DIFFERENTIATING USING THE DEFINITION OF THE DERIVATIVE OF A FUNCTION

According to the definition of a derivative if $f(x)=x$, then the derivative, which is denoted as $f'(x)$, will be:

$f'(x)=\lim_{ h \to 0}\frac{f(x+h)-f(x)}{h}$

Example 1

If $f(x)=x+7$, find $f'(x)$

Solution

$$f'(x)=\lim_{ h \to 0}\frac{f(x+h)-f(x)}{h}$$

First, determine $f(x+h)$

$f(x+h)=(x+h)+7$

Let now insert it into the formula

$$f'(x)=\lim_{h \to 0}\frac{((x+h)+7)-(x+7)}{h}$$

Now, let distribute the negative sign

$$f'(x)=\lim_{h \to 0}\frac{x+h+7-x-7)}{h}$$

$$f'(x)=\lim_{h \to 0}\frac{h}{h}$$

$$f'(x)=\lim_{h \to 0}1$$

Since there is no h in the expression, there is no need to apply the limit expression, therefore

$$f'(x)=1$$

Example 2

If $f(x)=x^2+9$, find the derivative of the function

Solution:

$f(x+h)=(x+h)^2+9$

$$f'(x)=\lim_{h \to 0}\frac{((x+h)^2+9)-(x^2+9)}{h}$$

Now, let distribute the negative sign

$$f'(x)=\lim_{h \to 0}\frac{(x+h)^2+9-x^2-9}{h}$$

Expanding the bracket

$$f'(x)=\lim_{h \to 0}\frac{x^2+2xh +h^2+9-x^2-9}{h}$$

$$f'(x)=\lim_{h \to 0}\frac{2xh +h^2}{h}$$

$$f'(x)=\lim_{h \to 0}\frac{h(2x+h)}{h}$$

$$f'(x)=\lim_{h \to 0}2x+h$$

Now, let's apply the limit expression

$$f'(x)=2x+0$$

$$f'(x)=2x$$

Example 3

If $f(x)=x^2+x+1$, find $f'(x)$

Solution:

$$f(x+h)=(x+h)^2+(x+h)+1$$

$f'(x)=\lim_{ h \to 0}\frac{(x+h)^2+(x+h)+1-(x^2+x+1)}{h}$

$f'(x)=\lim_{ h \to 0}\frac{x^2+2xh+h^2+x+h+1-(x^2+x+1)}{h}$

Now, let distribute the negative sign

$f'(x)=\lim_{ h \to 0}\frac{x^2+2xh+h^2+x+h+1-x^2-x-1}{h}$

$f'(x)=\lim_{ h \to 0}\frac{2xh+h^2+h}{h}$

$f'(x)=\lim_{ h \to 0}\frac{h(2x+h+1)}{h}$

$f'(x)=\lim_{ h \to 0}2x+h+1$

Now, let's apply the limit expression

$f'(x)=2x+0+1$

$f'(x)=2x+1$

There you have it, we will look at the product rule of differentiating next.

Meanwhile, You can join the discussion on telegram here.

Tags