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.
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