Question

How do you use the chain rule to differentiate `y=5/(2x^3+3x)`?

Answer 1

`dy/dx = -5((6x^2+3))/(2x^3+3x)^2`

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 = 5/(2x^3+3x)`, Then:

`{ ("Let "u=2x^3+3x, => , (du)/dx=6x^2+3), ("Then "y=5/u=u^-1, =>, dy/(du)=-u^-2=-5/u^2 ) :}`

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

`dy/dx = (-5/u^2)(6x^2+3)`
`:. dy/dx = -5(6x^2+3)/(2x^3+3x)^2`