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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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    I'd start with writing the function in the base e, and then go on.2012-11-06

2 Answers 2

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

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

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

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

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

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

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    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?2012-11-06
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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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    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 $\frac{d}{du} = 2^x \ln 2$.2012-11-06