# What is the derivative of sin(Pix) and the derivative of cos(Pix)?

Updated on April 27, 2012 Finding the derivative of sin(ax) can be done using the following result:

d/dx[sin(ax)] = acos(ax)

Let’s take a look at some examples of working out the derivative of sin(ax).

Example 1

Work out the derivative of sin(5x).

All you need to do is apply the above formula:

d/dx[sin(5x)] = 5cos(5x).

Example 2

Work out the derivative of sin(-4x).

Like example 1, all you need to do is apply the above formula:

d/dx[sin(-4x)] = -4cos(-4x)

Example 3

Work out the derivative of sin(πx).

This time you have Pi before x, but this makes no difference as Pi is just a number like the two examples shown above:

d/dx[sin(πx)] = πcos(πx)

Let’s move on to the derivative of cos(ax). Finding the derivative of cos(ax) can be done using the following result:

d/dx[cos(ax)] = -asin(ax)

Let’s take a look at some examples of working out the derivative of cos(ax).

Example 4

Work out the derivative of cos(2x).

All you need to do is apply the formula just stated:

d/dx[cos(2x)] = -2sin(2x).

Example 5

Work out the derivative of cos(-0.5x).

Like example 1, all you need to do is apply the above formula:

d/dx[cos(-0.5x)] = 0.5sin(-0.5x)

Note, the two negative signs on this last example make a positive coefficient of sine.

Example 6

Work out the derivative of cos(πx).

This time you have Pi before x, so like the example with the derivative of sinπx the formula for cos(ax) can be applied as Pi is just like any other number.

d/dx[cos(πx)] = -πsin(πx)

So to summarise, the derivative of sin(Pix) is Picos(Pix) and the derivative of cos(Pix) is –Pisin(Pix).

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• Jessee R

6 years ago from Gurgaon, India

you can use the mathmagic math editor to make your math hub more interesting!

:)

Rahul

• Jessee R

6 years ago from Gurgaon, India

Interesting approach... you could have taken a more simpler route of explanation of chain rule to underline the basics here as in

let us take an example

to find d/dx of sin(ax) where "a" is constant

let ax = t , we have

d(ax)/dx = dt/dx

this implies

adx/dx = dt/dx [a is a constant, moved out of derivative]

which implies

a = dt/dx --i. [dx/dx =1 ]

now as ax =t we have

dsin(ax)/dx = dsint/dx

by chain rule

dsint/dx = dsint/dt * dt/dx

[ df(t)/dx = df(t)/dt * dt/dx... by chain rule]

this implies

dsint/dx = cost * dt/dx

reinstating the assumptions we have

dsin(ax)/dx = cos(ax) * a = acos(ax)... solved

I just love calculus so couldn't stop giving my input.. hope you don't mind.... :)

Great hub!

working