anonymous
  • anonymous
f''(x) for f(x) =e ^-x^2
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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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!

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

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