There is a rule for differentiating these functions

\(\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left[{a}^{{u}}\right]}={\left({\ln{{a}}}\right)}\cdot{\left({a}^{{u}}\right)}\cdot{\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}}\)

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

\(\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left[{10}^{{x}}\right]}={\left({\ln{{10}}}\right)}\cdot{\left({10}^{{x}}\right)}\cdot{\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}}\)

\(\displaystyle\text{if }\ {u}={x}\ \text{ then, }\ {\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}}={1}\)

because of the power rule: \(\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left[{x}^{{n}}\right]}={n}\cdot{x}^{{{n}-{1}}}\)

so, back to our problem, \(\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left[{10}^{{x}}\right]}={\left({\ln{{10}}}\right)}\cdot{\left({10}^{{x}}\right)}\cdot{\left({1}\right)}\)

which simplifies to \(\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left[{10}^{{x}}\right]}={\left({\ln{{10}}}\right)}\cdot{\left({10}^{{x}}\right)}\)

This would work the same if u was something more complicated than x. A lot of calculus deals with the ability to relate the given problem to one of the rules of differentiation. Often we have to alter the way the problem looks before we can begin, however that was not the case with this problem.

\(\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left[{a}^{{u}}\right]}={\left({\ln{{a}}}\right)}\cdot{\left({a}^{{u}}\right)}\cdot{\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}}\)

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

\(\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left[{10}^{{x}}\right]}={\left({\ln{{10}}}\right)}\cdot{\left({10}^{{x}}\right)}\cdot{\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}}\)

\(\displaystyle\text{if }\ {u}={x}\ \text{ then, }\ {\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}}={1}\)

because of the power rule: \(\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left[{x}^{{n}}\right]}={n}\cdot{x}^{{{n}-{1}}}\)

so, back to our problem, \(\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left[{10}^{{x}}\right]}={\left({\ln{{10}}}\right)}\cdot{\left({10}^{{x}}\right)}\cdot{\left({1}\right)}\)

which simplifies to \(\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left[{10}^{{x}}\right]}={\left({\ln{{10}}}\right)}\cdot{\left({10}^{{x}}\right)}\)

This would work the same if u was something more complicated than x. A lot of calculus deals with the ability to relate the given problem to one of the rules of differentiation. Often we have to alter the way the problem looks before we can begin, however that was not the case with this problem.