The quotient rule questions with answers

Questions

1. Question 1Linear over linear

   Given that $y = \dfrac{2x + 1}{3x + 2}$, find $\dfrac{dy}{dx}$, simplifying your answer.

   Show the worked solution

   1. Step 1: Let u be the top and v the bottom$u = 2x + 1$$v = 3x + 2$
   2. Step 2: Differentiate each one$\dfrac{du}{dx} = 2$$\dfrac{dv}{dx} = 3$
   3. Step 3: Quotient rule: v times du/dx, minus u times dv/dx, all over v squared$\dfrac{dy}{dx} = \dfrac{v\dfrac{du}{dx} - u\dfrac{dv}{dx}}{v^{2}}$$\dfrac{dy}{dx} = \dfrac{(3x + 2) \cdot 2 - (2x + 1) \cdot 3}{(3x + 2)^{2}}$
   4. Step 4: Expand the top and collect like terms (leave the bottom as it is)$= \dfrac{1}{(3x + 2)^{2}}$
2. Question 2Linear over linear

   Given that $y = \dfrac{2x - 3}{x - 5}$, find $\dfrac{dy}{dx}$, simplifying your answer.

   Show the worked solution

   1. Step 1: Let u be the top and v the bottom$u = 2x - 3$$v = x - 5$
   2. Step 2: Differentiate each one$\dfrac{du}{dx} = 2$$\dfrac{dv}{dx} = 1$
   3. Step 3: Quotient rule: v times du/dx, minus u times dv/dx, all over v squared$\dfrac{dy}{dx} = \dfrac{v\dfrac{du}{dx} - u\dfrac{dv}{dx}}{v^{2}}$$\dfrac{dy}{dx} = \dfrac{(x - 5) \cdot 2 - (2x - 3) \cdot 1}{(x - 5)^{2}}$
   4. Step 4: Expand the top and collect like terms (leave the bottom as it is)$= \dfrac{-7}{(x - 5)^{2}}$
3. Question 3With a quadratic

   Given that $y = \dfrac{x + 2}{x^{2} - 3}$, find $\dfrac{dy}{dx}$, simplifying your answer.

   Show the worked solution

   1. Step 1: Let u be the top and v the bottom$u = x + 2$$v = x^{2} - 3$
   2. Step 2: Differentiate each one$\dfrac{du}{dx} = 1$$\dfrac{dv}{dx} = 2x$
   3. Step 3: Quotient rule: v times du/dx, minus u times dv/dx, all over v squared$\dfrac{dy}{dx} = \dfrac{v\dfrac{du}{dx} - u\dfrac{dv}{dx}}{v^{2}}$$\dfrac{dy}{dx} = \dfrac{(x^{2} - 3) \cdot 1 - (x + 2) \cdot 2x}{(x^{2} - 3)^{2}}$
   4. Step 4: Expand the top and collect like terms (leave the bottom as it is)$= \dfrac{-x^{2} - 4x - 3}{(x^{2} - 3)^{2}}$
4. Question 4With a quadratic

   Given that $y = \dfrac{x^{2}}{3x + 5}$, find $\dfrac{dy}{dx}$, simplifying your answer.

   Show the worked solution

   1. Step 1: Let u be the top and v the bottom$u = x^{2}$$v = 3x + 5$
   2. Step 2: Differentiate each one$\dfrac{du}{dx} = 2x$$\dfrac{dv}{dx} = 3$
   3. Step 3: Quotient rule: v times du/dx, minus u times dv/dx, all over v squared$\dfrac{dy}{dx} = \dfrac{v\dfrac{du}{dx} - u\dfrac{dv}{dx}}{v^{2}}$$\dfrac{dy}{dx} = \dfrac{(3x + 5) \cdot 2x - x^{2} \cdot 3}{(3x + 5)^{2}}$
   4. Step 4: Expand the top and collect like terms (leave the bottom as it is)$= \dfrac{3x^{2} + 10x}{(3x + 5)^{2}}$
5. Question 5With a square root

   Given that $y = \dfrac{2x + 1}{\sqrt{3x + 4}}$, find $\dfrac{dy}{dx}$, simplifying your answer.

   Show the worked solution

   1. Step 1: Let u be the top and v the bottom$u = 2x + 1$$v = \sqrt{3x + 4}$
   2. Step 2: Differentiate each one (v is a square root: use the chain rule)$\dfrac{du}{dx} = 2$$\dfrac{dv}{dx} = \dfrac{3}{2\sqrt{3x + 4}}$
   3. Step 3: Quotient rule: v times du/dx, minus u times dv/dx, all over v squared$\dfrac{dy}{dx} = \dfrac{v\dfrac{du}{dx} - u\dfrac{dv}{dx}}{v^{2}}$$\dfrac{dy}{dx} = \dfrac{(\sqrt{3x + 4}) \cdot 2 - (2x + 1) \cdot (\dfrac{3}{2\sqrt{3x + 4}})}{(\sqrt{3x + 4})^{2}}$
   4. Step 4: Multiply the top and the bottom by 2√(3x + 4) to clear the fraction on top, then simplify$= \dfrac{6x + 13}{2(3x + 4)^{\frac{3}{2}}}$
6. Question 6With a square root

   Given that $y = \dfrac{2x + 1}{\sqrt{x - 2}}$, find $\dfrac{dy}{dx}$, simplifying your answer.

   Show the worked solution

   1. Step 1: Let u be the top and v the bottom$u = 2x + 1$$v = \sqrt{x - 2}$
   2. Step 2: Differentiate each one (v is a square root: use the chain rule)$\dfrac{du}{dx} = 2$$\dfrac{dv}{dx} = \dfrac{1}{2\sqrt{x - 2}}$
   3. Step 3: Quotient rule: v times du/dx, minus u times dv/dx, all over v squared$\dfrac{dy}{dx} = \dfrac{v\dfrac{du}{dx} - u\dfrac{dv}{dx}}{v^{2}}$$\dfrac{dy}{dx} = \dfrac{(\sqrt{x - 2}) \cdot 2 - (2x + 1) \cdot (\dfrac{1}{2\sqrt{x - 2}})}{(\sqrt{x - 2})^{2}}$
   4. Step 4: Multiply the top and the bottom by 2√(x - 2) to clear the fraction on top, then simplify$= \dfrac{2x - 9}{2(x - 2)^{\frac{3}{2}}}$