CodeCogs

Differentiation of Simple Algebraic Functions from first principles

You're viewing an older version of this page (#3603). View the current version.

Differentiation from First Principles

It is sometimes required that Differentiation be carried out from first principles.

Consider the following equation

$y=Ax^2+Bx+c$

(1)

Let there be small increase in x of $\delta x$ and let the corresponding increase in y be $\delta y$ as shown on the graph

Rewriting the original equation

$y+\delta y=A(x+\delta x)^2+B(x+\delta x)+C$

(2)

Multiplying out

$y+\delta y=A(x^2+2x\delta x+\delta x^2)+B(x+\delta x)+C$

(3)

$=Ax^2+2Ax\delta x+A\delta x^2+2Bx+2B\delta x+C$

(4)

$\text {But we know that }\;\;\;y=Ax^2+Bx+C$

(5)

Subtracting equation (5) from Equation (4)

$\delta y=2Ax\delta x+\delta x^2+B\delta x$

(6)

As the value $\delta x$ is reduced and tends towards dx i.e. until it is infinitesimally small, the value of $\delta x ^2$ tends towards zero and can be neglected. $\delta y$ is now written as dy.

It can be seen from the diagram that the value of the tangent at x,y is $\displaystyle \frac{\delta y}{\delta x}$ and at the limit this is written as $\displaystyle \frac{dy}{dx}$

Referring to Equation (1) the value of the tangent is given by:

$\frac{dy}{dx}= 2Ax+B$

(7)

The Differential Coefficient (Gradient Function)

$\displaystyle \mathbf{\frac{dy}{dx}}$ is known as the Gradient

Function and represents the Derivative of y with respect of x. It is also known as the Differential Coefficient.

In the simplest case:

$\text{If}\;\;\;\;\;y=Kx^n+\text{Constant}$

(8)

$\text{Then}\;\;\;\; \frac{dy}{dx}=Knx^{n-1}$

(9)

Example 1

$\text{If}\;\;\;\;\;y=3x^5+2x^2-5x+6$

(10)

$\text{Then}\;\;\;\;\;\frac{dy}{dx}=5\times 3\;x^{5-1}+2\times 2\;x^{2-1}-5\;x^{1-1}$

(11)

$Or\;\;\;\;\frac{dy}{dx}=15\;x^4+4x-5$

(12)

Example 2

$\text{If} \;\;\;\;\;y=(3x^2-5)(x+1)$

(13)

It is not possible to differentiate the above equation term by term. It must either be multiplied out or treated as a product of two variables. ( See "Product Rule") The choice of which to use depends on which is the best solution for a particular equation.

Then this must be multiplied out to give:

$y=3x^3+3x^2-5x-5$

(14)

And So:

$\frac{dy}{dx}=9x^2+6x-5$

(15)

Example 3

Find the gradient of the curve $(y=6x^4+2x^3-5x+6)$ at x=1

To see the calculation please click on the red button

Example 4

If the distance traveled by a particle in time t is given by:-

$s=ut+\frac{1}{2}at^2$

(21)

Where u and a are constants show that the velocity at time t is $u +at$

Note. Velocity is the rate at which a particle moves in a straight line from a fixed point and is thus $\displaystyle \frac{ds}{dt}$

To see the working please click on the red button

Example 5

Differentiate with respect to x

$y=(5x+1)(x-3)$

(26)

And find the slope of the Tangent at $x=2 \;\;\;and\;\;\;x=1.4$

To see the workings please click on the red button

The Differentiation 0f a Product of two Functions of x

It is obvious, that by taking two simple factors such as 5 X 8 that the total increase in the product is Not obtained by multiplying together the increases of the separate factors and therefore the Differential Coefficient is not equal to the product of the d.c's of its factors.

$\text{If}\;\;\;\;\mathbf{y=uv}\;\;\;\;\;\;\text{Then}\;\;\;\;\mathbf{\frac{dy}{dx}=u\frac{dv}{dx}+v\frac{dv}{dx}}$

(27)

Proof of the above Equation

Please click on the red button to see the proof

Example 6

$\text{If}\;\;\;\;\;y=(x^3+1)(x^2+2)$

(34)

$\frac{dy}{dx}=(x^3+1)2x+(x^2+2)3x^2$

(35)

$=5x^4+6x^2+2x$

(36)

Example 7

$\text{If}\;\;\;\;\;y=\left ( x^2-4 \right )\left ( x^4+3 \right )$

(39)

To see solution please click on the red button

Example 8

$\text{If}\;\;\;\;\;y=x\left (x^2-1 \right )\left ( x^4+4 \right )$

(44)

To see solution please click on the red button

Example 9

$\text{If}\;\;\; y=x^n\;\left ( 1+\sqrt{x} \right )$

(47)

More...

Example 10

$\text{If}\;\;\;\;\;y=\left ( x^n+a^n \right )\left ( x^m+a^m \right )$

(51)

More...

The Differentiation of a Product of any number of Functions of x

The rule for finding the differential coefficient of a product of two functions of x can be extended to apply to the product of any finite numbers of functions of x

$\text{If} \;\;\;\;\;\;\;y=u\;v\;w$

(52)

Where u, v, w are all functions of x, then regarding this as the product of the two factors un and w:-

$\frac{dy}{dx}=uv\;\frac{dw}{dx}+w\;\frac{d(uv)}{dx}$

(53)

$=uv\:\frac{dw}{dx}+\left ( u\;\frac{dv}{dx}+v\;\frac{du}{dx} \right )$

(54)

$=uv\:\frac{dw}{dx}+uw\;\frac{dv}{dx}+vw\;\frac{du}{dx}$

(55)

And similarly for any finite number of factors.

Note An important result follows from the above rules. The differential coefficient of $y^2$ with respect to x can be considered to be the product of two factors each of x and hence is given by:-