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

Aug 10, 2016

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

#### Explanation:

(d f(x))/(d x)=?

$y = a \cdot b \text{ ; } y ' = a ' \cdot b + {b}^{'} \cdot a$

$\frac{d f \left(x\right)}{d x} = 2 x \cdot {\left(x - 5\right)}^{3} + 3 {\left(x - 5\right)}^{2} \cdot 1 \cdot {x}^{2}$

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

Aug 10, 2016

Step 1: Determine the derivative of ${\left(x - 5\right)}^{3}$ using the chain rule

Let $y = {u}^{3}$ and $u = x - 5$. The derivative of $y$ is $y ' = 3 {u}^{2}$ and the derivative of $u$ is $1$. Hence, the derivative of $y = {\left(x - 5\right)}^{3}$ is $y ' = 3 {\left(x - 5\right)}^{2} \times 1 = \textcolor{red}{3 {\left(x - 5\right)}^{2}}$

Step 2: Determine the derivative of $f \left(x\right)$ using the product rule

Let $f \left(x\right) = g \left(x\right) \times h \left(x\right)$. Then our derivative is given by $f ' \left(x\right) = g ' \left(x\right) \times h \left(x\right) + g \left(x\right) \times h ' \left(x\right)$.

The derivative of $g \left(x\right)$ is $2 x$. The derivative of $h \left(x\right)$, as mentioned in step $1$ is $3 {\left(x - 5\right)}^{2}$.

Hence, $f ' \left(x\right) = 2 x \times {\left(x - 5\right)}^{3} + {x}^{2} \times 3 {\left(x - 5\right)}^{2} = 2 x {\left(x - 5\right)}^{3} + 3 {x}^{2} {\left(x - 5\right)}^{2} = {\left(x - 5\right)}^{2} \left(2 x \left(x - 5\right) + 3 {x}^{2}\right) = {\left(x - 5\right)}^{2} \left(2 {x}^{2} - 10 x + 3 {x}^{2}\right) = {\left(x - 5\right)}^{2} \left(5 {x}^{2} - 10 x\right) = \textcolor{b l u e}{5 x {\left(x - 5\right)}^{2} \left(x - 2\right)}$

Hopefully this helps!