Phoebe Ware

2023-02-22

What is the derivative of ${10}^{x}$?

Brady Mays

Beginner2023-02-23Added 7 answers

There is a rule for differentiating these functions

$\frac{d}{dx}\left[{a}^{u}\right]=\left(\mathrm{ln}a\right)\cdot \left({a}^{u}\right)\cdot \frac{du}{dx}$

Notice that for our problem a=10 and u=x so let's plug in what we know.

$\frac{d}{dx}\left[{10}^{x}\right]=\left(\mathrm{ln}10\right)\cdot \left({10}^{x}\right)\cdot \frac{du}{dx}$

if $u=x$ thus, $\frac{du}{dx}=1$

because of the power rule: $\frac{d}{dx}\left[{x}^{n}\right]=n\cdot {x}^{n-1}$

Hence, back to our problem, $\frac{d}{dx}\left[{10}^{x}\right]=\left(\mathrm{ln}10\right)\cdot \left({10}^{x}\right)\cdot \left(1\right)$ which simplifies to $\frac{d}{dx}\left[{10}^{x}\right]=\left(\mathrm{ln}10\right)\cdot \left({10}^{x}\right)$

This would work the same if u was something more complicated than x.

A lot of calculus is concerned with the ability to relate a given problem to one of the differentiation rules. Often, we have to change the way the problem appears before we can begin, but this was not the case with this problem.

$\frac{d}{dx}\left[{a}^{u}\right]=\left(\mathrm{ln}a\right)\cdot \left({a}^{u}\right)\cdot \frac{du}{dx}$

Notice that for our problem a=10 and u=x so let's plug in what we know.

$\frac{d}{dx}\left[{10}^{x}\right]=\left(\mathrm{ln}10\right)\cdot \left({10}^{x}\right)\cdot \frac{du}{dx}$

if $u=x$ thus, $\frac{du}{dx}=1$

because of the power rule: $\frac{d}{dx}\left[{x}^{n}\right]=n\cdot {x}^{n-1}$

Hence, back to our problem, $\frac{d}{dx}\left[{10}^{x}\right]=\left(\mathrm{ln}10\right)\cdot \left({10}^{x}\right)\cdot \left(1\right)$ which simplifies to $\frac{d}{dx}\left[{10}^{x}\right]=\left(\mathrm{ln}10\right)\cdot \left({10}^{x}\right)$

This would work the same if u was something more complicated than x.

A lot of calculus is concerned with the ability to relate a given problem to one of the differentiation rules. Often, we have to change the way the problem appears before we can begin, but this was not the case with this problem.