Question

In Problems 21–32, find the derivative of each function at the given number.

$f\left(x\right)=\sin \left(x\right)$at$0$

Short answer

The derivative of the given function at $0$is equal to$1$.

Step-by-step solution

Step 1. Given information

Given function is $f\left(x\right)=\sin \left(x\right)$

We have to find the derivative of the given function at$0$

Step 2. Definition of the derivative

Let's say a function $y=f\left(x\right)$

The derivative of the function at c is defined as

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

Step 3. Finding the derivative

By substituting the given function in the derivative definition,

we will get:

$$\begin{gathered} f\text{'}\left(0\right)=\underset{x\to 0}{\lim }\frac{f\left(x\right)-f\left(0\right)}{x-0} \\ \Rightarrow f\text{'}\left(0\right)=\underset{x\to 0}{\lim }\frac{\sin \left(x\right)-\sin \left(0\right)}{x} \\ \Rightarrow f\text{'}\left(0\right)=\underset{x\to 0}{\lim }\frac{\sin \left(x\right)}{x} \end{gathered}$$

Step 4. Simplifying the limit

The expansion of $\sin \left(x\right)$is:

$\sin \left(x\right)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+........$

Now the limit is:

localid="1647086780847">f'(0)=limx→0x-x33!+x55!-x77!+.......x⇒f'(0)=limx→01-x23!+x45!-x67!+.......⇒f'(0)=1