Question

Find the derivative of $x^2-2$ at $x=10$ from the first principle.

Answer 1

Compute the derivative:

Given that, $f\left(x\right)=x^2-2$

As we know, the formula of the first principle,

$f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\left[\frac{f(x+h)-f\left(x\right)}{h}\right]$. That is,

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

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

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

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

$f\text{'}\left(x\right)=2x+0$

$f\text{'}\left(x\right)=2x$

So at $x=10$ the value is

$$\begin{gathered} f\text{'}\left(x\right)=2\times 10 \\ f\text{'}\left(x\right)=20 \end{gathered}$$

Hence, the required value is $f\text{'}\left(x\right)=20$.