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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0f''(x) for \[f(x)=e ^{x ^{2}}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[f'(x)=2xe ^{x^2}\] \[f''(x)=2x(2xe ^{x^2})+e ^{x^2}(2)\] just simplify!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0where did you get the 2?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0could you explain i understand how to find derivative just not really with e

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you know that the derivative of e^x is e^x right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the derivative of e^u ,where u = f(x), is e^u(f'(x))

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so like for y=(x^2+9)^4 it would be 4(x^2+9)^3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0derivative 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
 5 years ago
Best ResponseYou've already chosen the best response.0lim of (x^313x+12)/(x^314x+15) as x approaches 3 without using L hopital's rule. :( hope you can help me guys.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well, this is an easy one.. since, as you see, x=3 is a zero of both The numerator and the denominator, (x3) is a factor of both of them .. so just divide both of them by (x3), you will get: \[\lim_{x \rightarrow 3} {(x3)(x^2+3x4) \over (x3)(x^2+3x5)}\] (x3)/(x3) is 1, you just directly substitute in the new expression to get: \[\lim_{x \rightarrow 3}{x^2+3x4 \over x^2+3x5}={9+94 \over 9+95}=14/13\]
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