How do you find the derivative of $f (x) = x^{3} + 2 x^{2} + 4 x + 2$ $f (x) =x^3+2x^2+4x+2$ ?

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Answer 1 · 2016-05-24T16:33:48

$\frac{d}{d x} f (x) = 3 x^{2} + 4 x + 4$ $d/(d x) f(x)=3x^2+4x+4$

$\frac{d}{d x} f (x) = \frac{d}{d x} (x^{3} + 2 x^{2} + 4 x + 2)$ $d/(d x) f(x)=d/(d x) (x^3+2x^2+4x+2)$

$\frac{d}{d x} f (x) = 3 x^{2} + 4 x + 4 + 0$ $d/(d x) f(x)=3x^2+4x+4+0$

$\frac{d}{d x} f (x) = 3 x^{2} + 4 x + 4$ $d/(d x) f(x)=3x^2+4x+4$

How do you find the derivative of f(x)=x3+2x2+4x+2 f (x) =x^3+2x^2+4x+2?