How do you differentiate #y=(x+1)^2(2x-1)#?

2 Answers
Aug 13, 2017

Answer:

#dy/dx =4x(2x+1)#

Explanation:

#y=(x+1)^2(2x+1)#

Use the product rule and chain rule to differentiate #y#

Product rule:

#d/dx(pq) = qp' + pq'#

Chain rule:

#d/dx([f(x)]^n) = n[f(x)]^(n-1)f'(x)#

#dy/dx = 2(x+1)(2x-1) + 2(x+1)= 2(x+1)(2x-1+1) = 4x(2x+1)#

Aug 13, 2017

Answer:

#6x(x + 1)#

Explanation:

Actually, there are mainly 2 methods of solving your question,

  1. Either simplifying the equation and differentiating it

  2. Or differentiating first and then simplifying it.

Obviously, the first one is the easiest for beginners,

#rArr y = (x + 1)^2(2x - 1)#

Simplifying this, I get,
#rArr y = 2x^3 + 3x^2 - 1#

Now,
#rArr dy/dx = 6x^2 + 6x#

#therefore, dy/dx = 6x(x + 1)#

ENJOY MATHS !!!!!!!!!