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

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Answer 1 · 2017-08-13T14:57:06

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

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

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

Product rule:

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

Chain rule:

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

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

Answer 2 · 2017-08-13T15:05:50

$6x(x + 1)$

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

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 !!!!!!!!!

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