Question

question 2 of 14, step 1 of 1
use implicit differentiation to determine \\( \frac { d y } { d x } \\) for the equation \\( y e ^ { 3 x } + 5 y - 2 = 0 \\).
answer
\\( \frac { d y } { d x } = \\)

Explanation:

Step1: Differentiate both sides

Differentiate \(ye^{\frac{1}{x}}+5y - 2 = 0\) with respect to \(x\).
Using product rule \((uv)^\prime=u^\prime v+uv^\prime\) where \(u = y\), \(v=e^{\frac{1}{x}}\), and \((e^{\frac{1}{x}})^\prime=e^{\frac{1}{x}}\times(-\frac{1}{x^{2}})\), \((5y)^\prime = 5\frac{dy}{dx}\).
We get \(\frac{dy}{dx}e^{\frac{1}{x}}+y\times(-\frac{1}{x^{2}})e^{\frac{1}{x}}+5\frac{dy}{dx}=0\).

Step2: Solve for \(\frac{dy}{dx}\)

Factor out \(\frac{dy}{dx}\): \(\frac{dy}{dx}(e^{\frac{1}{x}} + 5)=\frac{y}{x^{2}}e^{\frac{1}{x}}\).
Then \(\frac{dy}{dx}=\frac{ye^{\frac{1}{x}}}{x^{2}(e^{\frac{1}{x}} + 5)}\).

Answer:

\(\frac{ye^{\frac{1}{x}}}{x^{2}(e^{\frac{1}{x}} + 5)}\)