Question

What’s the derivative of y=sec(3x^2)

Answer 1

By the chain rule, setting
`u=3x^2`, we have

`(\mathrm dy)/(\mathrm dx)=(\mathrm dy)/(\mathrm du)\cdot(\mathrm du)/(\mathrm dx)`

Since

`(\mathrm d(\sec u))/(\mathrm du)=\sec u\tan u`

`(\mathrm d(3x^2))/(\mathrm dx)=6x`

we end up with

`(\mathrm dy)/(\mathrm dx)=6x\sec3x^2\tan3x^2`