How do you differentiate #f(x) =xsec4x^3 # using the chain rule?

1 Answer
Oct 5, 2017

Answer:

# f'(x) = 12x^3sec(4x^3)tan(4x^3) + sec(4x^3) #

Explanation:

We seek #f'(x)# where:

# f(x) = xsec(4x^3) #

First we apply the product rule:

# f'(x) = (x)(d/dx sec(4x^3)) + (d/dx x)(sec(4x^3)) #
# \ \ \ \ \ \ \ \ \ = xd(d/dx sec(4x^3)) + (1)(sec(4x^3)) #
# \ \ \ \ \ \ \ \ \ = xd(d/dx sec(4x^3)) + sec(4x^3) #

Then, we apply the chain rule:

# f'(x) = x(sec(4x^3)tan(4x^3) d/dx (4x^3)) + sec(4x^3) #
# \ \ \ \ \ \ \ \ \ = xsec(4x^3)tan(4x^3) (12x^2) + sec(4x^3) #
# \ \ \ \ \ \ \ \ \ = 12x^3sec(4x^3)tan(4x^3) + sec(4x^3) #