# How do you find the derivative of  f(x)=x^3-5x^2+2x+8?

Jun 30, 2016

Use linearity of differentiation and the power rule to get $f ' \left(x\right) = 3 {x}^{2} - 10 x + 2$.

#### Explanation:

The power rule says that $\frac{d}{\mathrm{dx}} \left({x}^{n}\right) = n {x}^{n - 1}$ for any $n$. Therefore, $\frac{d}{\mathrm{dx}} \left({x}^{3}\right) = 3 {x}^{2}$, $\frac{d}{\mathrm{dx}} \left({x}^{2}\right) = 2 x$, $\frac{d}{\mathrm{dx}} \left(x\right) = 1 {x}^{0} = 1$, and $\frac{d}{\mathrm{dx}} \left(8\right) = \frac{d}{\mathrm{dx}} \left(8 {x}^{0}\right) = 0$.

Linearity of differentiation says that $\frac{d}{\mathrm{dx}} \left({a}_{1} {f}_{1} \left(x\right) + {a}_{2} {f}_{2} \left(x\right) + \setminus \cdots + {a}_{n} {f}_{n} \left(x\right)\right)$

$= {a}_{1} {f}_{1} ' \left(x\right) + {a}_{2} {f}_{2} ' \left(x\right) + \setminus \cdots + {a}_{n} {f}_{n} ' \left(x\right)$, where
${a}_{1} , {a}_{2} , \setminus \ldots , {a}_{n}$ are constants.

These facts lead to

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left({x}^{3} - 5 {x}^{2} + 2 x + 8\right)$

=d/dx(x^3)-5d/dx(x^2)+2d/dx(x)+d/dx(8) =3x^2-10x+2.