Question

find $\frac{dy}{dx}$ by implicit differentiation. $x^{4}-xy + y^{2}=4$

Explanation:

Step1: Differentiate each term

Differentiate $x^4 - xy + y^2=4$ term - by - term with respect to $x$.
The derivative of $x^4$ with respect to $x$ is $4x^3$ by the power rule $\frac{d}{dx}(x^n)=nx^{n - 1}$.
For the term $-xy$, use the product rule $\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}$, where $u=-x$ and $v = y$. So $\frac{d}{dx}(-xy)=-x\frac{dy}{dx}-y$.
For the term $y^2$, use the chain - rule. Let $u = y$, then $\frac{d}{dx}(y^2)=2y\frac{dy}{dx}$.
The derivative of the constant 4 with respect to $x$ is 0.
So we have $4x^3-(x\frac{dy}{dx}+y)+2y\frac{dy}{dx}=0$.

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

Expand the left - hand side: $4x^3 - x\frac{dy}{dx}-y + 2y\frac{dy}{dx}=0$.
Group the terms with $\frac{dy}{dx}$ together: $-x\frac{dy}{dx}+2y\frac{dy}{dx}=y - 4x^3$.
Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(2y - x)=y - 4x^3$.
Solve for $\frac{dy}{dx}$: $\frac{dy}{dx}=\frac{y - 4x^3}{2y - x}$.

Answer:

$\frac{y - 4x^3}{2y - x}$