Answer
$t^{\prime}(x)=-4^{-x+5} \ln (4)$
Work Step by Step
We have: $t(x)=4^{-x+5}$,
We differentiate both sides with respect to $x$.
$t^{\prime}(x)=\dfrac{d}{dx}[4^{-x+5}]$
Use rule: $\displaystyle \frac{d}{dx}[a^{u}]=a^u \ln (a) \frac{du}{dx}$
Now, $t^{\prime}(x)=4^{-x+5} \ln (4) \dfrac{d}{dx}[-x+5]$
So,
$t^{\prime}(x)=-4^{-x+5} \ln (4)$
