# Differetiation Calculus

Differetiation is refered to has the rate of change of a given function. Differetiation is the process of finding a derivative and a derivative is the result from differetiation

## Differetiation Notation

$\begin{array}{|ll|}\hline A Function& Leibniz notation\\ f\left(x\right)& \frac{d}{dx}\left(\mathrm{f\left(x\right)}\right)or\frac{\mathrm{df}}{dx}\\ y& \frac{d}{dx}\left(y\right)or\frac{\mathrm{dy}}{dx}\\ 4x + 9& \frac{d}{dx}\left(\mathrm{4x + 9}\right)\\ \hline\end{array}$

### Points to note

1. $\frac{d}{dx}\left(y\right)=\frac{\mathrm{dy}}{dx}$

2. The expression $\frac{d}{dx}\left(y\right)$ means differetiate with respect to x of y which is equal to the expression $\frac{\mathrm{dy}}{dx}$

## How to differetiate

Equation: y = ${\mathrm{ax}}^{3}$

1. multiply Co-oefficient by its power in this case $a×3=3a$

2. subtract power by 1 (one) in this case $3-1=2$

3. Merge the steps together therefore the combination will be ${\mathrm{3ax}}^{2}$

Hence $\frac{\mathrm{dy}}{dx}={\mathrm{3ax}}^{2}$

### Point to note

1. when a number is differetiated the result becomes a zero (0), E.G $\frac{\mathrm{dy}}{dx}7=0$

### Example 1

Differentiate each of the following with respect to x

(a) $y={\mathrm{4x}}^{5}$

(b) $y={x}^{2}+7$

(c) $y={\mathrm{3x}}^{4}+19x-2$

Solutions

(a)

$y={\mathrm{4x}}^{5}$

Answer: $\frac{\mathrm{dy}}{dx}={\mathrm{20x}}^{4}$

(b)

$y={x}^{2}+7$

$\frac{\mathrm{dy}}{dx}={\mathrm{2x}}^{1}+0$

Answer: $\frac{\mathrm{dy}}{dx}=\mathrm{2x}$

(c)

$y={\mathrm{3x}}^{4}+19x-2$

$\frac{\mathrm{dy}}{dx}={\mathrm{12x}}^{3}+19-0$

Answer: $\frac{\mathrm{dy}}{dx}={\mathrm{12x}}^{3}+19$

### Example 2

Find the equation of the normal to the curve $y={x}^{3}-{\mathrm{2x}}^{2}-4x+1$ at the point (-1, 2)

Solutions

$y={x}^{3}-{\mathrm{2x}}^{2}-4x+1$

$\frac{\mathrm{dy}}{dx}={\mathrm{3x}}^{2}-{\mathrm{4x}}^{1}-4+0$

$\frac{\mathrm{dy}}{dx}={\mathrm{3x}}^{2}-\mathrm{4x}-4$

(-1, 2)

X = -1

Substitute in the equation to find the gradient or dy/dx

$\frac{\mathrm{dy}}{dx}={\mathrm{3\left(-1\right)}}^{2}-\mathrm{4\left(-1\right)}-4$

$\frac{\mathrm{dy}}{dx}=3-\mathrm{\left(-4\right)}-4$

$\frac{\mathrm{dy}}{dx}=3+4-4$

$\frac{\mathrm{dy}}{dx}=7-4$

$\frac{\mathrm{dy}}{dx}=3$

Answer: $\mathrm{Gradient}=3$

Equation of the straight line y = mx + c

(-1, 2)

x = -1, y = 2 , m = 3, c = ?

2 = 3(-1) + c

2 = -3 + c

c = 2 + 3

c = 5

Therefore the equation of the normal is Answer: y = 3x + 5

### Example 3

Determine the equation of the normal to the curve

$y={\mathrm{2x}}^{2}-3x-2$

that passes through the point(3, 7).

Solutions

$y={\mathrm{2x}}^{2}-3x-2$

$\frac{\mathrm{dy}}{dx}={\mathrm{4x}}^{1}-3-0$

$\frac{\mathrm{dy}}{dx}=\mathrm{4x}-3$

(3, 7)

x = 3

$\frac{\mathrm{dy}}{dx}=\mathrm{4\left(3\right)}-3$

$\frac{\mathrm{dy}}{dx}=\mathrm{12}-3$

$\frac{\mathrm{dy}}{dx}=9$

$\mathrm{Gradient}=9$

y = mx + c

Find the constant c

(3, 7)

x = 3, y = 7

7 = 9(3) + c

7 = 27 + c

c = 7 - 27

c = -20

Equation of the normal Answer: y = 9x - 20

### Example 4

Find the equation of the normal to the curve $y={\mathrm{5x}}^{3}-{\mathrm{6x}}^{2}+2x+5$ at the point (1, 2)

Solutions

$y={\mathrm{5x}}^{3}-{\mathrm{6x}}^{2}+2x+5$

Find the derivative

$\frac{\mathrm{dy}}{dx}={\mathrm{15x}}^{2}-{\mathrm{12x}}^{1}+2+0$

$\frac{\mathrm{dy}}{dx}={\mathrm{15x}}^{2}-\mathrm{12x}+2$

Point (1, 2)

x = 1, y = 2

$\frac{\mathrm{dy}}{dx}={\mathrm{15\left(1\right)}}^{2}-\mathrm{12\left(1\right)}+2$

$\frac{\mathrm{dy}}{dx}=\mathrm{15}-\mathrm{12}+2$

$\frac{\mathrm{dy}}{dx}=3+2$

$\frac{\mathrm{dy}}{dx}=5$

y = mx + c

Point (1, 2)

x = 1, y = 2

2 = 5(1) + c

2 = 5 + c

c = 2 - 5

c = -3

Equation of the normal Answer: y = 5x - 3

## How to find coordinates using differetiation

Find the x coordinates of the point on the curve $y={\mathrm{2x}}^{3}-{\mathrm{3x}}^{2}-36x-3$ where the gradient is zero

Solution

$y={\mathrm{2x}}^{3}-{\mathrm{3x}}^{2}-36x-3$

$\frac{\mathrm{dy}}{dx}={\mathrm{6x}}^{2}-{\mathrm{6x}}^{1}-{\mathrm{36x}}^{0}-0$

$\frac{\mathrm{dy}}{dx}={\mathrm{6x}}^{2}-\mathrm{6x}-36$

$\frac{\mathrm{dy}}{dx}=$ Gradient

$\frac{\mathrm{dy}}{dx}=0$

$0={\mathrm{6x}}^{2}-\mathrm{6x}-36$

${\mathrm{6x}}^{2}-\mathrm{6x}-36=0$

${x}^{2}-x-6=0$

P = 6

S = -1

F = (-3 , 2)

${x}^{2}-\mathrm{3x}+\mathrm{2x}-6=0$

$x\left(x - 3\right)+2\left(x - 3\right)=0$

$\left(x - 3\right)\left(x + 2\right)=0$

$\left(x - 3\right)=0$

$\left(x + 2\right)=0$

x = 3 or x = -2

## How to find coordinates of the stationary points

### Stationary Points

Find the coordinates of the stationary points on the curve

$y={\mathrm{2x}}^{3}-{\mathrm{3x}}^{2}-12x+4$

1. The gradient at a stationary point is always 0 (Zero)

2. The first step to under take to find the coordinates at the stationary points is to differetiate the equation and then equate it to 0 (Zero)

3. Then after equating 0 (Zero) to the differetiated function, find the coordinates of x from the simplified equation (Quadratic equation)

$\frac{\mathrm{Dy}}{Dx}={\mathrm{6x}}^{2}-\mathrm{6x}-12$

$\frac{\mathrm{Dy}}{Dx}={x}^{2}-x-2$

${x}^{2}-x-2=0$

P = -2

S = -1

F = (-2, 1)

${x}^{2}- 2x\mathrm{+ x}-2=0$

$x\left(x- 2\right)\mathrm{+ 1\left(}x -2\right)=0$

$\left(x- 2\right)\left(x +1\right)=0$

$\left(x- 2\right)=0$

$\left(x +1\right)=0$

x = 2 or x = - 1

$y={\mathrm{2x}}^{3}-{\mathrm{3x}}^{2}-12x+4$

$y={\mathrm{2\left(2\right)}}^{3}-{\mathrm{3\left(2\right)}}^{2}-12\left(2\right)+4$

$y=2\left(8\right)-3\left(4\right)-12\left(2\right)+4$

$y=16-12-24+4$

$y=-20+4$

$y=-16$

$y={\mathrm{2x}}^{3}-{\mathrm{3x}}^{2}-12x+4$

$y={\mathrm{2\left(-1\right)}}^{3}-{\mathrm{3\left(-1\right)}}^{2}-12\left(-1\right)+4$

$y=\mathrm{2\left(-1\right)}-\mathrm{3\left(1\right)}-12\left(-1\right)+4$

$y=\mathrm{-2}-3+12+4$

$y=\mathrm{-5}+16$

$y=\mathrm{11}$

Therefore the coordinates of the stationary points are : (2, -16) and (-1, 11)