2 Differentiating a quotient

In this Section we consider functions of the form y = f ( x ) g ( x ) . To find the derivative of such a function we make use of the following Key Point:

Key Point 10

Quotient Rule

If y = f ( x ) g ( x ) then d y d x = g ( x ) d f d x − d g d x f ( x ) [ g ( x ) ] 2 ( or y ′ = g f ′ − g ′ f g 2 )

If y = u v  then d y d x = v d u d x − d v d x u v 2 ( or y ′ = v u ′ − v ′ u v 2 )

These are two equivalent versions of the quotient rule .

Example 10

Find the derivative of y = ln x x

Solution

Here f ( x ) = ln x and g ( x ) = x

∴ d f d x = 1 x and d g d x = 1

Hence d y d x = x 1 x − 1 ( ln x ) [ x ] 2 = 1 − ln x x 2

Task!

Obtain the derivative of y = sin x x 2

  1. using the formula for differentiating a product and
  2. using the formula for differentiating a quotient.
  1. Write y = x − 2 sin x then use the product rule to find d y d x :

    y = x − 2 sin x ∴ d y d x = ( − 2 x − 3 ) sin x + x − 2 cos x ∴ d y d x = − 2 sin x + x cos x x 3

  2. Now use the quotient rule instead to find d y d x :

    y = sin x x 2 ∴ d y d x = x 2 ( cos x ) − ( 2 x ) sin x ( x 2 ) 2 = x cos x − 2 sin x x 3

Exercise

Find the derivatives of the following:

  1. ( 2 x 3 − 4 x 2 ) ( 3 x 5 + x 2 )
  2. 2 x 3 + 4 x 2 − 4 x + 1
  3. x 2 + 2 x + 1 x 2 − 2 x + 1
  4. ( x 2 + 3 ) ( 2 x − 5 ) ( 3 x + 2 )
  5. ( 2 x + 1 ) ( 3 x − 1 ) x + 5
  6. ( ln x ) sin x
  7. ( ln x ) ∕ sin x
  8. e x ∕ x 2
  9. e x sin x cos 2 x
  1. 48 x 7 − 84 x 6 + 10 x 4 − 16 x 3
  2. 2 x 4 − 16 x 3 + 6 x 2 − 8 x + 16 ( x 2 − 4 x + 1 ) 2
  3. − 4 ( x + 1 ) ( x − 1 ) 3
  4. 24 x 3 − 33 x 2 + 16 x − 33
  5. 6 ( x 2 + 10 x + 1 ) ( x + 5 ) 2
  6. 1 x sin x + ( ln x ) cos x
  7. sin x 1 x − ( ln x ) cos x sin 2 x = cosec x ( 1 x − cot x ln x )
  8. x 2 e x − 2 x e x x 4 = ( x − 2 − 2 x − 3 ) e x
  9. cos 2 x ( e x sin x + e x cos x ) + 2 sin 2 x e x sin x cos 2 2 x

    = e x [ ( sin x + cos x ) sec 2 x + 2 sin x sin 2 x sec 2 2 x ]