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So I'm trying to solve this problem: Take the derivative of $2^{t^{3}}$

This is the relevant text from my textbook which makes sense to me.

enter image description here

The trick seems to convert anything in the form of $b^x$ to $e^{x\cdot lnb }$ because $b = e^{lnb}.$

So, then I think the derivative is (via chain rule and this above rule):

$$2^{t^{3}} \cdot \ln{2} \cdot \frac{d}{dt} (t^3)$$ $$=2^{t^{3}} \cdot \ln{2} \cdot 3t^2.$$

Is that right?

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    $\begingroup$ Yep, it is right...and the name of the huge theorem you're actually using is The Chain Rule. $\endgroup$ – DonAntonio Feb 5 '18 at 16:21
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    $\begingroup$ Small correction - did you mean to type $$\frac d{dt} t^3$$ in the first line? There is nothing to do with an $x$ in the question. The final result is correct though :) $\endgroup$ – John Doe Feb 5 '18 at 16:24
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Perhaps this is a way to see this:

$$y = 2^{t^3}$$ $$\ln(y) = t^3 \ln(2)$$

Taking the derivative with respect to $t$ on both sides leads to $$\frac{y’}{y} = 3t^2 \ln(2)$$ and so

$$y’ = 3yt^2 \ln(2) = 3 \ln(2) t^2 \cdot 2^{t^3}$$ and so your answer is correct.

To find this derivative, I relied on the chain rule after taking the logarithm to both sides and so this is an application of the chain rule. I have seen this referred to as the “logarithm rule” in some places.

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    $\begingroup$ @Jwan622 sort of I suppose but the differential operator is linear and $\ln(2)$ is a constant. $\endgroup$ – user328442 Feb 5 '18 at 16:36
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    $\begingroup$ @Jwan622 ah, my apologies. I’m using a logarithm property: $\ln(x^y) = y \ln(x)$ where $x = 2$ and $y = t^3$ in this case. $\endgroup$ – user328442 Feb 5 '18 at 16:41
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    $\begingroup$ @Jwan622 that is correct but stating that this is the product rule is a bit overkill. I’m simply using the fact that given a constant $c$ (which $\ln(2)$ in this case), we have $d/dx(c \cdot f(x)) = c \cdot f’(x).$ It is true that this follows from the product rule but this special case is easier to see from the definition of the derivative where one simply pulls the constant term out of the limit. So, my point is that there are two ways to see this but one does not need a machine gun (the product rule) to kill a spider. $\endgroup$ – user328442 Feb 5 '18 at 16:51
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    $\begingroup$ @Jwan622 do you know the limit definition of the derivative? $f’(x) = \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h}$ $\endgroup$ – user328442 Feb 5 '18 at 17:00
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    $\begingroup$ @Jwan622 excellent. Then if we apply that definition to the function $c \cdot f(x)$ where $c$ is some constant then we have $$\frac{d}{dx} (c \cdot f(x)) = \lim_{h \rightarrow 0} \frac{c \cdot f(x+h) - c \cdot f(x)}{h} = \lim_{h \rightarrow 0} c \cdot \frac{f(x+h) - f(x)}{h} = c \cdot f’(x).$$ $\endgroup$ – user328442 Feb 5 '18 at 17:08
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Formulaic approach via the substitution $u=t^3$:

$$\begin{align} {d \over dt}\left(2^{t^3}\right) &= {d \over dt}\left( 2^u \right) \\ &= 2^u\ln(2){du \over dt} \\ &= \ln(2)2^{\left(t^3\right)}\left(3t^2\right) \end{align}$$

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Consider the general case of $$ x = 2^{t^3} = {\rm e}^{t^3 \ln(2)} $$

Then from the chain rule you have

$$ \frac{{\rm d}x}{{\rm d}t} = \frac{{\rm d}}{{\rm d}t} \exp( t^3 \ln(2) ) = \exp( t^3 \ln(2) ) \frac{{\rm d}}{{\rm d}t} (t^3 \ln(2)) = \exp( t^3 \ln(2) ) (3 t^2 \ln(2)) = a^{t^3} 3 t^2 \ln(2) $$

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