Question

find the derivative of ( y = 8^{sqrt{t}} ) with respect to ( t ).
( \frac{dy}{dt} = )

Explanation:

Step1: Use the chain rule

The chain rule states that if \(y = a^{u(t)}\), then \(\frac{dy}{dt}=a^{u(t)}\ln a\cdot\frac{du}{dt}\). Here \(a = 8\) and \(u(t)=\sqrt{t}=t^{\frac{1}{2}}\).
So \(y = 8^{\sqrt{t}}\), then \(\frac{dy}{dt}=8^{\sqrt{t}}\ln 8\cdot\frac{d}{dt}(\sqrt{t})\).

Step2: Differentiate \(\sqrt{t}\)

We know that if \(u(t)=t^{n}\), then \(\frac{du}{dt}=nt^{n - 1}\). For \(n=\frac{1}{2}\), \(\frac{d}{dt}(t^{\frac{1}{2}})=\frac{1}{2}t^{\frac{1}{2}- 1}=\frac{1}{2}t^{-\frac{1}{2}}=\frac{1}{2\sqrt{t}}\).

Step3: Combine the results

Substitute \(\frac{d}{dt}(\sqrt{t})=\frac{1}{2\sqrt{t}}\) into \(\frac{dy}{dt}=8^{\sqrt{t}}\ln 8\cdot\frac{d}{dt}(\sqrt{t})\).
We get \(\frac{dy}{dt}=8^{\sqrt{t}}\ln 8\cdot\frac{1}{2\sqrt{t}}=\frac{8^{\sqrt{t}}\ln 8}{2\sqrt{t}}\).

Answer:

\(\frac{8^{\sqrt{t}}\ln 8}{2\sqrt{t}}\)