# How do you find the derivative of Y= x^2 ( 2x + 3 )?

Aug 4, 2015

#### Answer:

We get $y ' = 6 {x}^{2} + 6 x$

#### Explanation:

Method 1 Use the product ruel then simplify.

The product rule tells us that the derivative of a product of two function (I think of them as the First and the Second) is given by:

$\frac{d}{\mathrm{dx}} \left(F S\right) = F ' S + F S ' \text{ }$

(Because both addition and multiplication of functions are commutative, other orders are possible.)

So we get (including detail you might prefer to leave out eventually)

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

$y ' = \left[\frac{d}{\mathrm{dx}} \left({x}^{2}\right)\right] \left(2 x + 3\right) + {x}^{2} \left[\frac{d}{\mathrm{dx}} \left(2 x + 3\right)\right] \text{ }$

(usually we'll omit writing this step, but we need to DO this)

$y ' = \left[2 x\right] \left(2 x + 3\right) + {x}^{2} \left[2\right]$

$= 4 {x}^{2} + 6 x + 2 {x}^{2}$

$= 6 {x}^{2} + 6 x$

Method 2 Multiply first, the differentiate.

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

$y = 2 {x}^{3} + 3 {x}^{2} \text{ }$ (by algebra)

Now we do not need the product rule, just the sum and power and constant multiple ruel)

$y ' = 6 {x}^{2} + 6 x$

Two lessons:

We can use either method to get to the correct answer. (There are many paths to one destination.)

We can take control of how a problem is written. (Unless our tester has told us we must use a particular method -- that is sometimes done to test our knowledge of that method.)