Mastering the Chain, Product, and Quotient Rules for Differentiation

What the examiner is testing

The examiner is assessing your ability to correctly identify the structure of a function and apply the appropriate differentiation rule (chain, product, or quotient). Marks are awarded for setting up the correct rule, accurately differentiating the component parts, and simplifying the final expression.

The method

1. Identify the function's structure: Determine if the function is a product of two functions, a quotient of two functions, or a function nested inside another function.
2. Assign variables: For product rule, let \(u\) and \(v\) be the two functions being multiplied. For quotient rule, let \(u\) be the numerator and \(v\) be the denominator. For chain rule, let \(u\) be the inner function and \(y\) be a function of \(u\).
3. Differentiate the component parts: Find \(\frac{du}{dx}\) and \(\frac{dv}{dx}\) (for product/quotient) or \(\frac{du}{dx}\) and \(\frac{dy}{du}\) (for chain).
4. Substitute into the formula:
   • Product Rule: \(\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}\)
   • Quotient Rule: \(\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}\)
   • Chain Rule: \(\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}\)
5. Simplify the expression: Combine like terms and factorise where possible to present the derivative in its simplest form.

Worked example

Differentiate \(y = (3x^2 + 5)^4\) with respect to \(x\).

We identify this as a chain rule problem, where one function is nested inside another.

Let \(u = 3x^2 + 5\).
Then \(y = u^4\).

Step 1: Differentiate \(u\) with respect to \(x\).
$$ \frac{du}{dx} = \frac{d}{dx}(3x^2 + 5) = 6x $$

Step 2: Differentiate \(y\) with respect to \(u\).
$$ \frac{dy}{du} = \frac{d}{du}(u^4) = 4u^3 $$

Step 3: Apply the chain rule formula \(\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}\).
$$ \frac{dy}{dx} = (4u^3) \times (6x) $$

Step 4: Substitute \(u = 3x^2 + 5\) back into the expression.
$$ \frac{dy}{dx} = 4(3x^2 + 5)^3 \times 6x $$

Step 5: Simplify the expression.
$$ \frac{dy}{dx} = 24x(3x^2 + 5)^3 $$

Sanity check: The original function is a polynomial raised to a power. The derivative should have a lower power for the outer function and include the derivative of the inner function. Our result matches this structure.

Worked example: a harder one

A curve has the equation \(y = \frac{\sin(2x)}{e^{3x}}\). Find \(\frac{dy}{dx}\).

The obvious first move is to apply the quotient rule.
Let \(u = \sin(2x)\) and \(v = e^{3x}\).

First, we need \(\frac{du}{dx}\) and \(\frac{dv}{dx}\). These both require the chain rule.
For \(u = \sin(2x)\):
Let \(w = 2x\), so \(u = \sin(w)\).
\(\frac{dw}{dx} = 2\)
\(\frac{du}{dw} = \cos(w)\)
So, \(\frac{du}{dx} = \frac{du}{dw} \times \frac{dw}{dx} = \cos(2x) \times 2 = 2\cos(2x)\).

For \(v = e^{3x}\):
Let \(z = 3x\), so \(v = e^z\).
\(\frac{dz}{dx} = 3\)
\(\frac{dv}{dz} = e^z\)
So, \(\frac{dv}{dx} = \frac{dv}{dz} \times \frac{dz}{dx} = e^{3x} \times 3 = 3e^{3x}\).

Now, apply the quotient rule: \(\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}\).
$$ \frac{dy}{dx} = \frac{e^{3x}(2\cos(2x)) - \sin(2x)(3e^{3x})}{(e^{3x})^2} $$

Simplify the expression:
$$ \frac{dy}{dx} = \frac{2e^{3x}\cos(2x) - 3e^{3x}\sin(2x)}{e^{6x}} $$
Factor out \(e^{3x}\) from the numerator:
$$ \frac{dy}{dx} = \frac{e^{3x}(2\cos(2x) - 3\sin(2x))}{e^{6x}} $$
Cancel \(e^{3x}\) with a term in the denominator:
$$ \frac{dy}{dx} = \frac{2\cos(2x) - 3\sin(2x)}{e^{3x}} $$

This problem could also be approached using the product rule by rewriting \(y = \sin(2x)e^{-3x}\).
Let \(u = \sin(2x)\) and \(v = e^{-3x}\).
\(\frac{du}{dx} = 2\cos(2x)\) (as calculated before).
For \(v = e^{-3x}\):
Let \(z = -3x\), so \(v = e^z\).
\(\frac{dz}{dx} = -3\)
\(\frac{dv}{dz} = e^z\)
So, \(\frac{dv}{dx} = e^{-3x} \times (-3) = -3e^{-3x}\).

Apply the product rule: \(\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}\).
$$ \frac{dy}{dx} = \sin(2x)(-3e^{-3x}) + e^{-3x}(2\cos(2x)) $$
$$ \frac{dy}{dx} = -3e^{-3x}\sin(2x) + 2e^{-3x}\cos(2x) $$
Factor out \(e^{-3x}\):
$$ \frac{dy}{dx} = e^{-3x}(2\cos(2x) - 3\sin(2x)) $$
This is equivalent to the quotient rule result. The product rule approach often leads to fewer algebraic steps for this type of function.

Practice

1. Differentiate \(y = (5x^3 - 2x)^6\) with respect to \(x\).
2. Find \(\frac{dy}{dx}\) for \(y = x^2 \cos(4x)\).
3. Given \(y = \frac{e^{2x}}{x^3}\), find \(\frac{dy}{dx}\).
4. A curve has the equation \(y = \ln(\sec(x))\). Show that \(\frac{dy}{dx} = \tan(x)\).

Answers:
1. \(\frac{dy}{dx} = 6(5x^3 - 2x)^5 (15x^2 - 2)\)
2. \(\frac{dy}{dx} = 2x\cos(4x) - 4x^2\sin(4x)\)
3. \(\frac{dy}{dx} = \frac{e^{2x}(2x - 3)}{x^4}\)
4. For \(y = \ln(\sec(x))\):
Let \(u = \sec(x)\). Then \(y = \ln(u)\).
\(\frac{du}{dx} = \sec(x)\tan(x)\)
\(\frac{dy}{du} = \frac{1}{u}\)
Using the chain rule: \(\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}\)
$$ \frac{dy}{dx} = \frac{1}{u} \times \sec(x)\tan(x) $$
Substitute \(u = \sec(x)\):
$$ \frac{dy}{dx} = \frac{1}{\sec(x)} \times \sec(x)\tan(x) $$
$$ \frac{dy}{dx} = \tan(x) $$

The three mistakes that lose marks

1. Incorrectly applying the quotient rule numerator: A common error is to swap the terms in the numerator, leading to \(\frac{u\frac{dv}{dx} - v\frac{du}{dx}}{v^2}\) instead of \(\frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}\). This results in a sign error in the final answer. For example, differentiating \(\frac{x}{e^x}\) might yield \(\frac{e^x - xe^x}{(e^x)^2}\) (correct) but the error would give \(\frac{xe^x - e^x}{(e^x)^2}\).
2. Forgetting the chain rule within product/quotient rule: When \(u\) or \(v\) (or both) are composite functions, students often forget to apply the chain rule when finding \(\frac{du}{dx}\) or \(\frac{dv}{dx}\). For example, differentiating \(x^2\sin(3x)\) might lead to \(\frac{d}{dx}(\sin(3x)) = \cos(3x)\) instead of the correct \(3\cos(3x)\).
3. Algebraic simplification errors: After correctly applying the rules, mistakes are made in factoring, expanding, or cancelling terms. For instance, differentiating \(\frac{e^{2x}}{x}\) yields \(\frac{2xe^{2x} - e^{2x}}{x^2}\), but incorrect simplification might lead to \(\frac{2x - 1}{x}\) by cancelling \(e^{2x}\) from only one term in the numerator.

30-second recap

The chain rule handles functions within functions, multiplying the derivative of the outer function by the derivative of the inner. The product rule differentiates two multiplied functions: \(u'v + uv'\). The quotient rule differentiates one function divided by another: \(\frac{u'v - uv'}{v^2}\).