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

Question

15512 views around the world

You can reuse this answer
Creative Commons License

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#