A-Level Product, quotient and chain rules Practice Questions

Free A-Level Product, quotient and chain rules practice questions with full step-by-step worked solutions. Covers product-rule, differentiation, chain-rule, quotient-rule. Practise exam-style problems and check your method.

product-ruledifferentiationchain-rulequotient-rule
A-Level70 questionsStep-by-step solutions
Question 1
2 markseasy
Differentiate y=xexy=x e^{x} with respect to xx.
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Worked solution

  1. Label the factors and their derivatives

    u=x, v=ex, dudx=1, dvdx=exu=x,\ v=e^{x},\ \frac{du}{dx}=1,\ \frac{dv}{dx}=e^{x}

    Set up the pieces for the product rule.

  2. Apply the product rule

    dydx=(x)⋅(ex)+(ex)⋅(1)\frac{dy}{dx}=\left(x\right)\cdot \left(e^{x}\right)+\left(e^{x}\right)\cdot \left(1\right)

    Use u times v' plus v times u'.

  3. State the derivative

    dydx=(x+1)ex\frac{dy}{dx}=\left(x + 1\right) e^{x}

    This is the required derivative.

Answer
(x+1)ex\left(x + 1\right) e^{x}
Question 2
2 markseasy
Which method correctly describes how to differentiate y=ex2y=e^{x^{2}}?
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Worked solution

  1. Examine the overall structure of the function

    y=ex2y=e^{x^{2}}

    Look at how the expression is built from simpler parts.

  2. Match the structure to the correct rule

    f(g(x))⇒ chain rulef\left(g(x)\right)\Rightarrow\text{ chain rule}

    The form of the function selects the rule.

  3. Describe the correct method

    Use the chain rule\text{Use the chain rule}

    This matches the structure of the function.

Answer
Use the chain rule
Question 3
3 marksintermediate
Which method correctly describes how to differentiate y=x2+1y=\sqrt{x^{2} + 1}?
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Worked solution

  1. Examine the overall structure of the function

    y=x2+1y=\sqrt{x^{2} + 1}

    Look at how the expression is built from simpler parts.

  2. Describe the structure in words

    a function inside another function\text{a function inside another function}

    Name the way the parts are combined.

  3. Rule out using the power rule on its own

    not a single power of x\text{not a single power of }x

    The power rule alone cannot handle this structure.

  4. Recall what triggers each rule

    uv→product, uv→quotient, f(g)→chainuv\to\text{product},\ \tfrac{u}{v}\to\text{quotient},\ f(g)\to\text{chain}

    Match the observed form to the standard triggers.

  5. Match the structure to the correct rule

    f(g(x))⇒ chain rulef\left(g(x)\right)\Rightarrow\text{ chain rule}

    The form of the function selects the rule.

  6. Describe the correct method

    Use the chain rule\text{Use the chain rule}

    This matches the structure of the function.

Answer
Use the chain rule
Question 4
5 markshard
Which of the following is dydx\frac{dy}{dx} for y=sin⁡(x)xy=\frac{\sin{\left(x \right)}}{x}?
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Worked solution

  1. Write the function and recognise a quotient

    y=sin⁡(x)xy=\frac{\sin{\left(x \right)}}{x}

    It is one function divided by another, so the quotient rule applies.

  2. State the quotient rule

    ddx(uv)=vdudx−udvdxv2\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^{2}}

    The numerator is bottom times derivative of top, minus top times derivative of bottom.

  3. Label numerator and denominator

    u=sin⁡(x),v=xu=\sin{\left(x \right)},\quad v=x

    Identify the top (u) and the bottom (v).

  4. Differentiate the top and the bottom

    dudx=cos⁡(x),dvdx=1\frac{du}{dx}=\cos{\left(x \right)},\quad \frac{dv}{dx}=1

    Differentiate u and v separately.

  5. Apply the quotient rule

    dydx=(x)⋅(cos⁡(x))−(sin⁡(x))⋅(1)(x)2\frac{dy}{dx}=\frac{\left(x\right)\cdot \left(\cos{\left(x \right)}\right)-\left(\sin{\left(x \right)}\right)\cdot \left(1\right)}{\left(x\right)^{2}}

    Substitute the parts into the quotient-rule formula.

  6. Simplify the numerator

    dydx=xcos⁡(x)−sin⁡(x)(x)2\frac{dy}{dx}=\frac{x \cos{\left(x \right)} - \sin{\left(x \right)}}{\left(x\right)^{2}}

    Expand and collect like terms in the numerator.

  7. Write the derivative in simplest form

    dydx=xcos⁡(x)−sin⁡(x)x2\frac{dy}{dx}=\frac{x \cos{\left(x \right)} - \sin{\left(x \right)}}{x^{2}}

    Cancel any common factors to finish.

  8. Compare the result with the given options

    only one option equals dydx\text{only one option equals }\frac{dy}{dx}

    Eliminate the options that do not match.

  9. Evaluate the gradient at x=1 as a check

    dydx∣x=1=−sin⁡(1)+cos⁡(1)\left.\frac{dy}{dx}\right|_{x=1}=- \sin{\left(1 \right)} + \cos{\left(1 \right)}

    Substituting a value gives a quick numerical sanity check.

  10. Select the correct derivative

    dydx=xcos⁡(x)−sin⁡(x)x2\frac{dy}{dx}=\frac{x \cos{\left(x \right)} - \sin{\left(x \right)}}{x^{2}}

    This matches the first option.

Answer
xcos⁡(x)−sin⁡(x)x2\frac{x \cos{\left(x \right)} - \sin{\left(x \right)}}{x^{2}}
Question 5
8 markschallenging
Which of the following is dydx\frac{dy}{dx} for y=(5x−2)4y=\left(5 x - 2\right)^{4}?
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Worked solution

  1. Write the function and spot the composition

    y=(5x−2)4y=\left(5 x - 2\right)^{4}

    There is a function inside another function, so the chain rule applies.

  2. Substitute for the inner function

    u=5x−2 ⇒ y=u4u=5 x - 2\ \Rightarrow\ y=u^{4}

    Let u be the inside; then y is a simple function of u.

  3. State the chain rule

    dydx=dydu⋅dudx\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}

    Differentiate the outer function, then multiply by the derivative of the inner function.

  4. Differentiate the outer function

    dydu=4u3\frac{dy}{du}=4 u^{3}

    Differentiate y with respect to u.

  5. Differentiate the inner function

    dudx=5\frac{du}{dx}=5

    Differentiate u with respect to x.

  6. Apply the chain rule

    dydx=(4(5x−2)3)⋅(5)\frac{dy}{dx}=\left(4 \left(5 x - 2\right)^{3}\right)\cdot \left(5\right)

    Multiply the two derivatives and replace u by the inner function.

  7. Simplify the derivative

    dydx=20(5x−2)3\frac{dy}{dx}=20 \left(5 x - 2\right)^{3}

    Tidy the result into its simplest form.

  8. Compare the result with the given options

    only one option equals dydx\text{only one option equals }\frac{dy}{dx}

    Eliminate the options that do not match.

  9. Evaluate the gradient at x=1 as a check

    dydx∣x=1=540\left.\frac{dy}{dx}\right|_{x=1}=540

    Substituting a value gives a quick numerical sanity check.

  10. Evaluate the gradient at x=2 as a check

    dydx∣x=2=10240\left.\frac{dy}{dx}\right|_{x=2}=10240

    Substituting a value gives a quick numerical sanity check.

  11. Evaluate the gradient at x=3 as a check

    dydx∣x=3=43940\left.\frac{dy}{dx}\right|_{x=3}=43940

    Substituting a value gives a quick numerical sanity check.

  12. Evaluate the gradient at x=4 as a check

    dydx∣x=4=116640\left.\frac{dy}{dx}\right|_{x=4}=116640

    Substituting a value gives a quick numerical sanity check.

  13. Evaluate the gradient at x=5 as a check

    dydx∣x=5=243340\left.\frac{dy}{dx}\right|_{x=5}=243340

    Substituting a value gives a quick numerical sanity check.

  14. Evaluate the gradient at x=-1 as a check

    dydx∣x=−1=−6860\left.\frac{dy}{dx}\right|_{x=-1}=-6860

    Substituting a value gives a quick numerical sanity check.

  15. Select the correct derivative

    dydx=20(5x−2)3\frac{dy}{dx}=20 \left(5 x - 2\right)^{3}

    This matches the first option.

Answer
20(5x−2)320 \left(5 x - 2\right)^{3}

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