Question

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

Answer 1

f''(x) for \[f(x)=e ^{-x ^{2}}\]

Answer 2

Chain Rule?

Answer 3

\[f'(x)=-2xe ^{-x^2}\]
\[f''(x)=-2x(-2xe ^{-x^2})+e ^{-x^2}(-2)\]
just simplify!

Answer 4

where did you get the -2?

Answer 5

by product rule?

Answer 6

could you explain i understand how to find derivative just not really with e

Answer 7

you know that the derivative of e^x is e^x right?

Answer 8

i forgot that

Answer 9

the derivative of e^u ,where u = f(x), is e^u(f'(x))

Answer 10

so like for y=(x^2+9)^4
it would be 4(x^2+9)^3

Answer 11

derivative of exponential function is the same function multiplied by the derivative if the power.. as dichalao said:
\[e ^{f(x)}=f'(x)e ^{f(x)}\]
then I used the product rule which states that:
\[(f(x)g(x))'=f(x)g'(x)+f'(x)g(x)\]

Answer 12

lim of
(x^3-13x+12)/(x^3-14x+15) as x approaches 3

without using L hopital's rule. :( hope you can help me guys.

Answer 13

well, this is an easy one.. since, as you see, x=3 is a zero of both The numerator and the denominator, (x-3) is a factor of both of them .. so just divide both of them by (x-3), you will get:
\[\lim_{x \rightarrow 3} {(x-3)(x^2+3x-4) \over (x-3)(x^2+3x-5)}\]
(x-3)/(x-3) is 1, you just directly substitute in the new expression to get:
\[\lim_{x \rightarrow 3}{x^2+3x-4 \over x^2+3x-5}={9+9-4 \over 9+9-5}=14/13\]