anonymous
  • anonymous
f''(x) for f(x) =e ^-x^2
Mathematics
jamiebookeater
  • jamiebookeater
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
f''(x) for \[f(x)=e ^{-x ^{2}}\]
anonymous
  • anonymous
Chain Rule?
anonymous
  • anonymous
\[f'(x)=-2xe ^{-x^2}\] \[f''(x)=-2x(-2xe ^{-x^2})+e ^{-x^2}(-2)\] just simplify!

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
where did you get the -2?
anonymous
  • anonymous
by product rule?
anonymous
  • anonymous
could you explain i understand how to find derivative just not really with e
anonymous
  • anonymous
you know that the derivative of e^x is e^x right?
anonymous
  • anonymous
i forgot that
anonymous
  • anonymous
the derivative of e^u ,where u = f(x), is e^u(f'(x))
anonymous
  • anonymous
so like for y=(x^2+9)^4 it would be 4(x^2+9)^3
anonymous
  • 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: \[(f(x)g(x))'=f(x)g'(x)+f'(x)g(x)\]
anonymous
  • 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.
anonymous
  • 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: \[\lim_{x \rightarrow 3}{x^2+3x-4 \over x^2+3x-5}={9+9-4 \over 9+9-5}=14/13\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.