How to take the derivative of a power.

Question

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.

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?

Answer 1

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.

Answer 2

Formulaic approach via the substitution $u=t^3$:

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

Answer 3

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

Then from the chain rule you have

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