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

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Answer 1 · 2017-03-02T18:52:13

$6 x^{2} - 2 x + 3$ $6x^2-2x+3$

Derivative of any simple power is

$\frac{d}{d x} a x^{n} = n \cdot a x^{n - 1}$ $d/dx ax^n = n*ax^(n-1)$

and for polynomials,

$\frac{d}{d x} (a + b + c) = \frac{d}{d x} a + \frac{d}{d x} b + \frac{d}{d x} c$ $d/dx (a+b+c)=d/dxa + d/dxb + d/dxc$ .

For the above question,

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

$\frac{d}{d x} (2 x^{3}) = 6 x^{2}$ $d/dx(2x^3) = 6x^2$

$\frac{d}{d x} (- x^{2}) = - 2 x^{1} = - 2 x$ $d/dx(-x^2) = -2x^1=-2x$

$\frac{d}{d x} (3 x) = 3 x^{0} = 3$ $d/dx(3x)=3x^0=3$

so

$f'(x)=6x^2-2x+3$

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