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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?

Jwan622
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    Yep, it is right...and the name of the huge theorem you're actually using is The Chain Rule. – DonAntonio Feb 05 '18 at 16:21
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    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 :) – John Doe Feb 05 '18 at 16:24

3 Answers3

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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.

user328442
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  • don't you rely on the product rule after taking the log to both sides? – Jwan622 Feb 05 '18 at 16:35
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    @Jwan622 sort of I suppose but the differential operator is linear and $\ln(2)$ is a constant. – user328442 Feb 05 '18 at 16:36
  • I'm sorta having trouble seeing how you took the log of both sides. The right side of t^3ln(2) is confusing to me. I'm sure it's right though... I just don't see it. – Jwan622 Feb 05 '18 at 16:38
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    @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. – user328442 Feb 05 '18 at 16:41
  • dope, that's very clear now. So you're telling me that when you take the derivative with respect to t on both sides, on the third line, you're using the chain rule and not the product rule? The product rule makes sense since ln(2) is a constant and so the derivative of that is 0, and so all that's left is the derivative of $t^3 * ln(2)$ since the other part of the product rule comes out to 0. Is that not right? – Jwan622 Feb 05 '18 at 16:46
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    @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. – user328442 Feb 05 '18 at 16:51
  • got it, that's awesomely explained and super clear. Perhaps last question... where does that constant rule come from? Is it derived from the product rule? Can you show me quickly the way it's proved using limits? – Jwan622 Feb 05 '18 at 16:52
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    @Jwan622 do you know the limit definition of the derivative? $f’(x) = \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h}$ – user328442 Feb 05 '18 at 17:00
  • yeah I'm familiar with that definition using limits. That's kind of how I first learned about derivatives. – Jwan622 Feb 05 '18 at 17:02
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    @Jwan622 excellent. Then if we apply that definition to the function $c \cdot f(x)$ where $c$ is some constant then we have $$\frac{d}{dx} (c \cdot f(x)) = \lim_{h \rightarrow 0} \frac{c \cdot f(x+h) - c \cdot f(x)}{h} = \lim_{h \rightarrow 0} c \cdot \frac{f(x+h) - f(x)}{h} = c \cdot f’(x).$$ – user328442 Feb 05 '18 at 17:08
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    @AccidentalFourierTransform my bad. I wrote the answer in a haste – user328442 Feb 05 '18 at 19:18
  • I see, so its reliant on the limit rule with constants.... Is there a proof of the limit rule? I feel pretty convinced by it though. In math, do you constantly feel the need to verify proofs to yourself over and over again? – Jwan622 Feb 05 '18 at 23:40
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    @Jwan622 the limit rule used here is referred to as the “algebraic limit theorem.” It would rely on the definition of a limit and so I suspect that verifying it is unnecessary at the level of single variable calculus. I will often verify proofs beyond an unreasonable doubt but there is no need to be terribly rigorous at your level of study. – user328442 Feb 05 '18 at 23:58
  • thanks a lot for your help @user328442 – Jwan622 Feb 06 '18 at 00:05
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Formulaic approach via the substitution $u=t^3$:

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

gen-ℤ ready to perish
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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

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

John Alexiou
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