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

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(__π__x).

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(__π__x)] = __π__cos(__π__x)

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(__π__x).

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

d/dx[cos(__π__x)] = -__π__sin(__π__x)

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

## Comments

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Rahul

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!