use implicit differentiation to find $\frac{dy}{dx}$. $x^{4}+cos y = x^{3}y^{5}$ $\frac{dy}{dx}=square$

Explanation:

Step1: Differentiate both sides

Differentiate $x^{4}+\cos y=x^{3}y^{5}$ with respect to $x$. The derivative of $x^{4}$ with respect to $x$ is $4x^{3}$ by the power - rule. The derivative of $\cos y$ with respect to $x$ is $-\sin y\frac{dy}{dx}$ using the chain - rule. For the right - hand side, use the product rule $(uv)^\prime = u^\prime v+uv^\prime$ where $u = x^{3}$ and $v = y^{5}$. The derivative of $x^{3}$ is $3x^{2}$ and the derivative of $y^{5}$ is $5y^{4}\frac{dy}{dx}$. So, $4x^{3}-\sin y\frac{dy}{dx}=3x^{2}y^{5}+5x^{3}y^{4}\frac{dy}{dx}$.

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

Move all terms with $\frac{dy}{dx}$ to one side: $-\sin y\frac{dy}{dx}-5x^{3}y^{4}\frac{dy}{dx}=3x^{2}y^{5}-4x^{3}$.

Step3: Factor out $\frac{dy}{dx}$

Factor out $\frac{dy}{dx}$ on the left - hand side: $\frac{dy}{dx}(-\sin y - 5x^{3}y^{4})=3x^{2}y^{5}-4x^{3}$.

Step4: Solve for $\frac{dy}{dx}$

Divide both sides by $(-\sin y - 5x^{3}y^{4})$ to get $\frac{dy}{dx}=\frac{4x^{3}-3x^{2}y^{5}}{\sin y + 5x^{3}y^{4}}$.

Answer:

$\frac{4x^{3}-3x^{2}y^{5}}{\sin y + 5x^{3}y^{4}}$