# How do you find the derivative of f(x)=5x^3+12x^2-15?

$f ' \left(x\right) = 15 {x}^{2} + 24 x$

#### Explanation:

Given function:

$f \left(x\right) = 5 {x}^{3} + 12 {x}^{2} - 15$

differentiating above function w.r.t. $x$ as follows

$\setminus \frac{d}{\mathrm{dx}} f \left(x\right) = \setminus \frac{d}{\mathrm{dx}} \left(5 {x}^{3} + 12 {x}^{2} - 15\right)$

$f ' \left(x\right) = \setminus \frac{d}{\mathrm{dx}} \left(5 {x}^{3}\right) + \setminus \frac{d}{\mathrm{dx}} \left(12 {x}^{2}\right) - \setminus \frac{d}{\mathrm{dx}} \left(15\right)$

$= 5 \setminus \frac{d}{\mathrm{dx}} \left({x}^{3}\right) + 12 \setminus \frac{d}{\mathrm{dx}} \left({x}^{2}\right) - 0$

$= 5 \left(3 {x}^{2}\right) + 12 \left(2 x\right)$

$= 15 {x}^{2} + 24 x$

Jul 12, 2018

$f ' \left(x\right) = 15 {x}^{2} + 24 x$

#### Explanation:

Whenever we're trying to differentiate a polynomial, it helps to use the power rule.

In essence, with the power rule, the exponent becomes the coefficient, and the power is decremented by one. We get

$f ' \left(x\right) = 15 {x}^{2} + 24 x$

Recall that the derivative of a constant is zero.

Hope this helps!