Question

if (ze^{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.
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Explanation:

Step1: Differentiate both sides with respect to \(x\)

Differentiate \(ze^{5y}+y^{3}\cos(5x)=e^{4x}\) term - by - term.
For the left - hand side, use the product rule and chain rule. The derivative of \(ze^{5y}\) with respect to \(x\) is \(z\cdot5e^{5y}\frac{dy}{dx}+e^{5y}\frac{dz}{dx}\) (product rule \((uv)^\prime = u^\prime v+uv^\prime\) where \(u = z\), \(v = e^{5y}\)), and the derivative of \(y^{3}\cos(5x)\) with respect to \(x\) is \(3y^{2}\frac{dy}{dx}\cos(5x)-5y^{3}\sin(5x)\) (product rule and chain rule). The derivative of the right - hand side \(e^{4x}\) with respect to \(x\) is \(4e^{4x}\). So we have:
\(z\cdot5e^{5y}\frac{dy}{dx}+e^{5y}\frac{dz}{dx}+3y^{2}\cos(5x)\frac{dy}{dx}-5y^{3}\sin(5x)=4e^{4x}\)

Step2: Solve for \(\frac{dy}{dx}\)

Group the terms with \(\frac{dy}{dx}\) together:
\((5ze^{5y}+3y^{2}\cos(5x))\frac{dy}{dx}=4e^{4x}+5y^{3}\sin(5x)-e^{5y}\frac{dz}{dx}\)
Then \(\frac{dy}{dx}=\frac{4e^{4x}+5y^{3}\sin(5x)-e^{5y}\frac{dz}{dx}}{5ze^{5y}+3y^{2}\cos(5x)}\)

Answer:

\(\frac{4e^{4x}+5y^{3}\sin(5x)-e^{5y}\frac{dz}{dx}}{5ze^{5y}+3y^{2}\cos(5x)}\)