If f(x)=e^(x)lnx, can you calculate f " (x)?

Answer 1

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

Answer 2

f"(x) means "second derivative"

To get first derivative, use the product rule
derivative of first * second + derivative of second * first
e^x * ln x + 1/ x * e^x

Now, to get the second derivative, use the product rule on each part in the addition.
d/dx ( e^x * ln x) = e^x * ln x + 1/x * e^x
= (xe^x(ln x) + e^x) / x

ADDED to

d/dx (1/x * e^x) = -x^-2*e^x + e^x/x
= (-x^-1*e^x + e^x)/x

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

The answer is (e^x(xlnx + 2-x^-1)/x

Answer 3

of course...

first...

derivatives:

e^(x) = e^(x)
ln x = 1/x
1/x = x^(-1) = -x^(-2) = -1/x²

for the first derivative use the product rule...

f'(x) = e^(x) ln x + e^(x) 1/x

for the second derivative use the product rule again...

f''(x) = e^(x) ln x + e^(x) 1/x + e^(x) 1/x - e^(x) 1/x²

= e^(x) ln x + 2 e^(x) 1/x - e^(x) 1/x²

Answer 4

Ue the product rule.
f'(x) = e^x*x^(-1) + lnx*e^x
f"(x) =-e^x*x^(-2) +e^x*x^(-1)+lnx*e^x+e^x*x^(-1)
f"(x) =2e^x*x^(-1)-e^x*x^(-2)+lnx*e^x

Answer 5

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