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

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