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What is the derivative of $$y = 2^{3^{x^{2}}}$$ using the chain rule?

Please show step by step solution.

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what steps did you attempt on your own? –  Sasha Nov 6 '12 at 18:16
I'd start with writing the function in the base e, and then go on. –  TheClock Nov 6 '12 at 18:19
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2 Answers

up vote 1 down vote accepted

$$y = 2^{3^{x^{2}}}$$

$$y' = 2^{3^{x^{2}}}\ln2 (3^{x^{2}})'$$

$$y' = 2^{3^{x^{2}}}\ln2 (3^{x^{2}}\ln3)(x^2)'$$

$$y' = 2^{3^{x^{2}}}\ln2 (3^{x^{2}}\ln3)2x$$

$$y' = 2x\ln2 \ln3 2^{3^{x^{2}}}(3^{x^{2}})$$

$$y' = 2^{3^{x^{2}}}3^{x^{2}}2x\ln3\ln2 $$

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Why didn't you leave in the (x^2) between step 3 and 4 as you did with the (3^(x^2)) between steps 2 and 3? –  user44816 Nov 6 '12 at 18:34
@user44816 He used the chain rule again, so he differentiated $(3^{x^2})$ into $((3^{x^2}\ln 3) (x^2)')$, and then at the next step he differentiated the last $x^2$. –  Arthur Nov 6 '12 at 18:46
If you are leaving in the (3^(x^2)) when differentiating (derivative is (ln3)(x^2), why wouldn't you leave in the (x^2) when differentiating at the next step (becomes 2x)? What is the difference? –  user44816 Nov 6 '12 at 18:52
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Hint: for a function $a^u$, $$\frac{d}{du} = a^u \ln a \cdot u'$$

In your case, $a=2, u = 3^{x^2}$.

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Is $u=3^{x^2}$ not $u=3x^2$ –  Adi Dani Nov 6 '12 at 18:37
Yup, typo. Thanks. –  Joe Nov 6 '12 at 18:40
Shouldn't it be $\frac{d}{dx}$? If the derivative were with respect to $u$ then $u'=1$ and there's no point writing it. –  Daniel Littlewood Nov 6 '12 at 18:45
I replaced his function $3^{x^2}$ as $u$. If I were to write $\frac{d}{dx} a^{u}$, it would not make sense since you are not differentiating with respect to $u$. Also, consider the example with $u = x.$ Then $\frac{d}{du} = 2^x \ln 2$. –  Joe Nov 6 '12 at 18:47
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