How do you find the first and second derivatives of y=(2x^4-3x)/(4x-1) using the quotient rule?

1 Answer
Feb 29, 2016

First Derivative: dy/dx=(24x^4-8x^3+3)/((4x-1)^2)
Second Derivative: (d^2y)/dx^2=(192x^4-128x^3+24x^2-24)/(4x-1)^3

Explanation:

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)