Question

How do you use the chain rule to differentiate `f(x)=sin(x^2)/(x^4-3x)^4`?

Answer 1

`f'(x)=((2x^5-6x^2)cos(x^2)-(16x^2+12)sin(x^2))/(x^4-3x)^5`

Explanation:

This is a rational expression that will require us to use the quotient rule to differentiate. First let's rewrite the function as a product since the product rule is easier to use.

`f(x)=(sin(x^2))/(x^4-3x)^4` `= sin(x^2)(x^4-3x)^-4`

Recall that for a function of the form `f(x)=a*b`

The product rule for differentiation is as follows:
`f'(x)=a'b+ab'`

Also notice that in our function, `a` and `b` are both compositions. Recall that the derivative of a composition requires us to use the chain rule.

For a function of the form `f(x)=f(g(x))`, the derivative is `f'(x)=f'(g(x))*g'(x)`

So, applying the product rule and chain rule on our function, we have
`f'(x)=[cos(x^2)*2x]` `[(x^4-3x)^-4]`+`[sin(x^2)][-4(x^4-3x)^-5` `(4x^3-3)]`

`f'(x)=(2xcos(x^2))/(x^4-3x)^4`- `(4(4x^2-3)sin(x^2))/(x^4-3x)^5`

`f'(x)= (x^4-3x)/(x^4-3x)` `(2xcos(x^2))/(x^4-3x)^4`- `(4(4x^2-3)sin(x^2))/(x^4-3x)^5`

`f'(x)=((x^4-3x)2xcos(x^2)-4(4x^2-3)sin(x^2))/(x^4-3x)^5`

`f'(x)=((2x^5-6x^2)cos(x^2)-(16x^2+12)sin(x^2))/(x^4-3x)^5`