How do you use the chain rule to differentiate #r=sec2thetatan2theta#?

We're asked to find the derivative

#(dr)/(d theta) [r = sec(2theta)tan(2theta)]#

We can first use the product rule, which states

#d/(d theta) [uv] = v(du)/(d theta) + u(dv)/(d theta)#

where

• #u = sec(2theta)#

• #v = tan(2theta)#:

#r'(theta) = tan(2theta)d/(d theta) [sec(2theta)] + sec(2theta)d/(d theta) [tan(2theta)]#

Now, we use the chain rule for first the #sec(2theta)# term:

#d/(d theta)[sec(2theta)] = d/(du) [secu] (du)/(d theta)#

where

• #u = 2theta#

• #d/(du) [secu] = secu tanu#:

#r'(theta) = sec(2theta)tan(2theta)tan(2theta) d/(d theta)[2theta] + sec(2theta)d/(d theta) [tan(2theta)]#

The derivative of #2theta# is #2#:

#r'(theta) = 2sec(2theta)tan^2(2theta) + sec(2theta)d/(d theta) [tan(2theta)]#

Now we use the chain rule on the #tan(2theta)# term:

#d/(d theta) [tan(2theta)] = d/(du)[tanu] (du)/(d theta)#

where

• #u = 2theta#

• #d/(du) [tanu] = sec^2u#:

#r'(theta) = 2sec(2theta)tan^2(2theta) + sec(2theta)sec^2(2theta)d/(d theta) [2theta]#

The derivative of #2theta# is #2#:

#r'(theta) = 2sec(2theta)tan^2(2theta) + 2sec(2theta)sec^2(2theta)#

Simplifying..

#color(blue)(r'(theta) = 2sec(2theta)tan^2(2theta) + 2sec^3(2theta)#

We could also change everything into terms of #sin# and #cos#:

#r'(theta) = (2sin^2(2theta))/(cos^3(2theta)) + 2sec^3(2theta)#

#r'(theta) = (2sin^2(2theta))/(cos^3(2theta)) + 2/(cos^3(2theta))#

#color(blue)((2sin^2(2theta)+2)/(cos^3(2theta))#