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AI HL · AHL 2

Further Derivatives

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AHL 2.4

The product and quotient rules

AHL 2.4

The product and quotient rules

Target: I can differentiate a product of two functions with the product rule, \(u'v+uv'\), including when one factor needs the chain rule. (+4 more)

By the end of this section

Key points

Why a product needs its own rule

  • You cannot differentiate a product or a quotient factor by factor. \(\dfrac{d}{dx}(x^2\ln x)\) is not \(2x\times\dfrac1x\).
  • Expanding first works for some products, such as \(x^2(2x+1)\). It does not work for \(x^2\ln x\).
  • Check a product or quotient answer with a point: put a value of x in your derivative and compare with the calculator's gradient.

The product rule

  • If \(y=uv\), then \(\dfrac{dy}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}\). In short: \(\dfrac{d}{dx}(uv)=u'v+uv'\).
  • Write u, v, \(u'\), \(v'\) down first. Then substitute.
  • If a factor is a function of a function, such as \((3x+1)^4\), \(\sin 4x\), \(e^{5x-2}\) or \(\ln(2x+5)\), use the chain rule for its derivative.
  • With three factors, differentiate one at a time and keep the other two: \((uvw)'=u'vw+uv'w+uvw'\).
  • Factorise your answer when you can. It makes stationary points and signs easy to read.

The quotient rule

  • If \(y=\dfrac{u}{v}\), then \(\dfrac{dy}{dx}=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{v^2}\). In short: \(\left(\dfrac uv\right)'=\dfrac{u'v-uv'}{v^2}\).
  • The order on top matters: \(u'v\) first, then subtract \(uv'\).
  • When v is a power of a bracket, a factor of the bracket cancels from top and bottom. Cancel it before you expand.
  • For a root in the denominator, write it as a power. To remove a negative power, multiply top and bottom by the root.
  • Derivatives of trig functions follow: \(\dfrac{d}{dx}\tan x=\dfrac{1}{\cos^2x}\), found by writing \(\tan x=\dfrac{\sin x}{\cos x}\).

Using the derivative

  • The gradient at a point is \(f'(a)\). To find where the gradient has a given value, solve \(f'(x)=\) that value.
  • Stationary points: solve \(f'(x)=0\). \(e^{kx}\) is never 0, so only the other factor can be 0.
  • f is increasing where \(f'(x)>0\). Factorise \(f'\) and find its sign. A square is never negative.
  • A tangent parallel to a line has the same gradient as the line. Check each answer is on the curve's domain, for example \(x^2-5>0\) for \(\ln(x^2-5)\).
  • Trig derivatives use radians.

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
A power of x times sin x

Differentiate \(y=x^3\sin x\). Then find the gradient of the curve at \(x=1\) (x in radians).

Working
  1. Name the parts

    \(u=x^3,\; v=\sin x\)

    Pick u and v, then write down both derivatives before you use the rule.

  2. Their derivatives

    \(u'=3x^2,\; v'=\cos x\)

    Each is a rule you already know.

  3. The product rule

    \(\dfrac{dy}{dx}=u'v+uv'=3x^2\sin x+x^3\cos x\)

    Never multiply the two derivatives together. The rule adds two terms.

  4. Put in x = 1

    From the calculator (radians): \(3\sin 1+\cos 1=3.06\)

    Use radians for calculus with trig functions.

Video walkthrough coming
A product with a power of a linear factor

Differentiate \(f(x)=x^2(3x+1)^4\), giving your answer in factorised form. Then find \(f'(1)\).

Working
  1. Name the parts

    \(u=x^2,\; v=(3x+1)^4\)

    One factor is a function of a function.

  2. Derivatives

    \(u'=2x,\; v'=4(3x+1)^3\times 3=12(3x+1)^3\)

    The chain rule is for v′. Multiply by 3, the derivative of the bracket.

  3. The product rule

    \(f'(x)=2x(3x+1)^4+12x^2(3x+1)^3\)

    This is \(u'v+uv'\).

  4. Factorise

    \(f'(x)=2x(3x+1)^3\,[(3x+1)+6x]=2x(9x+1)(3x+1)^3\)

    Take out the common factor \(2x(3x+1)^3\).

  5. Evaluate

    \(f'(1)=2(10)(4^3)=1280\)

    Use the factorised form: the numbers are smaller.

Video walkthrough coming
A quotient simplified to one power

Differentiate \(y=\dfrac{x^2+4x-3}{(x+2)^2}\), giving your answer in the form \(\dfrac{a}{(x+2)^b}\), where a and b are integers. Write down a.

Working
  1. Name the parts

    \(u=x^2+4x-3,\; v=(x+2)^2\)

    u is the top and v is the bottom.

  2. Derivatives

    \(u'=2x+4,\; v'=2(x+2)\)

    The chain rule gives v′, and the derivative of \(x+2\) is 1.

  3. The quotient rule

    \(\dfrac{dy}{dx}=\dfrac{u'v-uv'}{v^2}=\dfrac{(2x+4)(x+2)^2-2(x+2)(x^2+4x-3)}{(x+2)^4}\)

    Write the rule with u′v first. The order matters in the subtraction.

  4. Cancel (x + 2)

    \(=\dfrac{(2x+4)(x+2)-2(x^2+4x-3)}{(x+2)^3}\)

    Every term on top has a factor \((x+2)\). Divide it out of top and bottom.

  5. Expand the top

    \(=\dfrac{2x^2+8x+8-2x^2-8x+6}{(x+2)^3}=\dfrac{14}{(x+2)^3}\)

    The x terms cancel. So \(a=14\) and \(b=3\).

Video walkthrough coming
A quotient with a root in the denominator

Find the gradient of the curve \(y=\dfrac{x+1}{\sqrt{2x+3}}\) at \(x=3\).

Working
  1. Write the root as a power

    \(u=x+1,\; v=(2x+3)^{1/2}\)

    A power is easier to differentiate than a root.

  2. Derivatives

    \(u'=1,\; v'=\tfrac12(2x+3)^{-1/2}\times 2=(2x+3)^{-1/2}\)

    The chain rule gives a factor 2.

  3. The quotient rule

    \(\dfrac{dy}{dx}=\dfrac{(2x+3)^{1/2}-(x+1)(2x+3)^{-1/2}}{2x+3}\)

    v squared is \(2x+3\).

  4. Tidy: multiply top and bottom by the root

    \(=\dfrac{(2x+3)-(x+1)}{(2x+3)^{3/2}}=\dfrac{x+2}{(2x+3)^{3/2}}\)

    This removes the negative power.

  5. Put in x = 3

    \(\dfrac{5}{9^{3/2}}=\dfrac{5}{27}=0.185\)

    \(9^{3/2}=27\).

Video walkthrough coming
Deriving the derivative of tan x

Use the quotient rule to show that if \(f(x)=\tan x\), then \(f'(x)=\dfrac{1}{\cos^2 x}\). Then find \(f'\!\left(\dfrac{\pi}{4}\right)\).

Working
  1. Write tan as a quotient

    \(\tan x=\dfrac{\sin x}{\cos x}\)

    Now it is a quotient of two functions you can differentiate.

  2. Parts and derivatives

    \(u=\sin x,\; v=\cos x,\; u'=\cos x,\; v'=-\sin x\)

    Watch the sign of v′.

  3. The quotient rule

    \(f'(x)=\dfrac{\cos x\cos x-\sin x(-\sin x)}{\cos^2x}=\dfrac{\cos^2x+\sin^2x}{\cos^2x}\)

    Minus a negative gives a plus.

  4. Use sin² + cos² = 1

    \(f'(x)=\dfrac{1}{\cos^2x}\)

    This is the result you were asked to show.

  5. Evaluate

    \(f'\!\left(\tfrac{\pi}{4}\right)=\dfrac{1}{(1/\sqrt2)^2}=2\)

    \(\cos\frac{\pi}{4}=\frac{1}{\sqrt2}\), and squaring gives \(\frac12\).

Video walkthrough coming
Where the gradient is zero

Find the coordinates of the point on the curve \(y=3xe^{-2x}\) where the gradient is zero. Give the y-coordinate.

Working
  1. Product rule

    \(u=3x,\; v=e^{-2x}\;\Rightarrow\; u'=3,\; v'=-2e^{-2x}\)

    The chain rule gives the \(-2\) in v′.

  2. The gradient

    \(\dfrac{dy}{dx}=3e^{-2x}-6xe^{-2x}=3e^{-2x}(1-2x)\)

    Factorise: it makes the zero easy to see.

  3. Set it to zero

    \(3e^{-2x}(1-2x)=0\)

    \(e^{-2x}\) is never 0, so only the bracket can be 0.

  4. Solve

    \(1-2x=0\;\Rightarrow\; x=0.5\)

    Now find y.

  5. The point

    \(y=3(0.5)e^{-1}=\dfrac{3}{2e}\), so \((0.5,\;0.552)\)

    Give the point as coordinates.

Video walkthrough coming
Showing a function is increasing

Show that \(f(x)=\dfrac{x^2}{x-2}\) is an increasing function for \(x>a\) and for \(x<b\), where a and b are constants to be found. Write down the larger of the two constants.

Working
  1. Differentiate

    \(f'(x)=\dfrac{2x(x-2)-x^2(1)}{(x-2)^2}=\dfrac{x^2-4x}{(x-2)^2}\)

    Quotient rule with \(u=x^2\), \(v=x-2\).

  2. Factorise

    \(f'(x)=\dfrac{x(x-4)}{(x-2)^2}\)

    Factorising shows the sign.

  3. The sign

    \((x-2)^2>0\), so the sign of \(f'\) is the sign of \(x(x-4)\)

    A square is positive for every x except 2, where f is not defined.

  4. Increasing means f′ > 0

    \(x(x-4)>0\) when \(x>4\) or \(x<0\)

    The product of two factors is positive when both are positive or both negative. So \(a=4\) and \(b=0\).

Video walkthrough coming