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

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Differetiation Notation

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

Points to note

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

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

How to differetiate

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

1. multiply Co-oefficient by its power in this case $a\times 3=\mathrm{3a}$

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

Point to note

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

Example 1

Differentiate each of the following with respect to x

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

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

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

Solutions

(a)

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

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

(b)

$y=x^2+7$

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

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

(c)

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

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

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

Example 2

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

Solutions

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

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

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

(-1, 2)

X = -1

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

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

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

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

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

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

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

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

that passes through the point(3, 7).

Solutions

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

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

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

(3, 7)

x = 3

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

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

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

$\mathrm{Gradient}=9$

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 $\mathrm{y}={\mathrm{5x}}^3-{\mathrm{6x}}^2+\mathrm{2x}+5$ at the point (1, 2)

Solutions

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

Find the derivative

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

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

Point (1, 2)

x = 1, y = 2

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

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

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

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

Therefore the gradient = 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

How to find coordinates of the stationary points