Explanation
Given y= $$e^{x^3} +\frac{logx}{x}$$
$$(\frac{dy}{dx})=(\frac{de^{x^3}}{dx})+(\frac{d(\frac{logx}{2})}{dx})\\=(\frac{de^{x^3}}{dx^3})\times(\frac{dx^3}{dx})+(\frac{1}{2})(\frac{dlogx}{dx})\\=3x^2e^{x^3}+(\frac{1}{2x})$$
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