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# The Product Rule and the Quotient Rule

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## The Product Rule and the Quotient Rule

The product rule is used when differentiating two functions that are being multiplied together. In some cases it will be possible to simply multiply them out.

**Example:
**

Differentiate y = x^{2}(x^{2} + 2x − 3).

Here y = x^{4} + 2x^{3} − 3x^{2} and so:

However functions like y = 2x(x^{2} + 1)^{5} and y = xe^{3x} are either more difficult or impossible to expand and so we need a new technique.

**The product rule states that for two functions, u and v.** If y = uv, then:

**For our example:**

y = 2x(x^{2} − 1)^{5}

u = 2x

v = (x^{2} − 1)^{5}

Therefore:

After factorising:

**Note:** After using the product rule you will normally be able to factorise the derivative and then you can find the stationary points.

**For our second example:**

y = xe^{3x}, find the turning point and sketch the graph.

u = x

v = e^{3x}

Therefore:

This means there is a stationary point when x = -1/3 (e^{3x} ≠ 0).

Also, when x = -1/3, y = -e^{-1}/3 = -0.123 (3sf).

By using the second derivative, which we find by using the product rule again, we can determine whether this is a maximum or a minimum.

when x = -1/3

Therefore there is a minimum at (-1/3, -0.123)

**To sketch the graph we know that:**

- When x = 0, y = 0
- There is a minimum at (-1/3, -0.123)
- As x → ∞, y → ∞
- As x → −∞, y → 0 (and is negative)

Therefore the graph looks like this:

The **quotient rule** is actually the product rule in disguise and is used when differentiating a fraction.

**The quotient rule states that for two functions, u and v, **

(See if you can use the product rule and the chain rule on y = uv^{-1} to derive this formula.)

**Example:**

Differentiate

Solution: