Expanding, splitting and increasing or decreasing Problem set B Worked solutions

Differentiate with respect to x. The other letters are constants.

1. (a) \(y=3px+\frac{4q}{x^3}\)

   • \(y=3px + \frac{4q}{x^{3}}\)
   • \(\frac{dy}{dx}=3p - \frac{12q}{x^{4}}\)
   • Letters other than x are constants.

2. (b) \(y=7q^2x^2-\frac{8p}{rx}\)

   • \(y=7q^{2}x^{2} - \frac{8p}{rx}\)
   • \(\frac{dy}{dx}=14q^{2}x + \frac{8p}{rx^{2}}\)
   • Letters other than x are constants.

Expand, then differentiate with respect to x.

1. (a) \(y=(kx)^2\)

   • \(y=(kx)^2=k^{2}x^{2}\)
   • \(\frac{dy}{dx}=2k^{2}x\)

2. (b) \(y=(5px)^2\)

   • \(y=(5px)^2=25p^{2}x^{2}\)
   • \(\frac{dy}{dx}=50p^{2}x\)

Expand, then find \(\frac{dy}{dx}\). The other letters are constants.

1. (a) \(y=(x+r)(x+k)\)

   • \(y=(x+r)(x+k)=x^{2} + (k+r)x + kr\)
   • \(\frac{dy}{dx}=2x + k + r\)

2. (b) \(y=(rx+k)(kx+r)\)

   • \(y=(rx+k)(kx+r)=krx^{2} + (k^{2}+r^{2})x + kr\)
   • \(\frac{dy}{dx}=2krx + k^{2} + r^{2}\)

Split each fraction into powers of x, then differentiate with respect to x.

1. (a) \(y=\frac{x-9}{x}\)

   • \(y=\frac{x-9}{x}=1 - \frac{9}{x}\)
   • \(\frac{dy}{dx}=\frac{9}{x^{2}}\)

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

   • \(y=\frac{2+5x}{x^2}=\frac{5}{x} + \frac{2}{x^{2}}\)
   • \(\frac{dy}{dx}=-\frac{5}{x^{2}} - \frac{4}{x^{3}}\)

3. (c) \(y=\frac{x^3+6x^4}{3x}\)

   • \(y=\frac{x^3+6x^4}{3x}=2x^{3} + \frac{1}{3}x^{2}\)
   • \(\frac{dy}{dx}=6x^{2} + \frac{2}{3}x\)

Find the values of x for which the function is decreasing.

1. (a) \(f(x)=3x^{2} - 18x + 4\)

   • \(f'(x)=6x - 18\)
   • \(6x - 18<0\)
   • \(6x - 18=0\Rightarrow x=3\)
   • The sign of the x term decides the side: \(x<3\)

2. (b) \(f(x)=-2x^{2} + 16x - 5\)

   • \(f'(x)=-4x + 16\)
   • \(-4x + 16<0\)
   • \(-4x + 16=0\Rightarrow x=4\)
   • The sign of the x term decides the side: \(x>4\)

Find the values of x for which the function is increasing.

1. (a) \(f(x)=x^{2} + 4x\)

   • \(f'(x)=2x + 4\)
   • \(2x + 4>0\Rightarrow x>-2\)

2. (b) \(f(x)=x^2+px+q\), where \(p\) and \(q\) are constants

   • \(f'(x)=2x+p\)
   • \(2x+p>0\Rightarrow x>-\frac{p}{2}\)
   • The constant q differentiates to 0, so it does not appear in the answer.

Differentiate \(w=7p+\frac{10}{p}\) with respect to \(p\).

• \(w=7p + \frac{10}{p}\)
• \(\frac{dw}{dp}=7 - \frac{10}{p^{2}}\)

If \(x^2-y=4x+1\), find \(\frac{dy}{dx}\).

• \(y=x^{2} - 4x - 1\)
• \(\frac{dy}{dx}=2x - 4\)

The thermal energy U joules of a sample of gas at temperature T kelvin is modelled by \(U=\frac52nRT\), where n and R are constants.

1. (a) Find the rate at which U increases with T.

   • \(U=\frac52nRT\)
   • \(\frac{dU}{dT}=\frac{5}{2}nR\)

2. (b) Find this rate when \(n=3\) and \(R=8.31\). Give the units and 3 significant figures.

   • \(\frac52\times3\times8.31=62.325\)
   • 62.3 joules per kelvin to 3 significant figures.

The noise level N from a loudspeaker at distance d metres is modelled by \(N=\frac{k}{d}\), where k is a constant. The fade rate is the rate of change of N with respect to d.

1. (a) Find an expression for the fade rate in terms of k and d.

   • \(N=kd^{-1}\)
   • \(\frac{dN}{dd}=-kd^{-2}=-\frac{k}{d^2}\)

2. (b) Find an expression for the fade rate in terms of N and k.

   • \(d=\frac{k}{N}\)
   • \(-\frac{k}{(k/N)^2}=-\frac{N^2}{k}\)

3. (c) Point P is one third of the way from the loudspeaker to a point Q. Find the value of (fade rate at P) ÷ (fade rate at Q).

   • \(\text{At Q: }-\frac{k}{d^2}.\ \text{At P (distance }\tfrac13d):\ -\frac{k}{(d/3)^2}=-\frac{9k}{d^2}\)
   • \(\text{ratio}=9\)

A model for the distance D km walked by a hiker after t hours is \(D=1+pt+3pt^2\), where p is a constant and \(t\geq0\).

1. (a) Find the rate of change of D with respect to t.

   • \(D=1+pt+3pt^2\)
   • \(\frac{dD}{dt}=p+6pt=p(1+6t)\)

2. (b) Explain why you would expect D to be an increasing function. What does this tell you about p?

   • The distance walked does not fall, so the gradient should be positive: \(\frac{dD}{dt}>0\).
   • \(p(1+6t)>0\)
   • For \(t\geq0\), \(1+6t\) is positive, so p must be positive.

Show that \(f(x)=2x^3+9x+1\) is increasing for all values of x.

• \(f'(x)=6x^{2} + 9\)
• \(x^2\geq0\Rightarrow 6x^2+9\geq9>0\)
• f′(x) is positive for all x, so f is increasing for all x.

The constant k is positive. Let \(f(x)=\frac{x^3+kx}{3x^2}\).

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

   • \(f(x)=\frac{x^3}{3x^2}+\frac{kx}{3x^2}=\frac{1}{3}x+\frac{k}{3}x^{-1}\)
   • \(f'(x)=\frac13-\frac{k}{3}x^{-2}=\frac13-\frac{k}{3x^2}\)

2. (b) Given that \(f'(3)=\frac19\), find the value of k.

   • \(f'(3)=\frac13-\frac{k}{27}=\frac19\)
   • \(\frac{k}{27}=\frac29\Rightarrow k=6\)