The chain rule Problem set A One question at a time

Find \(\frac{dy}{dx}\) for each.

1. (a) \(y=\left(4 x + 1\right)^{5}\)

   • \(u=4 x + 1,\ y=u^{5}\)
   • \(\frac{dy}{du}=5 u^{4},\ \frac{du}{dx}=4\)
   • \(\frac{dy}{dx}=20 \left(4 x + 1\right)^{4}\)

2. (b) \(y=\left(x^{3} - 2 x\right)^{4}\)

   • \(u=x^{3} - 2 x,\ y=u^{4}\)
   • \(\frac{dy}{du}=4 u^{3},\ \frac{du}{dx}=3 x^{2} - 2\)
   • \(\frac{dy}{dx}=\left(12 x^{2} - 8\right) \left(x^{3} - 2 x\right)^{3}\)

Find \(\frac{dy}{dx}\) for each.

1. (a) \(y=\left(x^{2} + 5\right)^{\frac{2}{3}}\)

   • \(u=x^{2} + 5,\ y=u^{\frac{2}{3}}\)
   • \(\frac{dy}{du}=\frac{2}{3 \sqrt[3]{u}},\ \frac{du}{dx}=2 x\)
   • \(\frac{dy}{dx}=\frac{4 x}{3 \sqrt[3]{x^{2} + 5}}\)

2. (b) \(y=\sqrt{3 x^{2} - x}\)

   • \(u=3 x^{2} - x,\ y=\sqrt{u}\)
   • \(\frac{dy}{du}=\frac{1}{2 \sqrt{u}},\ \frac{du}{dx}=6 x - 1\)
   • \(\frac{dy}{dx}=\frac{6 x - 1}{2\sqrt{3 x^{2} - x}}\)

Find \(\frac{dy}{dx}\) for each.

1. (a) \(y=\sin{\left(5 x \right)}\)

   • \(u=5 x,\ y=\sin{\left(u \right)}\)
   • \(\frac{dy}{du}=\cos{\left(u \right)},\ \frac{du}{dx}=5\)
   • \(\frac{dy}{dx}=5 \cos{\left(5 x \right)}\)

2. (b) \(y=\cos{\left(3 x - 2 \right)}\)

   • \(u=3 x - 2,\ y=\cos{\left(u \right)}\)
   • \(\frac{dy}{du}=- \sin{\left(u \right)},\ \frac{du}{dx}=3\)
   • \(\frac{dy}{dx}=- 3 \sin{\left(3 x - 2 \right)}\)

Find \(\frac{dy}{dx}\) for each.

1. (a) \(y=e^{5 x}\)

   • \(u=5 x,\ y=e^{u}\)
   • \(\frac{dy}{du}=e^{u},\ \frac{du}{dx}=5\)
   • \(\frac{dy}{dx}=5 e^{5 x}\)

2. (b) \(y=e^{x^{2} - 4 x}\)

   • \(u=x^{2} - 4 x,\ y=e^{u}\)
   • \(\frac{dy}{du}=e^{u},\ \frac{du}{dx}=2 x - 4\)
   • \(\frac{dy}{dx}=\left(2 x - 4\right) e^{x^{2} - 4 x}\)

Find \(\frac{dy}{dx}\) for each.

1. (a) \(y=\ln{\left(3 x \right)}\)

   • \(u=3 x,\ y=\ln{\left(u \right)}\)
   • \(\frac{dy}{du}=\frac{1}{u},\ \frac{du}{dx}=3\)
   • \(\frac{dy}{dx}=\frac{1}{x}\)

2. (b) \(y=\ln{\left(x^{2} + 4 \right)}\)

   • \(u=x^{2} + 4,\ y=\ln{\left(u \right)}\)
   • \(\frac{dy}{du}=\frac{1}{u},\ \frac{du}{dx}=2 x\)
   • \(\frac{dy}{dx}=\frac{2 x}{x^{2} + 4}\)

Find \(\frac{dy}{dx}\) for each. Each needs three derivatives multiplied.

1. (a) \(y=\cos^{2}{\left(2 x \right)}\)

   • \(y=u^2,\ u=\cos v,\ v=2x\)
   • \(\frac{dy}{dx}=2u^{1}\times\frac{du}{dv}\times\frac{dv}{dx}=2(\cos 2x)^{1}\times - \sin{\left(2 x \right)}\times2\)
   • \(\frac{dy}{dx}=- 4 \sin{\left(2 x \right)} \cos{\left(2 x \right)}\)

2. (b) \(y=e^{\sin{\left(3 x \right)}}\)

   • \(y=e^u,\ u=\sin v,\ v=3x\)
   • \(\frac{dy}{dx}=e^u\times\frac{du}{dv}\times\frac{dv}{dx}=e^{\sin 3x}\times(\cos{\left(3 x \right)})\times3\)
   • \(\frac{dy}{dx}=3 e^{\sin{\left(3 x \right)}} \cos{\left(3 x \right)}\)

3. (c) \(y=\sqrt{\ln{\left(5 x \right)}}\)

   • \(y=u^{\frac{1}{2}},\ u=\ln v,\ v=5x\)
   • \(\frac{dy}{dx}=\frac{1}{2}u^{- \frac{1}{2}}\times\frac{1}{v}\times5=\frac{1}{2}(\ln 5x)^{- \frac{1}{2}}\times\frac{1}{5x}\times5\)
   • \(\frac{dy}{dx}=\frac{1}{2 x \sqrt{\ln{\left(5 x \right)}}}\)

Find \(\frac{dy}{dx}\) for each. Name the inner function first.

1. (a) \(y=\frac{1}{e^{4 x}}\)

   • \(y=e^{-4x}=e^u,\ u=-4x\)
   • \(\frac{dy}{du}=e^u,\ \frac{du}{dx}=-4\)
   • \(\frac{dy}{dx}=- 4 e^{- 4 x}\)

2. (b) \(y=\sqrt{\cos{\left(x \right)}}\)

   • \(u=\cos{\left(x \right)},\ y=\sqrt{u}\)
   • \(\frac{dy}{du}=\frac{1}{2 \sqrt{u}},\ \frac{du}{dx}=- \sin{\left(x \right)}\)
   • \(\frac{dy}{dx}=\frac{- \sin{\left(x \right)}}{2\sqrt{\cos{\left(x \right)}}}\)

3. (c) \(y=\sin{\left(\sqrt{x} \right)}\)

   • \(u=\sqrt{x},\ y=\sin{\left(u \right)}\)
   • \(\frac{dy}{du}=\cos{\left(u \right)},\ \frac{du}{dx}=\frac{1}{2 \sqrt{x}}\)
   • \(\frac{dy}{dx}=\frac{\cos{\left(\sqrt{x} \right)}}{2 \sqrt{x}}\)

Find the gradient of the curve \(y=(e^x)^3+2x\) at the point where \(x=\ln 2\).

• \((e^x)^3=e^{3x}\), so \(\frac{dy}{dx}=3e^{3x}+2\)
• At \(x=\ln 2\): \(e^{3\ln 2}=2^3=8\)
• \(\frac{dy}{dx}=3(8)+2=26\)

The temperature of a drink, in °C, t minutes after it is poured is modelled by \(T=18+70e^{-0.05t}\). Find the rate of change of the temperature when \(t=10\). Give the units.

• \(u=-0.05t\), so \(\frac{dT}{dt}=70e^{u}\times(-0.05)=-3.5e^{-0.05t}\)
• At \(t=10\): \(-3.5e^{-0.5}\)
• From the calculator: -2.12 °C per minute

The height of the tide in a harbour, in metres, t hours after midnight is modelled by \(h=5+2\sin(0.5t)\), with the calculator in radians. Find the rate at which the height is changing at 02:00.

• \(u=0.5t\), so \(\frac{dh}{dt}=2\cos u\times 0.5=\cos(0.5t)\)
• At \(t=2\): \(\cos 1\)
• From the calculator (radians): 0.54 m per hour

Find the equation of the tangent to the curve \(y=\sqrt{x^2+16}\) at the point where \(x=3\). Write it as \(y=mx+c\).

• \(u=x^2+16\), \(y=u^{1/2}\): \(\frac{dy}{dx}=\frac12u^{-1/2}\times2x=\frac{x}{\sqrt{x^2+16}}\)
• At \(x=3\): \(y=5\) and gradient \(\frac{3}{5}\)
• \(y-5=\frac{3}{5}(x-3)\)
• \(y=\frac{3}{5}x+\frac{16}{5}\)

Find the coordinates of the point on the curve \(y=\ln(3x-4)\) where the tangent is parallel to the line \(y=2x-1\).

• \(u=3x-4\): \(\frac{dy}{dx}=\frac1u\times3=\frac{3}{3x-4}\)
• Parallel to \(y=2x-1\): gradient 2, so \(\frac{3}{3x-4}=2\)
• \(3x-4=\frac32\), so \(x=\frac{11}{6}\)
• \(y=\ln\left(3\times\frac{11}{6}-4\right)=\ln \frac{3}{2}=0.405\)

The line \(y=-4x+1\) is perpendicular to a tangent of the curve \(y=\ln(x+3)\). Find the equation of that tangent. Give \(y=mx+c\) with c to 3 significant figures.

• \(\frac{dy}{dx}=\frac{1}{x+3}\)
• Perpendicular to gradient \(-4\): tangent gradient \(\frac14\), so \(\frac{1}{x+3}=\frac14\)
• \(x=1\), \(y=\ln 4=1.39\)
• \(y-1.39=\frac14(x-1)\ \Rightarrow\ y=\frac{1}{4}x+1.14\)

Find the equation of the tangent to the curve \(y=\cos\left(\frac{2}{x}\right)\) at the point where \(x=\frac{4}{\pi}\). Give \(y=mx+c\) with m and c to 3 significant figures.

• \(u=\frac2x=2x^{-1}\), \(\frac{du}{dx}=-2x^{-2}\)
• \(\frac{dy}{dx}=-\sin u\times\left(-\frac{2}{x^2}\right)=\frac{2}{x^2}\sin\frac2x\)
• At \(x=\frac4\pi\): \(u=\frac\pi2\), \(y=0\), gradient \(\frac{2\pi^2}{16}=1.23\)
• \(y=0+1.23\left(x-\frac4\pi\right)\ \Rightarrow\ y=1.23x-1.57\)