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

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Answer 1 · 2014-09-23T00:44:08

$y'=(g(x)f'(x)-f(x)g'(x))/(g(x))^2$

$y=f(x)/g(x)=(2x^4-3x)/(4x-1)$

$f'(x)=8x^3-3$

$g'(x)=4$

$(g(x))^2=(4x-1)^2$

$y'=((4x-1)(8x^3-3)-(2x^4-3x)(4))/(4x-1)^2$

$y'=(32x^4-12x-8x^3+3-8x^4+12x)/(4x-1)^2$

Simplify for combining like terms.

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

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