all slides © Christine Crisp 4: The Quotient Rule all slides © Christine Crisp via TES

The following are examples of quotients: (b) (a) (c) (d) (c) can be divided out to form a simple function as there is a single polynomial term in the denominator. For the others we use the quotient rule.

The quotient rule gives us a way of differentiating functions which are divided. The rule is similar to the product rule. where u and v are functions of x. This rule can be derived from the product rule but it is complicated. If you want to go straight to the examples, click on the box below. Examples

We can develop the quotient rule by using the product rule! The problem now is that this v is not the same as the v of the product rule. That v is replaced by . So, becomes Simplifying Part of the 2nd term, , is the derivative of but with respect to x not v.

We use the chain rule: So, Then, Make the denominators the same by multiplying the numerator and denominator of the 1st term by v. Write with a common denominator:

e.g. 1 Differentiate to find . Solution: and We now need to simplify.

We could simplify the numerator by taking out the common factor x, but it’s easier to multiply out the brackets. We don’t touch the denominator. Multiplying out numerator: Now collect like terms: and factorise: We leave the brackets in the denominator as the factorised form is simpler.

Quotients can always be turned into products. e.g. can be written as However, differentiation is usually more awkward if we do this. In the quotient above, and ( both simple functions ) In the product , and ( v needs the chain rule )

SUMMARY To differentiate a quotient: Check if it is possible to divide out. If so, do it and differentiate each term. Otherwise use the quotient rule: If , where u and v are both functions of x

Exercise Use the quotient rule, where appropriate, to differentiate the following. Try to simplify your answers: 1. 2.

1. Solution: and

2. Solution: Divide out:

   

stop here and try some