Question #253962

Second order derivative of (log x-x) is


1
Expert's answer
2021-10-21T04:13:07-0400

as

logxx = 1

So

d2dx2(logxx)=ddx[ddx(logxx)]\cfrac{d^2}{dx^2} ( log_x x)= \cfrac{d}{dx}[\cfrac{d}{dx} (log_x x)]

Also,


ddx[ddx(logxx)]=ddx[ddx(1)]=0\cfrac{d}{dx}[\cfrac{d}{dx} (log_x x)]= \cfrac{d}{dx}[\cfrac{d}{dx} ( 1)] = 0


And hence the answer is 0

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