The derivative using the limit definition
Definition
The derivative of the function $ \displaystyle f$ at the point a is the limit when $ \displaystyle h\to 0$ of the function, if this limit exists.
We label it f´(a) and $\displaystyle f'(a)=\underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{f(a+h)-f(a)}}{h}$
When the function$ \displaystyle f$ is derivative on the point $ \displaystyle a$ then he is called differentiable in this point.
The derivative of the function $\displaystyle y=f(x)$ on the point $ \displaystyle a$ is labelled $\displaystyle y_{x}^{\acute{\ }}(a)$ and is read the derivative of $ \displaystyle y$ in relation with $ \displaystyle x$ on the point $ \displaystyle a$
The derivative $\displaystyle y_{x}^{\acute{\ }}(a)$ presents the rate of change of the $\displaystyle y$ with the growth of x at the point a.
Steps to find the derivative of a function on a point a
1. Compute $\displaystyle \text{f}(a)$
2. Compute $\displaystyle f(a+h)$
3. Find the difference $\displaystyle f(a+h)-f(a)$
4. We form the ratio $\displaystyle \frac{{f(a+h)-f(a)}}{h}$
5. We search for the limit of this ratio when $\displaystyle h\to 0$. If this limit exists then he is labelled $\displaystyle f'(a)$
TIP! See also Worked Examples – Derivatives
Example 1: Find the derivative of the function $\displaystyle f:y=2x$ on the point $\displaystyle a=2$
From the definition we know that $\displaystyle f'(a)=\underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{f(a+h)-f(a)}}{h}$
Firstly we find $\displaystyle f(a)$
$\displaystyle f(2)=2\cdot 2=4$
Secondly we find $\displaystyle f(a+h)$
$\displaystyle f(2+h)=2(2+h)=4+2h$
Then we find the difference $\displaystyle f(a+h)-f(a)$
$\displaystyle f(2+h)-f(2)=4+2h-4=2h$
We form the ratio $\displaystyle \frac{{f(a+h)-f(a)}}{h}$
$\displaystyle \frac{{f(2+h)-f(2)}}{h}=\frac{{4+2h-4}}{h}=\frac{{2h}}{h}=2$
Lastly we find the limit of the ratio: $\displaystyle \underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{f(2+h)-f(2)}}{h}=\underset{{h\to 0}}{\mathop{{\lim }}}\,2=2$
So, the derivative of the function on the point 2 is $\displaystyle f'(2)=2$
Another way to find the derivative of a function using the definition
We have $\displaystyle f'(a)=\underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{f(a+h)-f(a)}}{h}$
If we label $\displaystyle a+h=x$ from which $\displaystyle h=x-a$
$\displaystyle h\to 0$ becomes $\displaystyle x\to a$ and
$\displaystyle \frac{{f(a+h)-f(a)}}{h}$ becomes $\displaystyle \frac{{f(x)-f(a)}}{{x-a}}$
In conclusion: The function $\displaystyle f$ is derivative on the point $\displaystyle a$ only when exists $\displaystyle \underset{{x\to a}}{\mathop{{\lim }}}\,\frac{{f(x)-f(a)}}{{x-a}}$ and it’s labelled $\displaystyle f'(a)$
Example 2: Find the derivative of the function $\displaystyle f:y={{x}^{2}}-1$on the point $\displaystyle a$ if $\displaystyle a=3$.
Solution: We have $ \displaystyle f(x)={{x}^{2}}-1$ and $\displaystyle f(a)={{(3)}^{2}}-1=9-1=8$
Then we write $\displaystyle \underset{{x\to 3}}{\mathop{{\lim }}}\,\frac{{({{x}^{2}}-1)-8}}{{x-3}}=$
$\displaystyle \underset{{x\to 3}}{\mathop{{\lim }}}\,\frac{{{{x}^{2}}-1-8}}{{x-3}}=$
$ \displaystyle \underset{{x\to 3}}{\mathop{{\lim }}}\,\frac{{{{x}^{2}}-9}}{{x-3}}=$
$\displaystyle \underset{{x\to 3}}{\mathop{{\lim }}}\,\frac{{(x-3)(x+3)}}{{x-3}}=$
$ \displaystyle \underset{{x\to 3}}{\mathop{{\lim }}}\,(x+3)=3+3=6$
So $\displaystyle f'(3)=6$
Example 3: Find the derivatives of the functions using the limit definition.
a) $\displaystyle f(x)={{x}^{2}}+2x+1$ on the point $\displaystyle a=1$
Firstly we find $\displaystyle f(1)$
$\displaystyle f(1)={{(1)}^{2}}+2\cdot 1+1=4$
Secondly we find $\displaystyle f(1+h)$
$\displaystyle f(1+h)={{(1+h)}^{2}}+2(1+h)+1$
$\displaystyle f(1+h)=4+4h+{{h}^{2}}$
Then we find the difference $\displaystyle f(1+h)-f(1)$
$ \displaystyle f(1+h)-f(1)=4+4h+{{h}^{2}}-4$
$\displaystyle f(1+h)-f(1)=4h+{{h}^{2}}$
We form the ratio $\displaystyle \frac{{f(1+h)-f(1)}}{h}$
$\displaystyle \frac{{f(1+h)-f(1)}}{h}=\frac{{4h+{{h}^{2}}}}{h}=\frac{{h(4+h)}}{h}=4+h$
Lastly we find the limit of the ratio:
$\displaystyle \underset{{h\to 1}}{\mathop{{\lim }}}\,\frac{{f(1+h)-f(1)}}{h}=\underset{{h\to 1}}{\mathop{{\lim }}}\,4+h=5$
So the derivative of the function on the point 1 is $\displaystyle f'(1)=5$
b) $\displaystyle f(x)=\frac{{x+1}}{{x-1}}$ on the point $ \displaystyle a=2$
Firstly we find $\displaystyle f(2)$
$\displaystyle f(2)=\frac{{2+1}}{{2-1}}=\frac{3}{1}=3$
Secondly we find $\displaystyle f(2+h)$
$\displaystyle f(2+h)=\frac{{2+h+1}}{{2+h-1}}=\frac{{h+3}}{{h+1}}$
Then we find the difference $\displaystyle f(2+h)-f(2)$
$\displaystyle +h)-f(2)=\frac{{h+3}}{{h+1}}-3$
$\displaystyle =\frac{{h+3-3(h+1)}}{{h+1}}=\frac{{h+3-3h-3}}{{h+1}}=\frac{{-2h}}{{h+1}}$
We form the ratio $\displaystyle \frac{{f(2+h)-f(2)}}{h}$
$\displaystyle \frac{{f(2+h)-f(2)}}{h}=\frac{{\frac{{-2h}}{{h+1}}}}{h}$
Lastly we find the limit of the ratio: $\displaystyle \underset{{h\to 2}}{\mathop{{\lim }}}\,\frac{{f(2+h)-f(2)}}{h}=$
$\displaystyle \underset{{h\to 2}}{\mathop{{\lim }}}\,\frac{{\frac{{-2h}}{{h+1}}}}{h}=$
$\displaystyle \underset{{h\to 2}}{\mathop{{\lim }}}\,-\frac{{2h}}{{h+1}}\cdot \frac{1}{h}=$
$\displaystyle \underset{{h\to 2}}{\mathop{{\lim }}}\,-\frac{2}{{h+1}}=-\frac{2}{{2+1}}=-\frac{2}{3}$
So the derivative of the function on the point 2 is $\displaystyle f'(2)=-\frac{2}{3}$
c) $\displaystyle f(x)=\sqrt{{2x+3}}$ on the point $\displaystyle a=0$
Firstly we find $\displaystyle f(0)$
$\displaystyle f(0)=\sqrt{{2\cdot 0+3}}=\sqrt{3}$
Secondly we find $\displaystyle f(0+h)$ or $\displaystyle f(h)$
$\displaystyle f(h)=\sqrt{{2h+3}}$
Then we find the difference $\displaystyle f(h)-f(0)$
$ \displaystyle f(h)-f(0)=\sqrt{{2h+3}}-\sqrt{3}$
We form the ratio $ \displaystyle \frac{{f(h)-f(0)}}{h}$
$ \displaystyle \frac{{f(h)-f(0)}}{h}=\frac{{\sqrt{{2h+3}}-\sqrt{3}}}{h}$
$\displaystyle =\frac{{(\sqrt{{2h+3}}-\sqrt{3})(\sqrt{{2h+3}}+\sqrt{3})}}{{h(\sqrt{{2h+3}}+\sqrt{3})}}$
$\displaystyle =\frac{{2h}}{{h(\sqrt{{2h+3}}+\sqrt{3})}}$
$\displaystyle =\frac{2}{{(\sqrt{{2h+3}}+\sqrt{3})}}$
Lastly we find the limit of the ratio $\displaystyle \underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{f(h)-f(0)}}{h}$
$\displaystyle \underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{2}{{(\sqrt{{2h+3}}+\sqrt{3})}}=\frac{2}{{\sqrt{3}+\sqrt{3}}}=\frac{2}{{2\sqrt{3}}}=\frac{1}{{\sqrt{3}}}$
So the derivative of the function on the point 0 is $\displaystyle f'(0)=\frac{1}{{\sqrt{3}}}=\frac{{\sqrt{3}}}{3}$
