# Differentiation Methods

Remember that the gradient of a straight line $y=mx+b$ is m. The tangent to the line is the line itself, so the gradient of tangent is m everywhere along the line.

So if $y=mx, \cfrac{dy}{dx}=m$.

$\cfrac { d }{ dx } (kx)=k$

For a horizontal line in the form y=k, the gradient is zero.

So if $y=k, \cfrac{dy}{dx}=0$.

$\cfrac { d }{ dx } (k)=0$

In short, we could say that:

$\cfrac { d }{ dx } ({ x }^{ n })=n{ x }^{ n-1 }$

A more general way of writing the rule is:

$\cfrac { d }{ dx } ({ kx }^{ n })=kn{ x }^{ n-1 }$

OR

$\cfrac { d }{ dx } (kf\left( x \right) )=kf^{ ' }\left( x \right)$

Also, if there are several terms in an expression, we differentiate each one separately. We can write this as a rule:

$\cfrac { d }{ dx } (f\left( x \right) +g\left( x \right) )=f^{ ' }\left( x \right) +g^{ ' }\left( x \right)$

### Examples:

Differentiate each function:

1. $7x$

$\cfrac { d }{ dx } (7x)=7$

2. $y=4{x}^{7}$

$\cfrac { dy }{ dx } =4\times 7{ x }^{ 6 }=28{ x }^{ 6 }$

3. $f(x)={ x }^{ 4 }-{ x }^{ 3 }+5$

$f'(x)=4{ x }^{ 3 }-3{ x }^{ 2 }+0\\ f'(x)=4{ x }^{ 3 }-3{ x }^{ 2 }$

4. If $f(x)=2{ x }^{ 5 }-7{ x }^{ 3 }+5x-4$, evaluate $f'(-1)$

$f'(x)=10{ x }^{ 4 }-21{ x }^{ 2 }+5\\ f'(-1)=10{ (-1) }^{ 4 }-21{ (-1) }^{ 2 }+5\\ f'(-1)=-6$

5. Differentiate $S=2\pi {r}^{2}+2\pi rh$ with respect to r.

In this case, we are differentiating with r as the variable, so $\pi$ and h are constants.

$\cfrac { dS }{ dr } =2\pi (2r)+2\pi h\\ \\ \cfrac { dS }{ dr } =4\pi r+2\pi h$

## Composite Function Rule

composite function is a function composed of two or more other functions. For example, ${(3x^2 - 4)}^5$ is made up of a function $u^5$ where $u=3x^2 - 4$.

To differentiate a composite function, we need to use the result:

$\cfrac { dy }{ dx } =\cfrac { dy }{ du } \times \cfrac { du }{ dx }$

### Examples

1. Differentiate ${(5x+4)}^7$

2. Differentiate $\sqrt { 3-x }$

The derivative of a composite function is the product of two derivatives. One is the derivative of the function inside the brackets. The other is the derivative of the whole function.

$\cfrac { d }{ dx } [f\left( x \right) ]^{ n }=f^{ ' }\left( x \right) n[f\left( x \right) ]^{ n-1 }$

### Examples:

Differentiate:

1. ${ (8{ x }^{ 3 }-1) }^{ 5 }$

$\cfrac { dy }{ dx } =24{ x }^{ 2 }\cdot 5{ (8{ x }^{ 3 }-1) }^{ 4 }\\ \\ \cfrac { dy }{ dx } =120{ x }^{ 2 }{ (8{ x }^{ 3 }-1) }^{ 4 }$

2. $\cfrac { 1 }{ { (6x+1) }^{ 2 } }$

$y={ (6x+1) }^{ -2 }\\ y'=6\times -2{ (6x+1) }^{ -3 }\\ y'=-12{ (6x+1) }^{ -3 }\\ \\ y'=-\cfrac { 12 }{ { (6x+1) }^{ 3 } }$

## Product and Quotient Rule

Differentiating the product of two functions $y=uv$ gives the result:

$\cfrac { dy }{ dx } =u\cfrac { dv }{ dx } +v\cfrac { du }{ dx }$

If $y=uv, y'=u'v+v'u$

Differentiating the quotient of two function $y=\cfrac{u}{v}$ gives the result:

$\cfrac { dy }{ dx } =\cfrac { v\cfrac { du }{ dx } -u\cfrac { dv }{ dx } }{ { v }^{ 2 } }$

If $y=\cfrac{u}{v}, y'=\cfrac { vu'-uv' }{ v^2 }$

### Examples

1. Differentiate $2{ x }^{ 5 }{ (5x+3) }^{ 3 }$

$u=2{ x }^{ 5 }\quad and\quad v={ (5x+3 })^{ 3 }\\ u'=10{ x }^{ 4 }\quad and\quad v'=5\cdot 3{ (5x+3 })^{ 2 } \\ y'=u'v+v'u\\ y'=10{ x }^{ 4 }{ (5x+3) }^{ 3 }+5\cdot 3{ (5x+3) }^{ 2 }\cdot 2{ x }^{ 5 }\\ y'=10{ x }^{ 4 }{ (5x+3) }^{ 3 }+30{ x }^{ 5 }{ (5x+3) }^{ 2 }\\ y'=10{ x }^{ 4 }{ (5x+3) }^{ 2 }[(5x+3)+3x]\\ y'=10{ x }^{ 4 }{ (5x+3) }^{ 2 }(8x+3)$)

2. Differentiate $(3x-4) \sqrt(5-2x)$

3. Differentiate $\cfrac{3x-5}{5x+2}$
4. Differentiate $\cfrac{4x^3-5x+2}{x^3-1}$