Recall the generalized derivative rule for exponential functions:

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

In this exercise, we want to find the derivative of

\(\displaystyle{t}{\left({x}\right)}={4}^{{-{x}+{5}}}\)

To make use of the rule, denote the argument of the exponential function by u,i.e.

\(u=-x+5\)

Using the rule from Step 1, we get:

\(\displaystyle{t}'{\left({x}\right)}={4}^{{{u}}}{\ln{{4}}}{\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}}={4}^{{-{x}+{5}}}{\ln{{3}}}{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left(-{x}+{5}\right)}\rbrace=-{1}\)

\(\displaystyle=-{4}^{{-{x}+{5}}}{\ln{{4}}}\)