Solution:
Given a trigonometric function that is greater than power of 1 as shown below:
[tex]y=sin^2x\text{ ---- equation 1}[/tex]
To differentiate the function, we use the chain rule.
According to the chain rule,
[tex]\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}[/tex]
From equation 1, let
[tex]u=sin\text{ x --- equation 2}[/tex]
This implies that
[tex]\begin{gathered} y=u^2 \\ \Rightarrow\frac{dy}{du}=2u \end{gathered}[/tex]
From equation 2,
[tex]\begin{gathered} \begin{equation*} u=sin\text{ x} \end{equation*} \\ \Rightarrow\frac{du}{dx}=cos\text{ x} \end{gathered}[/tex]
[tex]\begin{gathered} Recall\text{ from the chain rule:} \\ \frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx} \\ \Rightarrow2u\times\text{cos x} \\ \frac{dy}{dx}=2ucos\text{ x} \\ but\text{ } \\ u=sin\text{ x} \\ \therefore\frac{dy}{dx}=2(sin\text{ }x)(cos\text{ }x) \end{gathered}[/tex]
