Find dydx- y-ex-ux-log x-

Given y =e^x +x^x log x on differentiating wrt x we get dy/dx =d/dx [ e^x +x^x log x] dy/dx =e^x +u^x log _eU 1/x ...

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Logarithmic Differentiation

Find dydx, ...

Question

Find $\frac{d y}{d x}$ , $y = e^{x} + u^{x} - log x$ .

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Solution

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Similar questions

y = e^{x} + x^{2} + log x

\frac{d y}{d x}

x^{y} = e^{x - y}

\frac{d y}{d x} = \frac{log x}{(log x + 1)^{2}}

x^{y} = e^{x - y}, x \neq y

\frac{d y}{d x} = \frac{log x}{(log x + 1)^{2}}

x^{y} = e^{x - y}

\frac{d y}{d x} = \frac{l o g x}{(1 - l o g x)^{2}}

x^{y} = e^{x - y}

\frac{d y}{d x}

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