Find f'(x) for f(x)=x^2-1/x 2e^x

Question

I get 2x-Lnx e^x is that right?

Answer 1

You need to add ( ) or else I can't tell if its 1/x times 2e^x or its 1/(x2e^x). But the derivative of -1/x or -x^-1 = - -1x^-2 = x^-2 or 1/(x^2)

f(x)= (x^2)-(1/x)+(2e^x)

f' = 2x + (x^-2) + (2e^x)

the anti-derivative or integral of 1/x is ln x, i think you got a bit confused there? check your text book for integral of 1/x. I'm pretty sure that the derivative of e^x is e^x.

Answer 2

You need to add ( ) or else I can't tell if its 1/x times 2e^x or its 1/(x2e^x). But the derivative of -1/x or -x^-1 = - -1x^-2 = x^-2 or 1/(x^2)

f(x)= (x^2)-(1/x)+(2e^x)

f' = 2x + (x^-2) + (2e^x)

the anti-derivative or integral of 1/x is ln x, i think you got a bit confused there? check your text book for integral of 1/x. I'm pretty sure that the derivative of e^x is e^x.