##### Derivative Rules

We know that if $\displaystyle y=f(x)$ then the derivative is defined to be $\displaystyle f'(x)=\underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{f(x+h)-f(x)}}{h}$

Some notions we use when we write the derivative are:

$\displaystyle y’=f'(x)=\frac{{df}}{{dx}}=\frac{{dy}}{{dx}}$

Determining the derivative of a function using the definition sometimes it  requires a lot of work and it’s easy to make mistakes so we are going to use some basic rules to find derivative of a function in a more simpler way.

### The Constant Rule

If $\displaystyle y=f(x)=c$ where c is a constant then the derivative is equal to zero. $\displaystyle f'(x)=f'(c)=0$

Example 1: Find the derivative of the functions.

a) $\displaystyle f(x)=12$ The function is a constant function so based on the rule the derivative would be zero. $\displaystyle f'(x)=(12)’=0$

b) $\displaystyle f(x)={{2}^{3}}$

Doing the power operation we obtain $\displaystyle f(x)={{2}^{3}}=8$ which is again a constant function  and its derivative would be zero.

$\displaystyle f'(x)=(8)’=0$

### The Power Function Rule

If $\displaystyle y=f(x)=a{{x}^{n}}$ where “a” is a constants and “n” a rational number the derivative is:

$\displaystyle f'(x)=an{{x}^{{n-1}}}$

### *It is also called the general rule for differentiation because by substituting the values of n we can get the first, second, third and nth derivative of any function.*

Example 2: Find the derivatives of the functions.

a) $\displaystyle y=x$

We see that in our function the constant a=1 and n=1 and by using the power rule we obtain:

$\displaystyle {y}’={{(x)}^{\prime }}=(1){{(x)}^{{1-1}}}$

$\displaystyle =(1){{x}^{0}}=1$

So $\displaystyle y’=1$

b) $\displaystyle y=2{{x}^{2}}$

We see that in our function a=2 and n=2 and by using the power rule we obtain:

$\displaystyle {y}’={{(2{{x}^{2}})}^{\prime }}=$
$\displaystyle 2\cdot 2{{x}^{{2-1}}}=$
$\displaystyle 4{{x}^{1}}=4x$
So $\displaystyle y’=4x$

c) $\displaystyle y=12{{x}^{{\frac{1}{4}}}}$

We see that a=12 and n=1/4 and by using the formula we obtain:

$\displaystyle {y}’=12\cdot \frac{1}{4}{{x}^{{(\frac{1}{4}-1)}}}$

$\displaystyle =3{{x}^{{-\frac{3}{4}}}}$

So $\displaystyle y’=3{{x}^{{-\frac{3}{4}}}}$

d) $\displaystyle y=\frac{2}{{{{x}^{2}}}}$

In this case we have to write our function as a power function so we can use the power rule:

$\displaystyle y=\frac{2}{{{{x}^{2}}}}=2\cdot {{x}^{{-2}}}$

Then we see that a=2 and n=-2 and by using the formula we obtain:

$\displaystyle {y}’=(2\cdot {{x}^{{-2}}}{)}’$

$\displaystyle =2\cdot (-2){{x}^{{-2+1}}}$

$\displaystyle =-4{{x}^{{-1}}}$

So $\displaystyle y’=-4{{x}^{{-1}}}$

### The Sum and Difference Rule

If $\displaystyle y$  is a function created by the sum or difference two functions $\displaystyle f(x)$ and $\displaystyle g(x)$
$\displaystyle y=f(x)\pm g(x)$

### The derivative of a sum is equal to the sum of derivatives.

$\displaystyle y’=(f(x)+g(x))’=f'(x)+g'(x)$

The derivative of a difference is equal to the difference of derivatives.
$\displaystyle y’=(f(x)-g(x))’=f'(x)-g'(x)$

Example 3: Find the derivatives of the functions.

a) $\displaystyle y=3{{x}^{3}}+2x$

We know that the derivative of a sum is equal to the sum of derivatives.

$\displaystyle {y}’={{(3{{x}^{3}}+2x)}^{\prime }}$

$\displaystyle ={{(3{{x}^{3}})}^{\prime }}+{{(2x)}^{\prime }}$

Then we apply the power rule at each of the terms.

$\displaystyle {y}’={{(3{{x}^{3}})}^{\prime }}+{{(2x)}^{\prime }}$

$\displaystyle =3\cdot 3{{x}^{{3-1}}}+2\cdot 1{{x}^{{1-1}}}$

$\displaystyle =9{{x}^{2}}+2{{x}^{0}}=9{{x}^{2}}+2$

So $\displaystyle y’=(3{{x}^{3}})’+(2x)’=9{{x}^{2}}+2$

b) $\displaystyle y={{x}^{2}}-2\sqrt{x}-1$

The derivative of a difference is equal to the difference of derivatives.

$\displaystyle {y}’={{({{x}^{2}}-2\sqrt{x}-1)}^{\prime }}$

$\displaystyle ={{({{x}^{2}})}^{\prime }}-{{(2\sqrt{x})}^{\prime }}-{{(1)}^{\prime }}$

Then we apply the power rule and constant rule at each of the terms and change the term with a root into a power term.

$\displaystyle y’=({{x}^{2}})’-(2{{x}^{{\frac{1}{2}}}})’-(1)’$

$\displaystyle {y}’=2{{x}^{{2-1}}}-2\cdot \frac{1}{2}{{x}^{{\frac{1}{2}-1}}}-0$

$\displaystyle =2x-{{x}^{{-\frac{1}{2}}}}$

So $\displaystyle y’=2x-{{x}^{{-\frac{1}{2}}}}$

c) $\displaystyle y=3{{x}^{{\frac{4}{3}}}}+x$

We know that the derivative of a sum is equal to the sum of derivatives.

$\displaystyle y’=(3{{x}^{{\frac{4}{3}}}}+x)’=(3{{x}^{{\frac{4}{3}}}})’+(x)’$

Then we apply the power rule to each term.

$\displaystyle y’=(3{{x}^{{\frac{4}{3}}}})’+(x)’=3\cdot \frac{4}{3}{{x}^{{\frac{4}{3}-1}}}+{{x}^{{1-1}}}=4{{x}^{{\frac{1}{3}}}}+1$

So $\displaystyle y’=4{{x}^{{\frac{1}{3}}}}+1$

### Product Rule

If $\displaystyle y$ is a function created by multiplying two functions f(x) and g(x) together.

$\displaystyle y=f(x)g(x)$ then:

The derivative of the product of two functions is equal with the derivative of the first function multiplied by the second plus the derivative of the second function multiplied by the first.
$\displaystyle y’=(f(x)g(x))’=f'(x)g(x)+g'(x)f(x)$

### *You can find it written in a simple way: $\displaystyle y’=f’g+g’f$ *

Example 4: Find the derivatives of the functions.

a) $\displaystyle y=(2x+1)(4x+3)$

Firstly we divide the function into two parts and find the derivative of each of them.

$\displaystyle f(x)=2x+1$ and its derivative is

$\displaystyle {f}'(x)={{(2x+1)}^{\prime }}$

$\displaystyle ={{(2x)}^{\prime }}+{{(1)}^{\prime }}=2$

$\displaystyle g(x)=4x+3$ and its derivative is
$\displaystyle {g}'(x)={{(4x+3)}^{\prime }}$

$\displaystyle ={{(4x)}^{\prime }}+{{(3)}^{\prime }}=4$.

Then we apply the Product Rule $\displaystyle y=f(x)g(x))$

$\displaystyle ={f}'(x)g(x)+{g}'(x)f(x)$

$\displaystyle y’=2\left( {4x+3} \right)+4\left( {2x+1} \right)$
$\displaystyle =8x+6+8x+4$
$\displaystyle =16x+10$
So $\displaystyle y’=16x+10$

b) $\displaystyle y=(3{{x}^{2}}+2)(3x+1)$

Firstly we divide the function into two parts and find the derivate of each of them by using the rules that we know.

$\displaystyle f(x)=3{{x}^{2}}+2$ and its derivate is $\displaystyle f'(x)=(3{{x}^{2}}+2)’=6x$

$\displaystyle g(x)=3x+1$ and its derivate is $\displaystyle g'(x)=(3x+1)’=3$

Then we apply the Product Rule $\displaystyle y’=(f(x)g(x))’=f'(x)g(x)+g'(x)f(x)$

$\displaystyle {y}’=6x(3x+1)+3(3{{x}^{2}}+2)$

$\displaystyle =18{{x}^{2}}+6x+9{{x}^{2}}+6$

$\displaystyle =27{{x}^{2}}+6x+6$

So $\displaystyle y’=27{{x}^{2}}+6x+6$

The Quotient Rule

If $\displaystyle y$ is a function created by dividing two functions $\displaystyle f\left( x \right)$ and $\displaystyle g\left( x \right)$ together $\displaystyle y=\frac{{f(x)}}{{g(x)}}$ then:
The derivative of the quotient of two functions is equal with the derivative of the first function multiplied by the second minus the derivative of the second function multiplied by the first and divided all by the square of the second function.

$\displaystyle {y}’=\left( {\frac{{f(x)}}{{g(x)}}} \right)=\frac{{{f}'(x)g(x)-{g}'(x)f(x)}}{{{{{\left[ {g(x)} \right]}}^{2}}}}$

*You can find it written in a simple way: $\displaystyle y’=\frac{{f’g-g’f}}{{{{g}^{2}}}}$

Example 5: Find the derivatives of the function.

$\displaystyle y=\frac{{2{{x}^{2}}+1}}{{3x+2}}$

Firstly we divide the function into two parts and find the derivative of each of them by using the rules that we know.

$\displaystyle f(x)=2{{x}^{2}}+1$ and its derivative is $\displaystyle f'(x)=(2{{x}^{2}}+1)’=4x$

$\displaystyle g(x)=3x+2$ and its derivative is $\displaystyle g'(x)=(3x+2)’=3$

Then we apply the quotient rule $\displaystyle {y}’=\left( {\frac{{f(x)}}{{g(x)}}} \right)=\frac{{{f}'(x)g(x)-{g}'(x)f(x)}}{{{{{\left[ {g(x)} \right]}}^{2}}}}$

$\displaystyle y’=\frac{{4x(3x+2)-3(2{{x}^{2}}+1)}}{{{{{(3x+2)}}^{2}}}}$

$\displaystyle =\frac{{12{{x}^{2}}+8x-6{{x}^{2}}-3}}{{{{{(3x+2)}}^{2}}}}$

$\displaystyle =\frac{{6{{x}^{2}}+8x-3}}{{9{{x}^{2}}+12x+4}}$

So $\displaystyle y’=\frac{{6{{x}^{2}}+8x-3}}{{9{{x}^{2}}+12x+4}}$

The Chain Rule
The chain rule is a formula to compute the derivative of a composite function.
If we have two functions $\displaystyle f(x)$ and $\displaystyle g(x)$ and $\displaystyle y=f(g(x))$ is a function of a function, then:

The derivative to a function of a function or a composite function is equal with the derivative of all the composite function multiplying the derivative of the second function.
$\displaystyle y’={{\left[ {f(g(x))} \right]}^{\prime }}={f}'(g(x)){g}'(x)$

Example 6: Find the derivatives of the functions.

a) $\displaystyle y=\sqrt{{2x+3}}$

Firstly we determine the two functions that make our composite function.

$\displaystyle f(x)=\sqrt{x}$  and $\displaystyle g(x)=2x+3$

Then we find the derivative by using the formula $\displaystyle y’={{\left[ {f(g(x))} \right]}^{\prime }}={f}'(g(x)){g}'(x)$

$\displaystyle y’=\left[ {\sqrt{{2x+3}}} \right]=$

$\displaystyle {{\left[ {\sqrt{{2x+3}}} \right]}^{‘}}(2x+3)’=$

$\displaystyle \frac{1}{{2\sqrt{{2x+3}}}}\cdot 2=$

$\displaystyle \frac{1}{{\sqrt{{2x+3}}}}$

So $\displaystyle y’=\frac{1}{{\sqrt{{2x+3}}}}$

b) $\displaystyle y={{(3{{x}^{2}}+1)}^{3}}$

Firstly we determine the two functions that make our composite function.

$\displaystyle f(x)={{x}^{3}}$ and  $\displaystyle g(x)=3{{x}^{2}}+1$

Then we find the derivative by using the formula $\displaystyle y’={{\left[ {f(g(x))} \right]}^{\prime }}={f}'(g(x)){g}'(x)$

$\displaystyle y’=\left[ {{{{(3{{x}^{2}}+1)}}^{3}}} \right]\left( {3{{x}^{2}}+1} \right)=$

$\displaystyle 3{{(3{{x}^{2}}+1)}^{{3-1}}}\cdot 6x=$

$\displaystyle 3{{(3{{x}^{2}}+1)}^{2}}\cdot 6x=$

$\displaystyle18x{{(3{{x}^{2}}+1)}^{2}}=$

So $\displaystyle y’=18x{{(3{{x}^{2}}+1)}^{2}}$