How do you use the quotient rule to differentiate $y = \frac{2 x^{4} - 3 x}{4 x - 1}$ $y=(2x^4-3x)/(4x-1)$ ?

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Answer 1 · 2018-07-19T22:17:08

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

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 · 2018-07-19T22:33:37

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

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!

How do you use the quotient rule to differentiate y=(2x^4-3x)/(4x-1)y=2x4−3x4x−1y=(2x^4-3x)/(4x-1)?