The chain rule Problem set B Worked solutions

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

1. (a) \(y=\left(5 x - 2\right)^{4}\)

   • \(u=5 x - 2,\ y=u^{4}\)
   • \(\frac{dy}{du}=4 u^{3},\ \frac{du}{dx}=5\)
   • \(\frac{dy}{dx}=20 \left(5 x - 2\right)^{3}\)

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

   • \(u=x^{2} + 3 x,\ y=u^{5}\)
   • \(\frac{dy}{du}=5 u^{4},\ \frac{du}{dx}=2 x + 3\)
   • \(\frac{dy}{dx}=\left(10 x + 15\right) \left(x^{2} + 3 x\right)^{4}\)

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

1. (a) \(y=\left(x^{2} + 7\right)^{\frac{3}{4}}\)

   • \(u=x^{2} + 7,\ y=u^{\frac{3}{4}}\)
   • \(\frac{dy}{du}=\frac{3}{4 \sqrt[4]{u}},\ \frac{du}{dx}=2 x\)
   • \(\frac{dy}{dx}=\frac{3 x}{2 \sqrt[4]{x^{2} + 7}}\)

2. (b) \(y=\sqrt{2 x^{2} + 5 x}\)

   • \(u=2 x^{2} + 5 x,\ y=\sqrt{u}\)
   • \(\frac{dy}{du}=\frac{1}{2 \sqrt{u}},\ \frac{du}{dx}=4 x + 5\)
   • \(\frac{dy}{dx}=\frac{4 x + 5}{2\sqrt{2 x^{2} + 5 x}}\)

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

1. (a) \(y=\cos{\left(4 x \right)}\)

   • \(u=4 x,\ y=\cos{\left(u \right)}\)
   • \(\frac{dy}{du}=- \sin{\left(u \right)},\ \frac{du}{dx}=4\)
   • \(\frac{dy}{dx}=- 4 \sin{\left(4 x \right)}\)

2. (b) \(y=\sin{\left(6 x - 1 \right)}\)

   • \(u=6 x - 1,\ y=\sin{\left(u \right)}\)
   • \(\frac{dy}{du}=\cos{\left(u \right)},\ \frac{du}{dx}=6\)
   • \(\frac{dy}{dx}=6 \cos{\left(6 x - 1 \right)}\)

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

1. (a) \(y=e^{- 3 x}\)

   • \(u=- 3 x,\ y=e^{u}\)
   • \(\frac{dy}{du}=e^{u},\ \frac{du}{dx}=-3\)
   • \(\frac{dy}{dx}=- 3 e^{- 3 x}\)

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

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

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

1. (a) \(y=\ln{\left(7 x \right)}\)

   • \(u=7 x,\ y=\ln{\left(u \right)}\)
   • \(\frac{dy}{du}=\frac{1}{u},\ \frac{du}{dx}=7\)
   • \(\frac{dy}{dx}=\frac{1}{x}\)

2. (b) \(y=\ln{\left(x^{2} + 9 \right)}\)

   • \(u=x^{2} + 9,\ y=\ln{\left(u \right)}\)
   • \(\frac{dy}{du}=\frac{1}{u},\ \frac{du}{dx}=2 x\)
   • \(\frac{dy}{dx}=\frac{2 x}{x^{2} + 9}\)

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

1. (a) \(y=\sin^{2}{\left(4 x \right)}\)

   • \(y=u^2,\ u=\sin v,\ v=4x\)
   • \(\frac{dy}{dx}=2u^{1}\times\frac{du}{dv}\times\frac{dv}{dx}=2(\sin 4x)^{1}\times \cos{\left(4 x \right)}\times4\)
   • \(\frac{dy}{dx}=8 \sin{\left(4 x \right)} \cos{\left(4 x \right)}\)

2. (b) \(y=e^{\cos{\left(5 x \right)}}\)

   • \(y=e^u,\ u=\cos v,\ v=5x\)
   • \(\frac{dy}{dx}=e^u\times\frac{du}{dv}\times\frac{dv}{dx}=e^{\cos 5x}\times(- \sin{\left(5 x \right)})\times5\)
   • \(\frac{dy}{dx}=- 5 e^{\cos{\left(5 x \right)}} \sin{\left(5 x \right)}\)

3. (c) \(y=\sqrt[3]{\ln{\left(2 x \right)}}\)

   • \(y=u^{\frac{1}{3}},\ u=\ln v,\ v=2x\)
   • \(\frac{dy}{dx}=\frac{1}{3}u^{- \frac{2}{3}}\times\frac{1}{v}\times2=\frac{1}{3}(\ln 2x)^{- \frac{2}{3}}\times\frac{1}{2x}\times2\)
   • \(\frac{dy}{dx}=\frac{1}{3 x \ln{\left(2 x \right)}^{\frac{2}{3}}}\)

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

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

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

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

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

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

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

Find the gradient of the curve \(y=(e^x)^4+3x\) at the point where \(x=\ln 2\).

• \((e^x)^4=e^{4x}\), so \(\frac{dy}{dx}=4e^{4x}+3\)
• At \(x=\ln 2\): \(e^{4\ln 2}=2^4=16\)
• \(\frac{dy}{dx}=4(16)+3=67\)

A phone battery charges so that its charge level, in per cent, after t hours is modelled by \(C=100-80e^{-0.2t}\). Find the rate at which the charge level is changing when \(t=5\). Give the units.

• \(u=-0.2t\), so \(\frac{dC}{dt}=-80e^{u}\times(-0.2)=16e^{-0.2t}\)
• At \(t=5\): \(16e^{-1}\)
• From the calculator: 5.89 per cent per hour

The height above the ground of a seat on a Ferris wheel, in metres, t minutes after it starts is modelled by \(h=10+6\cos(0.3t)\), with the calculator in radians. Find the rate at which the height is changing when \(t=4\).

• \(u=0.3t\), so \(\frac{dh}{dt}=-6\sin u\times 0.3=-1.8\sin(0.3t)\)
• At \(t=4\): \(-1.8\sin 1.2\)
• From the calculator (radians): -1.68 m per minute

Find the equation of the tangent to the curve \(y=\sqrt{x^2+24}\) at the point where \(x=5\). Write it as \(y=mx+c\).

• \(u=x^2+24\), \(y=u^{1/2}\): \(\frac{dy}{dx}=\frac12u^{-1/2}\times2x=\frac{x}{\sqrt{x^2+24}}\)
• At \(x=5\): \(y=7\) and gradient \(\frac{5}{7}\)
• \(y-7=\frac{5}{7}(x-5)\)
• \(y=\frac{5}{7}x+\frac{24}{7}\)

Find the coordinates of the point on the curve \(y=\ln(5x-2)\) where the tangent is parallel to the line \(y=4x\).

• \(u=5x-2\): \(\frac{dy}{dx}=\frac1u\times5=\frac{5}{5x-2}\)
• Parallel to \(y=4x\): gradient 4, so \(\frac{5}{5x-2}=4\)
• \(5x-2=\frac54\), so \(x=\frac{13}{20}\)
• \(y=\ln\left(5\times\frac{13}{20}-2\right)=\ln \frac{5}{4}=0.223\)

The line \(y=-5x+3\) is perpendicular to a tangent of the curve \(y=\ln(x-2)\). Find the equation of that tangent. Give \(y=mx+c\) with c to 3 significant figures.

• \(\frac{dy}{dx}=\frac{1}{x-2}\)
• Perpendicular to gradient \(-5\): tangent gradient \(\frac15\), so \(\frac{1}{x-2}=\frac15\)
• \(x=7\), \(y=\ln 5=1.61\)
• \(y-1.61=\frac15(x-7)\ \Rightarrow\ y=\frac{1}{5}x+0.209\)

Find the equation of the tangent to the curve \(y=\cos\left(\frac{3}{x}\right)\) at the point where \(x=\frac{6}{\pi}\). Give \(y=mx+c\) with m and c to 3 significant figures.

• \(u=\frac3x=3x^{-1}\), \(\frac{du}{dx}=-3x^{-2}\)
• \(\frac{dy}{dx}=-\sin u\times\left(-\frac{3}{x^2}\right)=\frac{3}{x^2}\sin\frac3x\)
• At \(x=\frac6\pi\): \(u=\frac\pi2\), \(y=0\), gradient \(\frac{3\pi^2}{36}=0.822\)
• \(y=0+0.822\left(x-\frac6\pi\right)\ \Rightarrow\ y=0.822x-1.57\)