Question

How do you use the chain rule to differentiate `y=cos^7(pit-16)`?

Answer 1

We have:

`y = cos^7(pit-16)`

The chain rule gives us:

`dy/dt = dy/(du) * (du)/dt`

In application this means that, for example:

`d/dx u^n = n u^(n-1) * (du)/dx`

So to differentiate our given function we require two application of the chain rule; one to deal with `cos^7X` and one to deal with `cos(Y)`, as follows:

`dy/dt = 7cos^6(pi t-16) d/dt(cos(pi t-16) )`
`" " = 7cos^6(pi t-16) (-sin(pi t -16) )(d/dt(pi t -16 ))`
`" " = 7cos^6(pi t-16) (-sin(pi t -16) )(pi)`
`" " = -7pi \ cos^6(pi t-16) \ sin(pi t -16)`

If you didn't follow that then I will show you the same result using direct substitution:

Let `u = pi t -16 => (du)/dt = pi`
Let `v = cos u => (dv)/(du) = -sinu`

Then:

`y = cos^7(pit-16)`
`\ \ = cos^7(u)`
`\ \ = v^7`

And so:

`(dy)/(dv) = 7v^6`

Then by the chain rule we have:

`dy/dt = dy/(dv) * (dv)/(du) * (du)/dt`
`" " = 7v^6 * (-sinu) * pi`
`" " = -7pi \ (cos u)^6 \ sinu`
`" " = -7pi \ (cos (pi t -16))^6 \ sin(pi t -16)`
`" " = -7pi \ cos^6(pi t -16) \ sin(pi t -16)`, as above