Question

webwork 2 - topics 4 - 5: problem 8 (1 point) if $xe^{5y}+y^{3}cos(5x)=e^{4x}$ defines $y$ implicitly as a differentiable function of $x$ then $\frac{dy}{dx}=$ note: your answer may contain both $x$ and $y$ variables

Explanation:

Step1: Differentiate both sides

Differentiate $xe^{5y}+y^{3}\cos(5x)=e^{4x}$ with respect to $x$. Using product - rule $(uv)^\prime = u^\prime v+uv^\prime$ and chain - rule.
The derivative of $xe^{5y}$ is $e^{5y}+x\cdot e^{5y}\cdot5\frac{dy}{dx}$ (product - rule with $u = x$, $v = e^{5y}$). The derivative of $y^{3}\cos(5x)$ is $3y^{2}\frac{dy}{dx}\cos(5x)-5y^{3}\sin(5x)$ (product - rule with $u = y^{3}$, $v=\cos(5x)$) and the derivative of $e^{4x}$ is $4e^{4x}$.
So, $e^{5y}+5xe^{5y}\frac{dy}{dx}+3y^{2}\cos(5x)\frac{dy}{dx}-5y^{3}\sin(5x)=4e^{4x}$.

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

Group the terms with $\frac{dy}{dx}$ on one side:
$5xe^{5y}\frac{dy}{dx}+3y^{2}\cos(5x)\frac{dy}{dx}=4e^{4x}-e^{5y} + 5y^{3}\sin(5x)$.
Factor out $\frac{dy}{dx}$:
$\frac{dy}{dx}(5xe^{5y}+3y^{2}\cos(5x))=4e^{4x}-e^{5y}+5y^{3}\sin(5x)$.

Step3: Solve for $\frac{dy}{dx}$

$\frac{dy}{dx}=\frac{4e^{4x}-e^{5y}+5y^{3}\sin(5x)}{5xe^{5y}+3y^{2}\cos(5x)}$.

Answer:

$\frac{4e^{4x}-e^{5y}+5y^{3}\sin(5x)}{5xe^{5y}+3y^{2}\cos(5x)}$