if (ze^{5y}+y^{3}cos(5x)=e^{4x}) defines (y) implicitly as a differentiable function of (x) then…

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)})