What is the derivative of #y=sec(3x^2)#?

1 Answer
Jul 9, 2016

By the chain rule:

Let #y = sec(u)# and #u = 3x^2#.

The derivative of #secx# can be found by the following proof:

#secx = 1/cosx#

#(1/cosx)' = ((0 xx cosx) - (1 xx -sinx))/(cosx)^2#

#(1/cosx)' = sinx/(cos^2x)#

#(secx)' = secx xx sinx/cosx#

#(secx)' = secxtanx#

The derivative of #3x^2# can be obtained using the power rule:

#(3x^2)' = 2 xx 3x^(2 - 1)#

#(3x^2)' = 6x#

The chain rule states that #dy/dx = dy/(du) xx (du)/dx#.

Hence, #dy/dx = secutanu xx 6x = 6xsec(3x^2)tan(3x^2)#

Practice exercises:

  1. Differentiate the following:

a) #cscx#

b) #cot(3x^2 + 5x + 1)#

c) #tan(e^(2x^2))#

Hopefully this helps, and good luck!