Question

How do you use the quotient rule to differentiate `y=(2x^4-3x)/(4x-1)`?

Answer 1

`f'(x)=(24x^4-8x^3+3)/(4x-1)^2`

Explanation:

The Quotient rle is given by
`(u/v)'=(u'v-uv')/v^2`
let us denote by
`u=2x^4-3x`

and

`v=4x-1`

so we get
`u'=8x^3-3`
and
`v'=4`

now we Can build the derivative:
`f'(x)=((8x^3-3)(4x-1)-(2x^4-3x)*4)/(4x-1)^2`

multiplying out the numerator and collecting like Terms we get

`f'(x)=(24x^4-8x^3+3)/(4x-1)^2`

Answer 2

`(24x^4-8x^3+4)/(4x-1)^2`

Explanation:

If we have a quotient of functions `f(x)` and `g(x)`, we can find the derivative with the Quotient Rule

`(f'(x)g(x)-f(x)g'(x))/(g(x))^2`

In our example, we essentially have

`f(x)=2x^4-3x=>f'(x)=8x^3-3` and

`g(x)=4x-1=>g'(x)=4`

We can plug in our expressions into the Quotient Rule to get

`((8x^3-3)(4x-1)-4(2x^4-3x))/(4x-1)^2`

With FOIL and some algebraic distribution, we can simplify this expression to get

`(24x^4-8x^3+4)/(4x-1)^2`

Hope this helps!