Question

use implicit differentiation to find \\( \frac { d y } { d x } \\) without first solving for \\( y \\).
\\( e ^ { 5 x y } = y ^ { 8 } \\)
\\( \frac { d y } { d x } = \\)
question help: video message instructor

Explanation:

Step1: Differentiate both sides

Differentiate \(e^{5xy}\) and \(y^{8}\) with respect to \(x\).
For \(e^{5xy}\), use the chain rule \((e^{u})^\prime=e^{u}\cdot u^\prime\) where \(u = 5xy\). Then \(u^\prime=5y + 5x\frac{dy}{dx}\) (using the product rule \((uv)^\prime=u^\prime v+uv^\prime\) with \(u = 5x\) and \(v = y\)). So \((e^{5xy})^\prime=e^{5xy}(5y + 5x\frac{dy}{dx})\).
For \(y^{8}\), use the chain rule \((y^{n})^\prime=ny^{n - 1}\frac{dy}{dx}\), so \((y^{8})^\prime=8y^{7}\frac{dy}{dx}\).
We get \(e^{5xy}(5y + 5x\frac{dy}{dx})=8y^{7}\frac{dy}{dx}\).

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

Expand the left - hand side: \(5ye^{5xy}+5xe^{5xy}\frac{dy}{dx}=8y^{7}\frac{dy}{dx}\).
Move all terms with \(\frac{dy}{dx}\) to one side: \(5xe^{5xy}\frac{dy}{dx}-8y^{7}\frac{dy}{dx}=- 5ye^{5xy}\).
Factor out \(\frac{dy}{dx}\): \(\frac{dy}{dx}(5xe^{5xy}-8y^{7})=-5ye^{5xy}\).

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

\(\frac{dy}{dx}=\frac{-5ye^{5xy}}{5xe^{5xy}-8y^{7}}=\frac{5ye^{5xy}}{8y^{7}-5xe^{5xy}}\).

Answer:

\(\frac{5ye^{5xy}}{8y^{7}-5xe^{5xy}}\)