f(x) = x^2 + √x
We can rewrite this as:
f(x) = x^2 + x^(1/2)
Therefore, the first derivative is
f'(x) = 2x + (1/2)*(x^(-1/2))
And the second derivative is
f''(x) = 2 + (-1/4)*(x^(-3/2))
Hope this helps :3
Find f"(x) for the function.
f(x)= x^2+√x
6 answers
f(x) = x^2 + x^(1/2)
f ' (x) = 2x + (1/2)x^(-1/2)
f '' (x) = 2 - (1/4)x^(-3/2)
or 2 - 1/(4x√x)
explanation:
(x^(-3/2)
= 1/x^(3/2)
= 1/(x^(1/2)^3
= 1/(√x√x√x)
= 1/(x√x)
f ' (x) = 2x + (1/2)x^(-1/2)
f '' (x) = 2 - (1/4)x^(-3/2)
or 2 - 1/(4x√x)
explanation:
(x^(-3/2)
= 1/x^(3/2)
= 1/(x^(1/2)^3
= 1/(√x√x√x)
= 1/(x√x)
from the ans choices i have, I choose 8x^(3/2)-1/4x^(3/2)
even though both Jai and I had the same answer, which is different from the one you picked ?
Where would the 8 even come from ????
Where would the 8 even come from ????
haha is it then 2x^(3/2)-1/x^(3/2)
I was doing final review for my exam and the ans to this question is given 8x^(3/2)-1/4x^(3/2) at the end of the answer key sheet. i might have to check with my prof. But thank you Reiny & Jai.