**First, I'll introduce the quotient rule.**

If we are given #f(x)=(g(x))/(h(x))#, then the derivative of #f(x)# is:

#" "f'(x)=(h(x)g'(x)-g(x)h'(x))/[h(x)]^2#

**Getting the first derivative:**

#[1]" "dy/dx=d/dx((2x^4-3x)/(4x-1))#

*Use the quotient rule.*

#[2]" "dy/dx=((4x-1)*d/dx(2x^4-3x)-(2x^4-3x)*d/dx(4x-1))/((4x-1)^2)#

*The derivative of #2x^4-3x# is simply #8x^3-3#. The derivative of #4x-1# is simply #4#. (I got these using power rule and difference rule)*

#[3]" "dy/dx=((4x-1)(8x^3-3)-(2x^4-3x)(4))/((4x-1)^2)#

*Simplify.*

#[4]" "dy/dx=((32x^4-12x-8x^3+3)-(8x^4-12x))/((4x-1)^2)#

#[5]" "dy/dx=(32x^4-12x-8x^3+3-8x^4+12x)/((4x-1)^2)#

#[6]" "color(red)(dy/dx=(24x^4-8x^3+3)/((4x-1)^2))#

**Getting the second derivative:**

#[1]" "(d^2y)/dx^2=d/dx[(24x^4-8x^3+3)/((4x-1)^2)]#

*Use the quotient rule.*

#[2]" "(d^2y)/dx^2=((4x-1)^2*d/dx(24x^4-8x^3+3)-(24x^4-8x^3+3)*d/dx[(4x-1)^2])/[(4x-1)^2]^2#

*The derivative of #24x^4-8x^3+3# is #96x^3-24x^2#. The derivative of #(4x-1)^2# is #32x-8#.*

#[3]" "(d^2y)/dx^2=((4x-1)^2(96x^3-24x^2)-(24x^4-8x^3+3)(32x-8))/(4x-1)^4#

*Simplify.*

#[4]" "(d^2y)/dx^2=((4x-1)^2(96x^3-24x^2)-(24x^4-8x^3+3)(8)(4x-1))/(4x-1)^4#

#[5]" "(d^2y)/dx^2=(cancel((4x-1))[(4x-1)(96x^3-24x^2)-(24x^4-8x^3+3)(8)])/(4x-1)^(cancel4 3)#

#[6]" "(d^2y)/dx^2=(384x^4-96x^3-96x^3+24x^2-192x^4+64x^3-24)/(4x-1)^3#

#[7]" "color(red)((d^2y)/dx^2=(192x^4-128x^3+24x^2-24)/(4x-1)^3)#