How do you find the derivative of $f(x)=(x+1)(x^2+2x-3)$ ?

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Answer 1 · 2017-07-02T19:28:54

$(df)/dx = 3x^2+6x-1$

We can multiply the two polynomials:

$f(x) = (x+1)(x^2+2x-3)$

$f(x) = x^3+2x^2-3x+x^2+2x-3$

$f(x) = x^3+3x^2-x-3$

Then differentiate:

$(df)/dx = 3x^2+6x-1$

Alternatively we can use the product rule:

$(df)/dx = d/dx (x+1) xx (x^2+2x-3)+ (x+1) xx d/dx(x^2+2x-3)$

$(df)/dx = (x^2+2x-3)+ (x+1) (2x+2)$

$(df)/dx = (x^2+2x-3)+ (2x^2+4x+2)$

$(df)/dx = 3x^2+6x-1$

How do you find the derivative of f(x)=(x+1)(x^2+2x-3)f(x)=(x+1)(x^2+2x-3)?