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
#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
#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))#
