Question

How do you use the chain rule to differentiate `y=tan(2x^3-1)`?

Answer 1

`dy/dx = 6x^2sec^2(2x^3-1)`

Explanation:

If you are studying maths, then you should learn the Chain Rule for Differentiation, and practice how to use it:

If `y=f(x)` then `f'(x)=dy/dx=dy/(du)(du)/dx`

I was taught to remember that the differential can be treated like a fraction and that the "`dx`'s" of a common variable will "cancel" (It is important to realise that `dy/dx` isn't a fraction but an operator that acts on a function, there is no such thing as "`dx`" or "`dy`" on its own!). The chain rule can also be expanded to further variables that "cancel", E.g.

`dy/dx = dy/(dv)(dv)/(du)(du)/dx` etc, or `(dy/dx = dy/color(red)cancel(dv)color(red)cancel(dv)/color(blue)cancel(du)color(blue)cancel(du)/dx)`

So with `y = tan(2x^3-1)`, Then:

`{ ("Let "u=2x^3-1, => , (du)/dx=6x^2), ("Then "y=tanu, =>, dy/(du)=sec^2u ) :}`

Using `dy/dx=(dy/(du))((du)/dx)` we get:

`dy/dx = (sec^2u)(6x^2)`
`:. dy/dx = 6x^2sec^2(2x^3-1)`