How do you differentiate  y=sec (3x^2 - x) using the chain rule?

Dec 11, 2015

$y ' = \left(6 x - 1\right) \sec \left(3 {x}^{2} - x\right) \tan \left(3 {x}^{2} - x\right)$

Explanation:

Note that according to the chain rule $\frac{d}{\mathrm{dx}} \left[\sec \left(u\right)\right] = u ' \sec \left(u\right) \tan \left(u\right)$.

Therefore

$\frac{d}{\mathrm{dx}} \left[\sec \left(3 {x}^{2} - x\right)\right] = \sec \left(3 {x}^{2} - x\right) \tan \left(3 {x}^{2} - x\right) \frac{d}{\mathrm{dx}} \left[3 {x}^{2} - x\right]$

Find the derivative.

$\frac{d}{\mathrm{dx}} \left[3 {x}^{2} - x\right] = 6 x - 1$

Plug this back in.

$y ' = \left(6 x - 1\right) \sec \left(3 {x}^{2} - x\right) \tan \left(3 {x}^{2} - x\right)$