question 14 of 15, step 1 of 1 use implicit differentiation to determine dy/dx for the equation ye^5x + 6y…

Explanation:

Step1: Differentiate each term

Differentiate $ye^{5x}$, $6y$ and $1$ with respect to $x$. For $ye^{5x}$, use product - rule $(uv)^\prime = u^\prime v+uv^\prime$ where $u = y$ and $v = e^{5x}$. The derivative of $y$ with respect to $x$ is $\frac{dy}{dx}$, and the derivative of $e^{5x}$ with respect to $x$ is $5e^{5x}$. So the derivative of $ye^{5x}$ is $\frac{dy}{dx}e^{5x}+5ye^{5x}$. The derivative of $6y$ with respect to $x$ is $6\frac{dy}{dx}$, and the derivative of the constant $1$ with respect to $x$ is $0$. So we have $\frac{dy}{dx}e^{5x}+5ye^{5x}+6\frac{dy}{dx}-0 = 0$.

Step2: Isolate $\frac{dy}{dx}$

Factor out $\frac{dy}{dx}$ from the terms containing it: $\frac{dy}{dx}(e^{5x}+6)=- 5ye^{5x}$. Then solve for $\frac{dy}{dx}$: $\frac{dy}{dx}=\frac{-5ye^{5x}}{e^{5x}+6}$.

Answer:

$\frac{-5ye^{5x}}{e^{5x}+6}$