Question

Find the derivative of ${\sin }^{2}x$ using first principles.

Answer 1

Compute the derivative:

The first principle states that:

$f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{f(x+h)-f\left(x\right)}{h}$

That is, we can find the derivative as

$f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{{\sin }^2(x+h)-{\sin }^{2}x}{h}$

$f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{\sin (x+h+x)-\sin (x+h-x)}{h}$ $\left[\therefore {\sin }^{2}A-{\sin }^{2}B=\sin (A+B)·\sin \left(A-B\right)\right]$

$$\begin{gathered} f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{\sin (2x+h)·\sin \left(h\right)}{h} \\ f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\sin (2x+h)·\underset{h\to 0}{\lim }\frac{\sin \left(h\right)}{h} \end{gathered}$$

$f\text{'}\left(x\right)=\sin (2x+0)\left[\therefore \underset{x\to 0}{\lim }\frac{\sin x}{x}=1\right]$

$$\begin{gathered} f\text{'}\left(x\right)=\sin 2x \\ f\text{'}\left(x\right)=2\sin \left(x\right)\cos \left(x\right) \end{gathered}$$

Hence, the required first derivative is $2\sin \left(x\right)\cos \left(x\right)$.