Question

differentiate ( f(t)=8^{-9 t+2} ).

( f^{prime}(t)= )

Explanation:

Step1: Use the chain rule

Let \(u = -9t + 2\), then \(y = 8^{u}\). The chain rule states that \(\frac{dy}{dt}=\frac{dy}{du}\cdot\frac{du}{dt}\).
First, find \(\frac{dy}{du}\) for \(y = 8^{u}\). The formula for the derivative of \(a^{x}\) is \(\frac{d}{dx}(a^{x})=a^{x}\ln a\). So, \(\frac{dy}{du}=8^{u}\ln 8\).
Second, find \(\frac{du}{dt}\) for \(u=-9t + 2\). Using the power rule \(\frac{d}{dt}(at + b)=a\), we get \(\frac{du}{dt}=-9\).

Step2: Multiply the two derivatives

Substitute back \(u=-9t + 2\) into \(\frac{dy}{du}\). Then \(\frac{dy}{dt}=\frac{dy}{du}\cdot\frac{du}{dt}=8^{-9t + 2}\ln 8\cdot(-9)\).

Answer:

\(-9\ln 8\cdot8^{-9t + 2}\)