What is the derivative of #y= sin(tan 2x)#?

1 Answer
Jan 5, 2016

#dy/dx = 2cos(tan2x)sec^2(2x)#

Explanation:

We need to apply the chain rule twice.

Recall that the chain rule states, if we have some function #f(g(x))#, the derivative of #f# with respect to #x# is equal to the derivative of #f# with respect to #g#, multiplied by the derivative of #g# with respect to #x#.

So in this case, the derivative #dy/dx# will equal the derivative of #sin(tan 2x)# with respect to #tan 2x# (basically, treat #tan 2x# as a whole variable) times the derivative of #tan 2x# with respect to #x#.

Derivative of #sin# is just #cos#:

#dy/dx = cos(tan 2x) * d/dx[tan 2x]#

Derivative of #tan# is #sec^2#. However, we need to apply the chain rule again, meaning this time we will just pull the derivative of #2x# out. (which is just #2#)

#dy/dx = cos(tan 2x) * sec^2(2x) * 2#