What is the derivative of #sec^2(4x)#?

1 Answer
T Michael B. · Stefan V.
Dec 1, 2016

The final answer is: #(dy)/(dx)=8sec^2(4x)tan(4x)#

Explanation:

This is a composite function consisting of three separate functions, so we use the chain rule, starting with the "outside" function and working to the "inside" function. We must first recall that

#y=sec^2(4x)# is mathematical shorthand for:

#y=(sec(4x))^2#, so the outermost function is #y=u^2#.

Thus we have #(dy)/(dx)=2sec(4x)*sec(4x)tan(4x)*4#, where the second function was #secu# and the final function was #4x#.

We can now simplify this, giving us:

#(dy)/(dx)=8sec^2(4x)tan(4x)#

Related questions
See all questions in Derivatives of y=sec(x), y=cot(x), y= csc(x)
Impact of this question
13385 views around the world
You can reuse this answer
Creative Commons License