Calculus

Differentiation

A powerful tool, especially in physics.

The function y = f(x) can be plotted as shown here. Any point on this curve, such as Point A will have a specific gradient, that is a specific slope, that is the measure of the change of y with respect to x. It is also the same slope as the gradient of the tangent drawn on curve at position A.

This gradient, or slope, is very close to the gradient of the line AB where B is a point close to A. The slope of AB is equal to the change in y, ∂y, divided by the change in x, ∂x. This approximation improves as B gets closer to A.

The actual value of the slope is expressed as the value of ∂y/∂x as ∂x approaches 0. Once this is reached it is customary to change the notation of ∂y/∂x to be dy/dx. Sometimes you may see this written as d f(x)/dy.

Finally this whole process is written as

Lim ∂y/∂x = dy/dx

∂x → 0

Example

If you consider y to be distance travelled and x to be time, then dy/dt is the velocity.

Also if you consider y to be the velocity and x again to be time, then this time dy/dt will be acceleration.

In fact what we have done could be written as d(dx/dt)/dt.

I will develop this example further once we have considered a few more general factors concerning differential calculus.

Specific Function

This time we will consider an actual function

y = 4x2-2x+1

Consider the two points A(x1,y1) and B(x2,y2)

Draw perpendicular from B and horizontal from A which intersects at point E(x2,y1)

So BE is the change in y, ∂y

and AE is the change in x, ∂x

We want to determine the tangent at A, the line AM.

∂y/∂x is the slope of line AB

this approaches slope of AM as as B moves towards A.

Lim ∂y/∂x = dy/dx

∂x → 0

∂y/∂x = (y2-y1) / (x2-x1)

= ((4x22-2x2+1)-(4x12-2x1+1)) / (x2-x1)

= (4x22-4x12-2x2+2x1+1-1) / (x2-x1)

= (4(x2-x1)(x2+x1)-2(x2-x1) / (x2-x1)

= 4(x2+x1)-2

as point B approaches point A then x2→x1

so x2→x1→x

∴ dy/dx = 8x-2

Calculus p2

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