The product rule questions with answers

Questions

1. Question 1A power of x times a power of a bracket

   Given that $y = x^{2}(x - 2)^{4}$, find $\dfrac{dy}{dx}$, giving your answer in factorised form.

   Show the worked solution

   1. Step 1: Let u be the first factor and v the second$u = x^{2}$$v = (x - 2)^{4}$
   2. Step 2: Differentiate each one (use the chain rule for v)$\dfrac{du}{dx} = 2x$$\dfrac{dv}{dx} = 4(x - 2)^{3}$
   3. Step 3: Product rule: u times dv/dx, plus v times du/dx$\dfrac{dy}{dx} = u\dfrac{dv}{dx} + v\dfrac{du}{dx}$$\dfrac{dy}{dx} = x^{2} \cdot 4(x - 2)^{3} + (x - 2)^{4} \cdot 2x$
   4. Step 4: Take out the common factors (the lower power of each bracket), then simplify what is left$= 2x(x - 2)^{3}(3x - 2)$
2. Question 2A power of x times a power of a bracket

   Given that $y = x(3x - 1)^{3}$, find $\dfrac{dy}{dx}$, giving your answer in factorised form.

   Show the worked solution

   1. Step 1: Let u be the first factor and v the second$u = x$$v = (3x - 1)^{3}$
   2. Step 2: Differentiate each one (use the chain rule for v)$\dfrac{du}{dx} = 1$$\dfrac{dv}{dx} = 9(3x - 1)^{2}$
   3. Step 3: Product rule: u times dv/dx, plus v times du/dx$\dfrac{dy}{dx} = u\dfrac{dv}{dx} + v\dfrac{du}{dx}$$\dfrac{dy}{dx} = x \cdot 9(3x - 1)^{2} + (3x - 1)^{3} \cdot 1$
   4. Step 4: Take out the common factors (the lower power of each bracket), then simplify what is left$= (3x - 1)^{2}(12x - 1)$
3. Question 3Two brackets

   Given that $y = (x + 2)(3x + 2)^{2}$, find $\dfrac{dy}{dx}$, giving your answer in factorised form.

   Show the worked solution

   1. Step 1: Let u be the first factor and v the second$u = x + 2$$v = (3x + 2)^{2}$
   2. Step 2: Differentiate each one (use the chain rule for v)$\dfrac{du}{dx} = 1$$\dfrac{dv}{dx} = 6(3x + 2)$
   3. Step 3: Product rule: u times dv/dx, plus v times du/dx$\dfrac{dy}{dx} = u\dfrac{dv}{dx} + v\dfrac{du}{dx}$$\dfrac{dy}{dx} = (x + 2) \cdot 6(3x + 2) + (3x + 2)^{2} \cdot 1$
   4. Step 4: Take out the common factors (the lower power of each bracket), then simplify what is left$= (3x + 2)(9x + 14)$
4. Question 4Two brackets

   Given that $y = (x + 2)(2x + 1)^{3}$, find $\dfrac{dy}{dx}$, giving your answer in factorised form.

   Show the worked solution

   1. Step 1: Let u be the first factor and v the second$u = x + 2$$v = (2x + 1)^{3}$
   2. Step 2: Differentiate each one (use the chain rule for v)$\dfrac{du}{dx} = 1$$\dfrac{dv}{dx} = 6(2x + 1)^{2}$
   3. Step 3: Product rule: u times dv/dx, plus v times du/dx$\dfrac{dy}{dx} = u\dfrac{dv}{dx} + v\dfrac{du}{dx}$$\dfrac{dy}{dx} = (x + 2) \cdot 6(2x + 1)^{2} + (2x + 1)^{3} \cdot 1$
   4. Step 4: Take out the common factors (the lower power of each bracket), then simplify what is left$= (2x + 1)^{2}(8x + 13)$
5. Question 5With a square root

   Given that $y = x\sqrt{x + 3}$, find $\dfrac{dy}{dx}$, giving your answer in factorised form.

   Show the worked solution

   1. Step 1: Let u be the first factor and v the second$u = x$$v = \sqrt{x + 3}$
   2. Step 2: Differentiate each one (v is a square root: use the chain rule)$\dfrac{du}{dx} = 1$$\dfrac{dv}{dx} = \dfrac{1}{2\sqrt{x + 3}}$
   3. Step 3: Product rule: u times dv/dx, plus v times du/dx$\dfrac{dy}{dx} = u\dfrac{dv}{dx} + v\dfrac{du}{dx}$$\dfrac{dy}{dx} = x \cdot \dfrac{1}{2\sqrt{x + 3}} + \sqrt{x + 3} \cdot 1$
   4. Step 4: Write it as one fraction and simplify the top$= \dfrac{3x + 6}{2\sqrt{x + 3}}$
6. Question 6With a square root

   Given that $y = x\sqrt{2x + 1}$, find $\dfrac{dy}{dx}$, giving your answer in factorised form.

   Show the worked solution

   1. Step 1: Let u be the first factor and v the second$u = x$$v = \sqrt{2x + 1}$
   2. Step 2: Differentiate each one (v is a square root: use the chain rule)$\dfrac{du}{dx} = 1$$\dfrac{dv}{dx} = \dfrac{1}{\sqrt{2x + 1}}$
   3. Step 3: Product rule: u times dv/dx, plus v times du/dx$\dfrac{dy}{dx} = u\dfrac{dv}{dx} + v\dfrac{du}{dx}$$\dfrac{dy}{dx} = x \cdot \dfrac{1}{\sqrt{2x + 1}} + \sqrt{2x + 1} \cdot 1$
   4. Step 4: Write it as one fraction and simplify the top$= \dfrac{3x + 1}{\sqrt{2x + 1}}$