Question

double differentiation of y=e^x.log^x​

Answer 1

Explanation:

y=e^x.log^x

dy/dx = e^x d/dx (logx) + (logx)d/dx e^x

= e^x (1/x) + log x (e^x)

= e^x + log x e^x

= e^x (log x + 1/x)

now again take differentiation

d/dx [e^x (log x + 1/x)]

e^x d/dy (x/x + log x)

e^x ( 1+ log x)

hope it helps......

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