Hard A-Level Product, quotient and chain rules Questions

Challenging, exam-style A-Level Product, quotient and chain rules questions with worked solutions. Stretch yourself on the hardest product-rule, differentiation, quotient-rule, chain-rule problems.

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A-Level34 questionsStep-by-step solutions
Question 1
8 markschallenging
Which of the following is dydx\frac{dy}{dx} for y=(5x−2)4y=\left(5 x - 2\right)^{4}?
Show worked solution

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}
Question 2
8 markschallenging
Which of the following is dydx\frac{dy}{dx} for y=x2cos⁡(x)y=x^{2} \cos{\left(x \right)}?
Show worked solution

Worked solution

  1. Write the function and recognise a product

    y=x2cos⁡(x)y=x^{2} \cos{\left(x \right)}

    The right-hand side is a product of two functions of x, so the product rule applies.

  2. State the product rule

    ddx(uv)=udvdx+vdudx\frac{d}{dx}\left(uv\right)=u\frac{dv}{dx}+v\frac{du}{dx}

    The derivative of a product is the first times the derivative of the second, plus the second times the derivative of the first.

  3. Label the two factors

    u=x2,v=cos⁡(x)u=x^{2},\quad v=\cos{\left(x \right)}

    Choosing u and v clearly keeps the working organised.

  4. Differentiate each factor

    dudx=2x,dvdx=−sin⁡(x)\frac{du}{dx}=2 x,\quad \frac{dv}{dx}=- \sin{\left(x \right)}

    Differentiate u and v separately using standard rules.

  5. Apply the product rule

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

    Substitute u, v and their derivatives into the product-rule formula.

  6. Expand the brackets

    dydx=−x2sin⁡(x)+2xcos⁡(x)\frac{dy}{dx}=- x^{2} \sin{\left(x \right)} + 2 x \cos{\left(x \right)}

    Multiply everything out to prepare for collecting like terms.

  7. Simplify to obtain the derivative

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

    Collect like terms and take out common factors to reach the simplest form.

  8. Write the derivative in factored form

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

    A factored form helps to compare with the options.

  9. 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.

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

    dydx∣x=1=−sin⁡(1)+2cos⁡(1)\left.\frac{dy}{dx}\right|_{x=1}=- \sin{\left(1 \right)} + 2 \cos{\left(1 \right)}

    Substituting a value gives a quick numerical sanity check.

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

    dydx∣x=2=42cos⁡(π4+2)\left.\frac{dy}{dx}\right|_{x=2}=4 \sqrt{2} \cos{\left(\frac{\pi}{4} + 2 \right)}

    Substituting a value gives a quick numerical sanity check.

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

    dydx∣x=3=6cos⁡(3)−9sin⁡(3)\left.\frac{dy}{dx}\right|_{x=3}=6 \cos{\left(3 \right)} - 9 \sin{\left(3 \right)}

    Substituting a value gives a quick numerical sanity check.

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

    dydx∣x=4=8cos⁡(4)−16sin⁡(4)\left.\frac{dy}{dx}\right|_{x=4}=8 \cos{\left(4 \right)} - 16 \sin{\left(4 \right)}

    Substituting a value gives a quick numerical sanity check.

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

    dydx∣x=5=10cos⁡(5)−25sin⁡(5)\left.\frac{dy}{dx}\right|_{x=5}=10 \cos{\left(5 \right)} - 25 \sin{\left(5 \right)}

    Substituting a value gives a quick numerical sanity check.

  15. Select the correct derivative

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

    This matches the first option.

Answer
x(−xsin⁡(x)+2cos⁡(x))x \left(- x \sin{\left(x \right)} + 2 \cos{\left(x \right)}\right)
Question 3
8 markschallenging
Which of the following is dydx\frac{dy}{dx} for y=sin⁡(x2)y=\sin{\left(x^{2} \right)}?
Show worked solution

Worked solution

  1. Write the function and spot the composition

    y=sin⁡(x2)y=\sin{\left(x^{2} \right)}

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

  2. Substitute for the inner function

    u=x2 ⇒ y=sin⁡(u)u=x^{2}\ \Rightarrow\ y=\sin{\left(u \right)}

    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=cos⁡(u)\frac{dy}{du}=\cos{\left(u \right)}

    Differentiate y with respect to u.

  5. Differentiate the inner function

    dudx=2x\frac{du}{dx}=2 x

    Differentiate u with respect to x.

  6. Apply the chain rule

    dydx=(cos⁡(x2))⋅(2x)\frac{dy}{dx}=\left(\cos{\left(x^{2} \right)}\right)\cdot \left(2 x\right)

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

  7. Simplify the derivative

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

    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=2cos⁡(1)\left.\frac{dy}{dx}\right|_{x=1}=2 \cos{\left(1 \right)}

    Substituting a value gives a quick numerical sanity check.

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

    dydx∣x=2=4cos⁡(4)\left.\frac{dy}{dx}\right|_{x=2}=4 \cos{\left(4 \right)}

    Substituting a value gives a quick numerical sanity check.

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

    dydx∣x=3=6cos⁡(9)\left.\frac{dy}{dx}\right|_{x=3}=6 \cos{\left(9 \right)}

    Substituting a value gives a quick numerical sanity check.

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

    dydx∣x=4=8cos⁡(16)\left.\frac{dy}{dx}\right|_{x=4}=8 \cos{\left(16 \right)}

    Substituting a value gives a quick numerical sanity check.

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

    dydx∣x=5=10cos⁡(25)\left.\frac{dy}{dx}\right|_{x=5}=10 \cos{\left(25 \right)}

    Substituting a value gives a quick numerical sanity check.

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

    dydx∣x=−1=−2cos⁡(1)\left.\frac{dy}{dx}\right|_{x=-1}=- 2 \cos{\left(1 \right)}

    Substituting a value gives a quick numerical sanity check.

  15. Select the correct derivative

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

    This matches the first option.

Answer
2xcos⁡(x2)2 x \cos{\left(x^{2} \right)}
Question 4
8 markschallenging
Which of the following is dydx\frac{dy}{dx} for y=2x+3x2+1y=\frac{2 x + 3}{x^{2} + 1}?
Show worked solution

Worked solution

  1. Write the function and recognise a quotient

    y=2x+3x2+1y=\frac{2 x + 3}{x^{2} + 1}

    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=2x+3,v=x2+1u=2 x + 3,\quad v=x^{2} + 1

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

  4. Differentiate the top and the bottom

    dudx=2,dvdx=2x\frac{du}{dx}=2,\quad \frac{dv}{dx}=2 x

    Differentiate u and v separately.

  5. Apply the quotient rule

    dydx=(x2+1)⋅(2)−(2x+3)⋅(2x)(x2+1)2\frac{dy}{dx}=\frac{\left(x^{2} + 1\right)\cdot \left(2\right)-\left(2 x + 3\right)\cdot \left(2 x\right)}{\left(x^{2} + 1\right)^{2}}

    Substitute the parts into the quotient-rule formula.

  6. Simplify the numerator

    dydx=−2x2−6x+2(x2+1)2\frac{dy}{dx}=\frac{- 2 x^{2} - 6 x + 2}{\left(x^{2} + 1\right)^{2}}

    Expand and collect like terms in the numerator.

  7. Write the derivative in simplest form

    dydx=2(−x2−3x+1)x4+2x2+1\frac{dy}{dx}=\frac{2 \left(- x^{2} - 3 x + 1\right)}{x^{4} + 2 x^{2} + 1}

    Cancel any common factors to finish.

  8. Write the derivative in factored form

    dydx=−2(x2+3x−1)(x2+1)2\frac{dy}{dx}=- \frac{2 \left(x^{2} + 3 x - 1\right)}{\left(x^{2} + 1\right)^{2}}

    A factored form helps to compare with the options.

  9. 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.

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

    dydx∣x=1=−32\left.\frac{dy}{dx}\right|_{x=1}=- \frac{3}{2}

    Substituting a value gives a quick numerical sanity check.

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

    dydx∣x=2=−1825\left.\frac{dy}{dx}\right|_{x=2}=- \frac{18}{25}

    Substituting a value gives a quick numerical sanity check.

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

    dydx∣x=3=−1750\left.\frac{dy}{dx}\right|_{x=3}=- \frac{17}{50}

    Substituting a value gives a quick numerical sanity check.

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

    dydx∣x=4=−54289\left.\frac{dy}{dx}\right|_{x=4}=- \frac{54}{289}

    Substituting a value gives a quick numerical sanity check.

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

    dydx∣x=5=−326\left.\frac{dy}{dx}\right|_{x=5}=- \frac{3}{26}

    Substituting a value gives a quick numerical sanity check.

  15. Select the correct derivative

    dydx=2(−x2−3x+1)x4+2x2+1\frac{dy}{dx}=\frac{2 \left(- x^{2} - 3 x + 1\right)}{x^{4} + 2 x^{2} + 1}

    This matches the first option.

Answer
2(−x2−3x+1)x4+2x2+1\frac{2 \left(- x^{2} - 3 x + 1\right)}{x^{4} + 2 x^{2} + 1}
Question 5
8 markschallenging
Which of the following is dydx\frac{dy}{dx} for y=(x2+1)exy=\left(x^{2} + 1\right) e^{x}?
Show worked solution

Worked solution

  1. Write the function and recognise a product

    y=(x2+1)exy=\left(x^{2} + 1\right) e^{x}

    The right-hand side is a product of two functions of x, so the product rule applies.

  2. State the product rule

    ddx(uv)=udvdx+vdudx\frac{d}{dx}\left(uv\right)=u\frac{dv}{dx}+v\frac{du}{dx}

    The derivative of a product is the first times the derivative of the second, plus the second times the derivative of the first.

  3. Label the two factors

    u=x2+1,v=exu=x^{2} + 1,\quad v=e^{x}

    Choosing u and v clearly keeps the working organised.

  4. Differentiate each factor

    dudx=2x,dvdx=ex\frac{du}{dx}=2 x,\quad \frac{dv}{dx}=e^{x}

    Differentiate u and v separately using standard rules.

  5. Apply the product rule

    dydx=(x2+1)⋅(ex)+(ex)⋅(2x)\frac{dy}{dx}=\left(x^{2} + 1\right)\cdot \left(e^{x}\right)+\left(e^{x}\right)\cdot \left(2 x\right)

    Substitute u, v and their derivatives into the product-rule formula.

  6. Expand the brackets

    dydx=x2ex+2xex+ex\frac{dy}{dx}=x^{2} e^{x} + 2 x e^{x} + e^{x}

    Multiply everything out to prepare for collecting like terms.

  7. Simplify to obtain the derivative

    dydx=(x2+2x+1)ex\frac{dy}{dx}=\left(x^{2} + 2 x + 1\right) e^{x}

    Collect like terms and take out common factors to reach the simplest form.

  8. Write the derivative in factored form

    dydx=(x+1)2ex\frac{dy}{dx}=\left(x + 1\right)^{2} e^{x}

    A factored form helps to compare with the options.

  9. 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.

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

    dydx∣x=1=4e\left.\frac{dy}{dx}\right|_{x=1}=4 e

    Substituting a value gives a quick numerical sanity check.

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

    dydx∣x=2=9e2\left.\frac{dy}{dx}\right|_{x=2}=9 e^{2}

    Substituting a value gives a quick numerical sanity check.

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

    dydx∣x=3=16e3\left.\frac{dy}{dx}\right|_{x=3}=16 e^{3}

    Substituting a value gives a quick numerical sanity check.

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

    dydx∣x=4=25e4\left.\frac{dy}{dx}\right|_{x=4}=25 e^{4}

    Substituting a value gives a quick numerical sanity check.

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

    dydx∣x=5=36e5\left.\frac{dy}{dx}\right|_{x=5}=36 e^{5}

    Substituting a value gives a quick numerical sanity check.

  15. Select the correct derivative

    dydx=(x2+2x+1)ex\frac{dy}{dx}=\left(x^{2} + 2 x + 1\right) e^{x}

    This matches the first option.

Answer
(x2+2x+1)ex\left(x^{2} + 2 x + 1\right) e^{x}

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