## anonymous 5 years ago f''(x) for f(x) =e ^-x^2

1. anonymous

f''(x) for $f(x)=e ^{-x ^{2}}$

2. anonymous

Chain Rule?

3. anonymous

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

4. anonymous

where did you get the -2?

5. anonymous

by product rule?

6. anonymous

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

7. anonymous

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

8. anonymous

i forgot that

9. anonymous

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

10. anonymous

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

11. anonymous

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: $(f(x)g(x))'=f(x)g'(x)+f'(x)g(x)$

12. anonymous

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.

13. anonymous

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: $\lim_{x \rightarrow 3}{x^2+3x-4 \over x^2+3x-5}={9+9-4 \over 9+9-5}=14/13$