Question

How do you differentiate `f(x)=csc(1/x^3)` using the chain rule?

Answer 1

`f'(x) = (3csc(1/x^3)cot(1/x^3))/x^4`

Explanation:

f 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 = csc(1/x^3)`, Then:

`{ ("Let "u=1/x^3, => , (du)/dx=-3/x^4), ("Then "y=cscu, =>, dy/(du)=-cscucotu ) :}`

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

`\ \ \ \ \ dy/dx = (-cscucotu)(-3/x^4)`
`\ \ \ \ \ \ \ \ \ \ = (3cscucotu)/x^4`
`\ \ \ \ \ \ \ \ \ \ = (3csc(1/x^3)cot(1/x^3))/x^4`