The chain rule questions with answers

Questions

1. Question 1A power of ax + b

   Given that $y = 3(2x + 3)^{6}$, find $\dfrac{dy}{dx}$.

   Show the worked solution

   1. Step 1: Let u be the expression inside the bracket, and differentiate it$u = 2x + 3$$\dfrac{du}{dx} = 2$
   2. Step 2: Now y = 3uⁿ: differentiate with respect to u (bring the power down, take 1 off)$\dfrac{dy}{du} = 18u^{5}$
   3. Step 3: Chain rule: multiply the two, then write u in terms of x again$\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}$$\dfrac{dy}{dx} = 18(2x + 3)^{5} \times 2$$= 36(2x + 3)^{5}$
2. Question 2A power of ax + b

   Given that $y = (x + 7)^{3}$, find $\dfrac{dy}{dx}$.

   Show the worked solution

   1. Step 1: Let u be the expression inside the bracket, and differentiate it$u = x + 7$$\dfrac{du}{dx} = 1$
   2. Step 2: Now y = uⁿ: differentiate with respect to u (bring the power down, take 1 off)$\dfrac{dy}{du} = 3u^{2}$
   3. Step 3: Chain rule: multiply the two, then write u in terms of x again$\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}$$\dfrac{dy}{dx} = 3(x + 7)^{2} \times 1$$= 3(x + 7)^{2}$
3. Question 3Negative powers and roots

   Given that $y = \dfrac{4}{\sqrt{3x - 3}}$, find $\dfrac{dy}{dx}$.

   Show the worked solution

   1. Step 1: Write y as a power of the bracket: 1 over a power is a negative power, and a square root is the power ½$y = 4(3x - 3)^{-\frac{1}{2}}$
   2. Step 2: Let u be the expression inside the bracket, and differentiate it$u = 3x - 3$$\dfrac{du}{dx} = 3$
   3. Step 3: Now y = 4uⁿ: differentiate with respect to u (bring the power down, take 1 off)$\dfrac{dy}{du} = -2u^{-\frac{3}{2}}$
   4. Step 4: Chain rule: multiply the two, then write u in terms of x again$\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}$$\dfrac{dy}{dx} = -2(3x - 3)^{-\frac{3}{2}} \times 3$$= -6(3x - 3)^{-\frac{3}{2}}$
   5. Step 5: Write the answer with positive indices$= -\dfrac{6}{(3x - 3)^{\frac{3}{2}}}$
4. Question 4Negative powers and roots

   Given that $y = \dfrac{1}{(4x - 3)^{3}}$, find $\dfrac{dy}{dx}$.

   Show the worked solution

   1. Step 1: Write y as a power of the bracket: 1 over a power is a negative power, and a square root is the power ½$y = (4x - 3)^{-3}$
   2. Step 2: Let u be the expression inside the bracket, and differentiate it$u = 4x - 3$$\dfrac{du}{dx} = 4$
   3. Step 3: Now y = uⁿ: differentiate with respect to u (bring the power down, take 1 off)$\dfrac{dy}{du} = -3u^{-4}$
   4. Step 4: Chain rule: multiply the two, then write u in terms of x again$\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}$$\dfrac{dy}{dx} = -3(4x - 3)^{-4} \times 4$$= -12(4x - 3)^{-4}$
   5. Step 5: Write the answer with positive indices$= -\dfrac{12}{(4x - 3)^{4}}$
5. Question 5A power of a quadratic

   Given that $y = \dfrac{1}{x^{2} + 4x - 3}$, find $\dfrac{dy}{dx}$.

   Show the worked solution

   1. Step 1: Write y as a power of the bracket: 1 over a power is a negative power, and a square root is the power ½$y = (x^{2} + 4x - 3)^{-1}$
   2. Step 2: Let u be the expression inside the bracket, and differentiate it$u = x^{2} + 4x - 3$$\dfrac{du}{dx} = 2x + 4$
   3. Step 3: Now y = uⁿ: differentiate with respect to u (bring the power down, take 1 off)$\dfrac{dy}{du} = -u^{-2}$
   4. Step 4: Chain rule: multiply the two, then write u in terms of x again$\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}$$\dfrac{dy}{dx} = -1(x^{2} + 4x - 3)^{-2} \times (2x + 4)$$= -(2x + 4)(x^{2} + 4x - 3)^{-2}$
   5. Step 5: Write the answer with positive indices$= -\dfrac{2x + 4}{(x^{2} + 4x - 3)^{2}}$
6. Question 6A power of a quadratic

   Given that $y = (x^{2} - 2x + 6)^{4}$, find $\dfrac{dy}{dx}$.

   Show the worked solution

   1. Step 1: Let u be the expression inside the bracket, and differentiate it$u = x^{2} - 2x + 6$$\dfrac{du}{dx} = 2x - 2$
   2. Step 2: Now y = uⁿ: differentiate with respect to u (bring the power down, take 1 off)$\dfrac{dy}{du} = 4u^{3}$
   3. Step 3: Chain rule: multiply the two, then write u in terms of x again$\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}$$\dfrac{dy}{dx} = 4(x^{2} - 2x + 6)^{3} \times (2x - 2)$$= 4(2x - 2)(x^{2} - 2x + 6)^{3}$