\frac{y(e^{(xy)})}{(x(e^{(xy)}-4y^{3}))}$ question help: video 1 video 2

Explanation:

Step1: Differentiate both sides

Differentiate \(e^{xy}\) using the chain rule and product rule. The derivative of \(e^{u}\) with respect to \(x\) is \(e^{u}\cdot\frac{du}{dx}\), where \(u = xy\). So \(\frac{d}{dx}(e^{xy})=e^{xy}\cdot(y + x\frac{dy}{dx})\).

Differentiate \(y^{4}\) with respect to \(x\) using the chain rule: \(\frac{d}{dx}(y^{4}) = 4y^{3}\frac{dy}{dx}\).

So we have \(e^{xy}(y + x\frac{dy}{dx})=4y^{3}\frac{dy}{dx}\).

Step2: Expand and collect \(\frac{dy}{dx}\) terms

Expand the left - hand side: \(ye^{xy}+xe^{xy}\frac{dy}{dx}=4y^{3}\frac{dy}{dx}\).

Move all terms with \(\frac{dy}{dx}\) to one side: \(xe^{xy}\frac{dy}{dx}-4y^{3}\frac{dy}{dx}=-ye^{xy}\).

Factor out \(\frac{dy}{dx}\): \(\frac{dy}{dx}(xe^{xy}-4y^{3})=-ye^{xy}\).

Step3: Solve for \(\frac{dy}{dx}\)

Divide both sides by \((xe^{xy}-4y^{3})\) to get \(\frac{dy}{dx}=\frac{-ye^{xy}}{xe^{xy}-4y^{3}}=\frac{ye^{xy}}{4y^{3}-xe^{xy}}\).

Answer:

\(\frac{dy}{dx}=\frac{ye^{xy}}{4y^{3}-xe^{xy}}\)