Home > Calculus > Product Rule

How do you find the derivative of #y=2x(3x-1)(4-2x)#?

Answer 1

Emilia Aguilar

#(dy)/(dx)=-12x^3+4x^2+28x-8#

#y=2x(3x-1)(4-2x)#

#y=(6x^2-2x)(4-2x)#

#(dy)/(dx)=(6x^2-2x)(-2x)+(4-2x)(12x-2)#

#(dy)/(dx)=-12x^3+4x^2+48x-8-24x+4x#

#(dy)/(dx)=-12x^3+4x^2+28x-8#

The product rule can be found in

#y=uv#

#(dy)/(dx)=uv'+vu'#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Charles Arocha

To find the derivative of ( y = 2x(3x-1)(4-2x) ), you can use the product rule and the chain rule. The derivative is:

[ \frac{dy}{dx} = 2(3x-1)(4-2x) + 2x[3(4-2x) - (3x-1)2] ]

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.