Differentiating powers, trig, e^x and ln x Problem set B Worked solutions

Find the derivative of each function.

1. (a) \(f(x)=x^{7/4}\)

   • \(f(x)=x^{7/4}\)
   • \(f'(x)=\frac{7}{4}\,x^{7/4-1}\)
   • \(f'(x)=\frac{7}{4}x^{3/4}\)

2. (b) \(f(x)=15x^{-2/5}\)

   • \(f(x)=15x^{-2/5}\)
   • \(f'(x)=15\times (-\frac{2}{5})\,x^{-2/5-1}\)
   • \(f'(x)=-6x^{-7/5}\)

Differentiate by rewriting as a power of x.

1. (a) \(f(x)=8\sqrt[5]{x}\)

   • \(f(x)=8\sqrt[5]{x}=8x^{1/5}\)
   • \(f'(x)=\frac{8}{5}x^{-4/5}\)

2. (b) \(f(x)=-\dfrac{14}{\sqrt[3]{x}}\)

   • \(f(x)=-\dfrac{14}{\sqrt[3]{x}}=-14x^{-1/3}\)
   • \(f'(x)=\frac{14}{3}x^{-4/3}\)

Find \(\dfrac{dy}{dx}\) by splitting the fraction into powers of x.

1. (a) \(y=\dfrac{10x^3+3}{x^{3/2}}\)

   • \(y=10x^{3/2} + 3x^{-3/2}\quad\text{(split into powers)}\)
   • \(\frac{dy}{dx}=15x^{1/2} - \frac{9}{2}x^{-5/2}\)

2. (b) \(y=\dfrac{9+4x}{\sqrt[3]{x}}\)

   • \(y=9x^{-1/3} + 4x^{2/3}\quad\text{(split into powers)}\)
   • \(\frac{dy}{dx}=-3x^{-4/3} + \frac{8}{3}x^{-1/3}\)

Differentiate. The angle is in radians.

1. (a) \(f(x)=4\sin x + 7\cos x\)

   • \(f'(x)=4\cos x - 7\sin x\)

2. (b) \(f(x)=3 + \frac{1}{4}\sin x - 5\cos x\)

   • \(f'(x)=\frac{1}{4}\cos x + 5\sin x\)

Differentiate.

1. (a) \(f(x)=-6\tan x + 2\)

   • \(f'(x)=-\frac{6}{\cos^2 x}\)

2. (b) \(f(x)=x^{3} + \tan x\)

   • \(f'(x)=3x^{2} + \frac{1}{\cos^2 x}\)

Find \(y'\).

1. (a) \(y=8\ln x\)

   • \(y'=\frac{8}{x}\)

2. (b) \(y=-3\ln x + 5e^x\)

   • \(y'=-\frac{3}{x} + 5e^x\)

Use a law of logarithms first, then find \(y'\).

1. (a) \(y=5\ln x^{3}\)

   • \(y=5\ln x^{3}=15\ln x\)
   • \(y'=\frac{15}{x}\)

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

   • \(y=-2\ln x^{-4}=8\ln x\)
   • \(y'=\frac{8}{x}\)

The volume of water in a tank filling slowly is modelled by \(V=12t^{3/2}\) litres, where t is the time in minutes since the tap was opened (t > 0).

1. (a) Find \(\frac{dV}{dt}\).

   • \(V=12t^{3/2}\)
   • \(\frac{dV}{dt}=18t^{1/2}\)

2. (b) Find the rate of change of volume when t = 9.

   • \(\frac{dV}{dt}\) at t = 9: 54 litres per minute

The temperature of an oven is modelled by \(T=180+12\sin t-9\cos t\) degrees Celsius, where t is the time in minutes after it was switched on. The angle is in radians.

1. (a) Find \(\frac{dT}{dt}\).

   • \(T=180 + 12\sin t - 9\cos t\)
   • \(\frac{dT}{dt}=12\cos t + 9\sin t\)

2. (b) Find the rate of change of temperature when t = 1.5.

   • Calculator in radian mode.
   • \(\frac{dT}{dt}\) at t = 1.5: 9.83 degrees per minute

A laser pointer on a stand is 40 cm from a wall and turns through an angle of \(\theta\) radians. The dot on the wall is at a distance \(x=40\tan\theta\) centimetres from the point nearest the stand.

1. (a) Find \(\frac{dx}{d\theta}\).

   • \(x=40\tan \theta\)
   • \(\frac{dx}{d\theta}=\frac{40}{\cos^2 \theta}\)

2. (b) Find the rate of change of x when \(\theta\) = 0.7.

   • Calculator in radian mode.
   • \(\frac{dx}{d\theta}\) at \(\theta\) = 0.7: 68.4 cm per radian

The size of a fish population, in hundreds, is modelled by \(N=30+8\ln w^2\), where w is the number of weeks since a pond was stocked (w > 0).

1. (a) Use a law of logarithms to write \(N\) without a power inside the logarithm, then find \(\frac{dN}{dw}\).

   • \(\ln w^{2}=2\ln w\)
   • \(N=30 + 16\ln w\)
   • \(\frac{dN}{dw}=\frac{16}{w}\)

2. (b) Find the rate of change when w = 5.

   • \(\frac{dN}{dw}\) at w = 5: 3.2 hundred fish per week

The curve \(y=f(x)\) has equation \(f(x)=\dfrac{x^2-12x}{\sqrt{x}}\), for x > 0.

1. (a) Find \(f'(x)\).

   • \(f(x)=x^{3/2} - 12x^{1/2}\)
   • \(f'(x)=\frac{3}{2}x^{1/2} - 6x^{-1/2}\)

2. (b) Find the value of x at the stationary point of the curve, where \(f'(x)=0\).

   • \(f'(x)=0:\ \frac{3}{2}x^{1/2} - 6x^{-1/2}=0\)
   • Multiply by \(x^{3/2}\) or solve with the calculator: x = 4

The curve \(y=e^x-6\ln x\) is defined for x > 0.

1. (a) Find \(\dfrac{dy}{dx}\).

   • \(\frac{dy}{dx}=e^x - \frac{6}{x}\)

2. (b) Find the x-coordinate of the point where the gradient is zero. Give the answer to 3 significant figures.

   • \(e^x-\frac{6}{x}=0\)
   • Solve: x = 1.43 (the only positive solution)