Question

How do you differentiate $f\left(x\right)={\cos }^{2}x$?

Answer 1

Step 1: Differentiate the function

As we know that, if $f$ and $g$ are both differentiable and $F\left(x\right)$ is the composite function defined by $F\left(x\right)=f\left(g\left(x\right)\right)$ then $F$ is differentiable and $F\text{'}$ is given by the product $F\text{'}\left(x\right)=f\text{'}\left(g\left(x\right)\right)\times g\text{'}\left(x\right)$.

Here, $f\text{'}\left(g\left(x\right)\right)$ is the differentiation of outer function and $g\text{'}\left(x\right)$ is differentiation of inner function.

Differentiate using the chain rule, which states that $\frac{d}{dx}\left[f\left(g\left(x\right)\right)\right]$ is $f\text{'}\left(g\left(x\right)\right)·g\text{'}\left(x\right)$.

Here,

$f\left(x\right)=x^2$

and

$g\left(x\right)=\cos \left(x\right)$

$$\begin{gathered} \therefore \frac{dg\left(x\right)}{dx}=\frac{d\cos \left(x\right)}{dx} \\ =-\sin \left(x\right) \end{gathered}$$

Step 2: Use Power Rule to differentiate it.

Now, we will differentiate further by using the power rule.

$\Rightarrow$ $\frac{d}{dx}\left[x^n\right]=n·x^{n-1}$

Here, $n=2$

$\Rightarrow \frac{d{x}^2}{dx}=2x$

Step 3. Find the derivative

Given $f\left(x\right)={\cos }^{2}x$

We know that if $F\left(x\right)=f\left(g\left(x\right)\right)$ then $F\text{'}\left(x\right)=f\text{'}\left(g\left(x\right)\right)\times g\text{'}\left(x\right)$ that is $f\text{'}\left(g\left(x\right)\right)·g\text{'}\left(x\right)$

$$\begin{gathered} \therefore \frac{df\left(x\right)}{dx}=\frac{d{\cos }^2\left(x\right)}{dx} \\ =2\cos \left(x\right)\frac{d\cos \left(x\right)}{dx} \\ =2\cos \left(x\right)\left(-\sin \left(x\right)\right) \\ =-2\sin \left(x\right)\cos \left(x\right) \\ =-\sin \left(2x\right)\left[\because \sin \left(2x\right)=2\sin \left(x\right)\cos \left(x\right)\right] \end{gathered}$$

Hence, the differentiate value of $f\left(x\right)={\cos }^{2}x$ is $-\sin \left(2x\right)$.