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 ...

You visited us 2 times! Enjoying our articles? Unlock Full Access!

Logarithmic Differentiation

Find dydx, ...

Question

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

Open in App

Solution

Suggest Corrections

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}

Join BYJU'S Learning Program