# The derivative using the limit definition

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### 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 $\displaystyle f'(a)$

### 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 $\displaystyle f'(0)=\frac{1}{{\sqrt{3}}}=\frac{{\sqrt{3}}}{3}$