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$$e^{x^{3}}+7x$$ Here is what I have tried so far, $$y'=(e^{x^{3}})'+(7x)'$$ $$y'=3e^{x^{2}}e^x+7$$ $$y'=3e^{x^{3}+x}+7$$

I am supposed to find the second derivative but I believe I have already made a mistake. I am not sure how I would apply the chain rule to the $e^{x^{3}}$. I checked Wolframalpha and the derivative of $e^{x^{3}}$ was $$3e^{x^{3}}x^2$$ Did I leave out a step when I applied chain rule? I went over the problem a couple of times and couldn't catch my mistake.

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The derivative of $e^{x^3}$ is $e^{x^3}\cdot 3x^2$, by the chain rule. Indeed, $f(x)=e^x$ has $f'(x)=e^x$ and $g(x)=x^3$ yields $g'(x)=3x^2$. So $(f\circ g)'(x)=f'(g(x))g'(x)$. –  1015 Mar 14 '13 at 2:37
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5 Answers

up vote 3 down vote accepted

$$\frac d{dx}\left(e^{x^3}\right) = e^{x^3}\cdot \bigl(x^3\bigr)' = e^{x^3}\bigl(3x^2\bigr) = 3e^{x^3}x^2$$

This is by the chain rule, and for the exponential function, this gives us $$\frac d{dx}\left(e^{f(x)}\right) = e^{f(x)}f'x$$

Then for the second derivative, we can use the product rule (and the chain rule again):

So $$\frac {d^2}{dx}\left(e^{x^3}\right) = \frac{d}{dx}\Bigl(3e^{x^3}x^2\Bigr) = \Bigl(3e^{x^3}\Bigr)\Bigl(x^2\Bigr)' + x^2\Bigl(3e^{x^3}\Bigr)' = 6xe^{x^3} + 9x^4e^{x^3}$$

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Is $\frac d{dx}\left(e^{f(x)}\right) = e^{f(x)}f'x$ something I should memorize or is it derived from another rule? –  Kot Mar 14 '13 at 2:50
    
It is derived from the chain rule $g(x) = e^x, \; f(x) = x^3, \;g(f(x)) = e^{f(x)} = e^{x^3}$. But it handy to make the connection with how this works for $e^{f(x)}$, as noted, since the derivative of the exponential function $e^x = e^x$. –  amWhy Mar 14 '13 at 2:54
    
Ahh, I understand now. The $e$ was throwing me off. Thank you for the explanation! –  Kot Mar 14 '13 at 3:00
    
You're welcome...I found it really handy to "memorize" (but first understand): $\frac d{dx}\left(e^{f(x)}\right) = e^{f(x)}f'x$, just for these sorts of problems, because the derivative of the exponential function is unlike that of any other function: generally a no-brainer for $e^x$, and fairly straightforward for deriving $e^{f(x)}$ –  amWhy Mar 14 '13 at 3:04
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let T be the second derivative of the function ,

then ,

T = $ 3 e^{x^3} [ 3x^4 + 2x ] $

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$+^{+^{+^{+}}}$ –  B. S. Jun 28 '13 at 5:42
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Hint:

$$\left(e^{x^3}\right)'=3x^2e^{x^3}\;,\;\;\left(3x^2e^{x^3}\right)'=6xe^{x^3}+9x^4e^{x^3}=3xe^{x^3}\left(2+3x^3\right)$$

I think you can now handle the summand $\,7x\,$

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Let

$ f(x) = e^x $ and $ g(x) = x^3 $, so that your first term could be written $ f(g(x)) $. Then the derivative of the first term is

$ f'(g(x)) g'(x) = e^{g(x)} (3x^2) = e^{x^3} 3x^2 $

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Did you push the "Step-by-step solution" button in WolframAlpha? If you're accessing WolframAlpha from within Mathematica, you get as many pushes as you like - not just the three free ones. Here's the result inside Mathematica:

enter image description here

I think this makes it reasonably clear where the $3x^2$ came from.

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