Product, quotient and chain rules Worked Solutions — A-Level Maths

Fully worked, step-by-step solutions to A-Level Product, quotient and chain rules questions. See exactly how to solve problems on product-rule, differentiation, chain-rule, quotient-rule.

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

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
Differentiate y=xsin⁡(x)y=x \sin{\left(x \right)} with respect to xx.

Worked solution

  1. Label the factors and their derivatives

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

    Set up the pieces for the product rule.

  2. Apply the product rule

    dydx=(x)⋅(cos⁡(x))+(sin⁡(x))⋅(1)\frac{dy}{dx}=\left(x\right)\cdot \left(\cos{\left(x \right)}\right)+\left(\sin{\left(x \right)}\right)\cdot \left(1\right)

    Use u times v' plus v times u'.

  3. State the derivative

    dydx=xcos⁡(x)+sin⁡(x)\frac{dy}{dx}=x \cos{\left(x \right)} + \sin{\left(x \right)}

    This is the required derivative.

Answer
xcos⁡(x)+sin⁡(x)x \cos{\left(x \right)} + \sin{\left(x \right)}
Question 3
2 markseasy
Differentiate y=x2exy=x^{2} e^{x} with respect to xx.

Worked solution

  1. Label the factors and their derivatives

    u=x2, v=ex, dudx=2x, dvdx=exu=x^{2},\ v=e^{x},\ \frac{du}{dx}=2 x,\ \frac{dv}{dx}=e^{x}

    Set up the pieces for the product rule.

  2. Apply the product rule

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

    Use u times v' plus v times u'.

  3. State the derivative

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

    This is the required derivative.

Answer
x(x+2)exx \left(x + 2\right) e^{x}
Question 4
2 markseasy
Differentiate y=(3x+2)5y=\left(3 x + 2\right)^{5} with respect to xx.

Worked solution

  1. Identify the inner and outer functions and differentiate each

    u=3x+2, dudx=3, dydu=5u4u=3 x + 2,\ \frac{du}{dx}=3,\ \frac{dy}{du}=5 u^{4}

    Set up the pieces for the chain rule.

  2. Apply the chain rule

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

    Multiply dy/du by du/dx.

  3. State the derivative

    dydx=15(3x+2)4\frac{dy}{dx}=15 \left(3 x + 2\right)^{4}

    This is the required derivative.

Answer
15(3x+2)415 \left(3 x + 2\right)^{4}
Question 5
2 markseasy
Differentiate y=e3xy=e^{3 x} with respect to xx.

Worked solution

  1. Identify the inner and outer functions and differentiate each

    u=3x, dudx=3, dydu=euu=3 x,\ \frac{du}{dx}=3,\ \frac{dy}{du}=e^{u}

    Set up the pieces for the chain rule.

  2. Apply the chain rule

    dydx=(e3x)⋅(3)\frac{dy}{dx}=\left(e^{3 x}\right)\cdot \left(3\right)

    Multiply dy/du by du/dx.

  3. State the derivative

    dydx=3e3x\frac{dy}{dx}=3 e^{3 x}

    This is the required derivative.

Answer
3e3x3 e^{3 x}

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