# Find dy/dx for y = (4x^2 - 2) / (x^3 - 1)

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

We will use the quotient rule to find the derivative.

Let y= uv such that:

u= 4x^2 -2 ==> u' = 8x

v= x^3 -1 ==> v' = 3x^2

Then we know that:

dy/dx = ( u'v - uv')/v^2

=> dy/dx = ( 8x(x^3-1)-(4x^2-2)*3x^2 / (x^3-1)^2

==> dy/dx = ( 8x^4 - 8x) - 12x^4 + 6x^2)/ (x^3-1)^2

**==> dy/dx = (-4x^4 +6x^2 - 8x) / (x^3-1)^2**

We have to find the derivative of y = (4x^2 - 2) / (x^3 - 1)

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

=> (4x^2 - 2)*(x^3 - 1)^-1

Use the product rule for differentiation

y' = [(4x^2 - 2)]'*(x^3 - 1)^-1 + (4x^2 - 2)*[(x^3 - 1)^-1]'

=> (8x)*(x^3 - 1)^-1 + (4x^2 - 2)*-1*[(x^3 - 1)^-2]*3x^2

=> [(8x)*(x^3 - 1) + (4x^2 - 2)*-1*3x^2]/[(x^3 - 1)^2]

=> [(8x^4 - 8x - 12x^4 + 6x^2]/[(x^3 - 1)^2]

=> [(-4x^4 + 6x^2 - 8x]/[(x^3 - 1)^2]

**The value of dy/dx = [(-4x^4 + 6x^2 - 8x]/[(x^3 - 1)^2]**