Differentiate products and quotients of functions using the product and quotient rules

Calculus – Differentiating Products and Quotients (IGCSE Additional Mathematics 0606)

1. What a derived function is (Syllabus 14.1)

• A derived function (or derivative) of \(f(x)\) gives the *instantaneous* rate of change of \(f\) with respect to the variable \(x\).
• In the IGCSE syllabus a full limit definition is not required – you only need the intuitive idea “gradient of the curve at a point”.

2. Common notations (Syllabus 14.2)

• \(f'(x)\) or \(\displaystyle\frac{df}{dx}\) – the notation used most often in the exam.
• \(\delta y/\delta x\) – used for small‑increment problems.
• \(dy\) and \(dx\) – the differential symbols that appear in the table of symbols; they are useful when working with \(\delta y\approx y'(x)\,\delta x\).

3. Standard derivatives (required for the rules – Syllabus 14.3)

Keep this table handy when applying the product or quotient rule.

4. Product rule (Syllabus 14.4)

Derivation (brief)

1. Write \(y=u\,v\).
2. For a small change \(\Delta x\): \(\Delta y = u(x+\Delta x)v(x+\Delta x)-u(x)v(x)\).
3. Expand, discard terms containing \((\Delta x)^{2}\), divide by \(\Delta x\) and let \(\Delta x\to0\).
4. The limit gives the product rule above.

When to use it

• Both \(u\) and \(v\) must be differentiable on the interval considered.
• If more than two factors are present, apply the rule repeatedly (e.g. \(\frac{d}{dx}(abc)=a'b c + a b'c + a b c'\)).
• Sometimes it is quicker to rewrite a quotient as a product with a negative exponent and then use the product rule (see the tip under the Quotient rule).

Example 1 – Simple product

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

1. \(u(x)=3x^{2}+2\;\Rightarrow\;u'(x)=6x\).
2. \(v(x)=\sin x\;\Rightarrow\;v'(x)=\cos x\).
3. Apply the rule: \(\displaystyle y'=6x\sin x+(3x^{2}+2)\cos x\).
4. Gradient at \(x=\frac{\pi}{4}\):

   \[

   y'\!\left(\tfrac{\pi}{4}\right)=6\!\left(\tfrac{\pi}{4}\right)\!\sin\!\tfrac{\pi}{4}

   +\bigl(3(\tfrac{\pi}{4})^{2}+2\bigr)\!\cos\!\tfrac{\pi}{4}.

   \]

Example 2 – Three factors

Differentiate \(y=x^{2}\,e^{x}\,\ln x\).

1. First treat \(x^{2}e^{x}\) as a single product:

   \((x^{2}e^{x})' = 2x e^{x}+x^{2}e^{x}\).

2. Now apply the product rule with \(u=(x^{2}e^{x})\) and \(v=\ln x\):

   \[

   y'=(2x e^{x}+x^{2}e^{x})\ln x + x^{2}e^{x}\cdot\frac{1}{x}

   =(2x e^{x}+x^{2}e^{x})\ln x + x e^{x}.

   \]

Alternative method – rewrite a quotient as a product

Example: \(y=\dfrac{\sin x}{x^{3}} = \sin x\;x^{-3}\).

• \(y' = (\sin x)'\,x^{-3} + \sin x\,(x^{-3})'\).
• Using the table, \((\sin x)'=\cos x\) and \((x^{-3})'=-3x^{-4}\).
• Hence \(\displaystyle y'=\frac{\cos x}{x^{3}}-\frac{3\sin x}{x^{4}}\).

5. Quotient rule (Syllabus 14.4)

Derivation (brief)

1. Write \(y=\dfrac{u}{v}\;\Rightarrow\;u=yv\).
2. Differentiate the right‑hand side using the product rule: \(u'=y'v+yv'\).
3. Solve for \(y'\): \(\displaystyle y'=\frac{u'v-uv'}{v^{2}}\).

When to use it

• Both numerator \(u\) and denominator \(v\) must be differentiable.
• The denominator must never be zero in the domain of the question.
• Tip: If rewriting \(\dfrac{u}{v}=u\,v^{-1}\) makes the algebra simpler, do so and apply the product rule together with the power rule.

Example 3 – Algebraic quotient

Find \(y'\) for \(y=\dfrac{x^{3}}{\sqrt{x+1}}\).

1. \(u=x^{3},\;u'=3x^{2}\); \(v=(x+1)^{1/2},\;v'=\tfrac12(x+1)^{-1/2}\).
2. Apply the rule:

   \[

   y'=\frac{3x^{2}(x+1)^{1/2}-x^{3}\bigl[\tfrac12(x+1)^{-1/2}\bigr]}{(x+1)}.

   \]

3. Simplify:

   \[

   y'=\frac{6x^{2}(x+1)-x^{3}}{2\,(x+1)^{3/2}}.

   \]

Example 4 – Trigonometric quotient

Differentiate \(y=\dfrac{\tan x}{x^{2}}\).

1. \(u=\tan x,\;u'=\sec^{2}x\); \(v=x^{2},\;v'=2x\).
2. Quotient rule:

   \[

   y'=\frac{\sec^{2}x\;x^{2}-\tan x\;2x}{x^{4}}

   =\frac{x^{2}\sec^{2}x-2x\tan x}{x^{4}}.

   \]

3. Reduce:

   \[

   y'=\frac{\sec^{2}x}{x^{2}}-\frac{2\tan x}{x^{3}}.

   \]

Normal line (Syllabus 14.5)

If the tangent at \((x{0},y{0})\) has gradient \(m=y'(x_{0})\), the normal line has gradient \(-\dfrac{1}{m}\) and its equation is

\[

y-y{0}=-\frac{1}{\,y'(x{0})\,}\,(x-x_{0}).

\]

6. Stationary points (Syllabus 14.6 & 14.9)

A stationary point occurs where the derivative is zero: \(y'(x)=0\). After finding the points, classify them using either:

• First‑derivative test – examine the sign of \(y'\) on either side of the point.
• Second‑derivative test – compute \(y''\); if \(y''>0\) the point is a minimum, if \(y''<0\) it is a maximum, and if \(y''=0\) the test is inconclusive.

Example 5 – Using the second‑derivative test

Let \(y = x\,e^{-x}\).

1. Product rule: \(y' = e^{-x}+x(-e^{-x}) = e^{-x}(1-x).\)
2. Stationary point: set \(y'=0\Rightarrow 1-x=0\Rightarrow x=1.\)
3. Second derivative:

   \[

   y'' = \frac{d}{dx}\bigl[e^{-x}(1-x)\bigr]

   = -e^{-x}(1-x)-e^{-x}=e^{-x}(x-2).

   \]

   At \(x=1\), \(y''=e^{-1}(1-2)=-e^{-1}<0\), so the point \((1,e^{-1})\) is a maximum.

7. Rates of change & small increments (Syllabus 14.7)

When a quantity is the product or quotient of two time‑dependent quantities, the instantaneous rate of change follows directly from the product or quotient rule.

• Product example: If \(A(t)\) and \(B(t)\) are lengths, the area \(S(t)=A(t)B(t)\) changes at

  \[

  \frac{dS}{dt}=A'(t)B(t)+A(t)B'(t).

  \]

• Quotient example: For a volume \(V(t)=\dfrac{M(t)}{\rho(t)}\),

  \[

  \frac{dV}{dt}= \frac{M'(t)\,\rho(t)-M(t)\,\rho'(t)}{\rho(t)^{2}}.

  \]

• Small‑increment (linear) approximation: With a small change \(\delta x\),

  \[

  \delta y \approx y'(x)\,\delta x.

  \]

  Combine this with the product or quotient rule to estimate \(\delta y\) for complicated expressions.

8. Exam‑style checklist (Syllabus 14.4–14.7)

• Identify clearly which functions are \(u\) (numerator) and \(v\) (denominator) before differentiating.
• Check that any denominator is non‑zero in the domain of the question.
• After applying a rule, simplify: factor common terms, reduce powers, combine like terms – this often gains marks.
• If the algebra looks messy, consider rewriting a quotient as a product with a negative exponent.
• For tangent or normal lines, use the formulas

  \[

  y-y{0}=y'(x{0})(x-x_{0}),\qquad

  y-y{0}=-\frac{1}{y'(x{0})}(x-x_{0}).

  \]

• When asked for stationary points, follow the three‑step process:

  1. Find \(y'\) (product/quotient rule as required).
  2. Solve \(y'=0\) for \(x\).
  3. Classify using the first‑ or second‑derivative test.

• For rate‑of‑change questions, write the physical quantities as functions of time, differentiate with the appropriate rule, and substitute the given values.