Answer :
Given: The function below
[tex]y=(e^{cos(\frac{t}{9})})^4[/tex]To Determine: The derivative of the given function
Solution
Let us apply chain rule
[tex]\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}[/tex]The equation becomes
[tex]y=e^{4cos(\frac{t}{9})}[/tex][tex]\begin{gathered} u=4cos(\frac{t}{9}) \\ x=\frac{t}{9} \\ \frac{dx}{dt}=\frac{1}{9} \\ u=4cosx \\ \frac{du}{dx}=-4sinx \\ \frac{du}{dt}=\frac{du}{dx}\times\frac{dx}{dt} \\ \frac{du}{dt}=-4sinx\times\frac{1}{9} \\ \frac{du}{dt}=-\frac{4}{9}sin(\frac{1}{9}) \end{gathered}[/tex][tex]\begin{gathered} y=e^u \\ \frac{dy}{du}=e^u \\ \frac{dy}{dt}=\frac{dy}{du}\times\frac{du}{dt} \\ \frac{dy}{dt}=e^u\times-\frac{4}{9}sin(\frac{1}{9}) \\ \frac{dy}{dt}=e^{4cos(\frac{t}{9})}\times-\frac{4}{9}sin(\frac{1}{9}) \\ \frac{dy}{dt}=-\frac{4}{9}sin(\frac{t}{9})e^{cos(\frac{t}{9})} \end{gathered}[/tex]Hence,
[tex]\frac{dy}{dt}=-\frac{4}{9}sin(\frac{t}{9})e^{cos(\frac{t}{9})}[/tex]