Question

find $\frac{dy}{dx}$ by implicit differentiation
$2 + 5x=sin(xy^{4})$
$\frac{dy}{dx}=$

Explanation:

Step1: Differentiate both sides

Differentiate $2 + 5x$ and $\sin(xy^{4})$ with respect to $x$. The derivative of $2+5x$ is $5$. For $\sin(xy^{4})$, use the chain - rule. Let $u = xy^{4}$, then $\frac{d}{dx}\sin(u)=\cos(u)\cdot\frac{du}{dx}$. And $\frac{du}{dx}=y^{4}+4xy^{3}\frac{dy}{dx}$ by the product - rule. So $\frac{d}{dx}\sin(xy^{4})=\cos(xy^{4})(y^{4}+4xy^{3}\frac{dy}{dx})$. We get the equation $5=\cos(xy^{4})(y^{4}+4xy^{3}\frac{dy}{dx})$.

Step2: Expand the right - hand side

$5 = y^{4}\cos(xy^{4})+4xy^{3}\cos(xy^{4})\frac{dy}{dx}$.

Step3: Isolate $\frac{dy}{dx}$

First, move the term without $\frac{dy}{dx}$ to the left - hand side: $5 - y^{4}\cos(xy^{4})=4xy^{3}\cos(xy^{4})\frac{dy}{dx}$. Then, divide both sides by $4xy^{3}\cos(xy^{4})$ to solve for $\frac{dy}{dx}$.

Answer:

$\frac{5 - y^{4}\cos(xy^{4})}{4xy^{3}\cos(xy^{4})}$