# Differentiation

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## Differentiating x to the power of something

1) If y = xn, dy/dx = nxn-1

2) If y = kxn, dy/dx = nkxn-1(where k is a constant- in other words a number)

Therefore to differentiate x to the power of something you bring the power down to in front of the x, and then reduce the power by one.

#### Examples

If y = x4, dy/dx = 4x3
If y = 2x4, dy/dx = 8x3
If y = x5 + 2x-3, dy/dx = 5x4 - 6x-4

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## Finding the Gradient of a Curve

A formula for the gradient of a curve can be found by differentiating the equation of the curve.

#### Example

What is the gradient of the curve y = 2x3 at the point (3,54)?
dy/dx = 6x2
When x = 3, dy/dx = 6× 9 = 54

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### Notation

There are a number of ways of writing the derivative. They are all essentially the same:

(1) If y = x2, dy/dx = 2x
This means that if y = x2, the derivative of y, with respect to x is 2x.

(2) d (x2) = 2x
dx
This says that the derivative of x2 with respect to x is 2x.

(3) If f(x) = x2, f'(x) = 2x
This says that is f(x) = x2, the derivative of f(x) is 2x.

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## Differentiation From 1st Principle

#### Example

Find the formula for the gradient of the graph y = x² .
dy  =    lim      (x + dx)² - x²
dx      dx®0           dx

=   lim     x² + 2xdx + (dx)² - x²
dx®0             dx
=   lim      2xdx + (dx)²
dx®0           dx
The dx on the denominator cancels with those on the numerator.

Therefore dy/dx =
lim      2x + dx
dx®0
When dx becomes zero, dy/dx = 2x.

Therefore the gradient of y = x² is 2x.
For example, at the point (2, 4), the gradient is 2x = 4 .

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## Diferentiation From 1st Principle

A is any point, (x, y). To find the gradient at A, we need to find the gradient of the tangent at A. Let B be a point which is just a little further along the graph. The gradient of the chord AB is approximately the gradient of A. If the horizontal distance between A and B is called dx ("delta" x) and the vertical distance between A and B is called dy, the coordinates of B are (x + dx, y + dy).
From the Coordinate Geometry section, we know that the gradient of a straight line joining two points is:

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## Diferentiation From 1st Principle (Continued)

y2 - y1, where the two points are (x1, y1) and (x2, y2)
x2 - x1
In this case, the two points are (x, y) and (x + dx, y + dy). So substituting these values into the formula, the gradient of the chord is:
y + dy - y  =  dy     (pronounced "delta y by delta x")
x + dx - x      dx

This is the gradient of the chord. The gradient of the curve is the gradient of the chord when the chord has no length- i.e. when it is a tangent. This will happen when dx = 0 .
The gradient of the curve is therefore:
lim       ( dy )
dx®0    ( dx )
This basically means that the gradient is dy/dx as dx approaches or "tends to" (®) zero.

We can rewrite the coordinates of (x, y) as (x, f(x)) and the coordinates of (x + dx, y + dy) as (x + dx, f(x + dx)), since y is a function of x (y = f(x)).
So the gradient of the curve is:
lim      (y + dy - y)
dx®0   (x + dx - x)

since y = f(x) and y + dy = f(x + dx):
lim      f(x + dx) - f(x)
dx®0           dx

This is denoted by dy/dx ("dee y by dee x"). dy/dx is known as the derivative of y with respect to x.

So, in summary,
dy   =     lim      f(x + dx) - f(x)
dx        dx®0           dx

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