Further Derivatives: exam review Exam review One question at a time

Differentiate each function with respect to \(x\). Simplify your answers where you can. [6]

1. (a) (a) \(f(x)=\frac{6x^{3}-4x}{\sqrt{x}}\) [3]

   • M1 for splitting into separate powers: \(6x^{\frac{5}{2}} - 4 \sqrt{x}\)
   • A1 for one correct term of the derivative
   • A1 for \(15x^{\frac{3}{2}} - \frac{2}{\sqrt{x}}\)

2. (b) (b) \(g(x)=4\ln\left(x^{5}\right)+3\tan x\) [3]

   • M1 for using \(\ln x^{5}=5\ln x\)
   • A1 for \(\frac{20}{x}\)
   • A1 for \(3\sec^{2}x\)

Differentiate with respect to \(x\). [6]

1. (a) (a) \(y=\sqrt{9 - 2x^{2}}\) [2]

   • M1 for the chain rule: \(\frac{1}{2}(9 - 2x^{2})^{- \frac{1}{2}}\) multiplied by the derivative of the bracket
   • A1 for \(- \frac{2x}{\sqrt{9 - 2x^{2}}}\)

2. (b) (b) \(y=e^{\sin(3x)}\) [2]

   • M1 for \(e^{\sin(3x)}\times\) the derivative of \(\sin(3x)\)
   • A1 for \(3e^{\sin(3x)} \cos(3x)\)

3. (c) (c) Find the gradient of the curve \(y=(e^{x})^{3}+5x\) at the point where \(x=\ln 2\). [2]

   • M1 for \(\frac{dy}{dx}=3e^{3x}+5\)
   • A1 for \(3\times 2^{3}+5=29\)

Differentiate with respect to \(x\). Show the rule you use. [6]

1. (a) (a) \(y=x^{2}e^{-3x}\) [3]

   • M1 for the product rule \(uv'+vu'\)
   • A1 for \(u'=2x\) and \(v'=-3e^{-3x}\)
   • A1 for \(xe^{-3x}(2-3x)\)

2. (b) (b) \(y=\frac{e^{x}}{(2x+1)^{2}}\) [3]

   • M1 for the quotient rule \(\frac{vu'-uv'}{v^{2}}\)
   • A1 for \(u'=e^{x}\) and \(v'=4(2x+1)\)
   • A1 for \(\frac{e^{x}(2x-3)}{(2x+1)^{3}}\) (simplified)

The curve \(C\) has equation \(y=e^{2x}-5x\). [7]

1. (a) (a) Find \(\frac{dy}{dx}\). [1]

   • A1 for \(2e^{2x} - 5\)

2. (b) (b) The tangent to \(C\) at the point P is parallel to the line \(y=3x+4\). Find the exact \(x\)-coordinate of P. [3]

   • M1 for setting \(2e^{2x} - 5=3\)
   • A1 for \(e^{2x}=4\)
   • A1 for \(x=\ln(2)\)

3. (c) (c) Find the equation of the normal to \(C\) at the point where \(x=0\). Give your answer in the form \(y=mx+c\). [3]

   • A1 for gradient of the tangent \(-3\) at \(x=0\) and \(y=1\)
   • M1 for the negative reciprocal \(\frac{1}{3}\)
   • A1 for \(y=\frac{1}{3}x+1\)

The function \(f\) is defined by \(f(x)=xe^{-2x}\). [8]

1. (a) (a) Find \(f'(x)\). [2]

   • M1 for the product rule
   • A1 for \(f'(x)=(1 - 2x)e^{-2x}\)

2. (b) (b) Find the exact coordinates of the stationary point of the graph of \(y=f(x)\). [2]

   • M1 for setting \(f'(x)=0\) (the exponential is never zero)
   • A1 for \(x=\frac{1}{2}\ \text{and}\ y=\frac{1}{2e}\)

3. (c) (c) Find \(f''(x)\). [2]

   • M1 for differentiating \((1 - 2x)e^{-2x}\) using the product rule
   • A1 for \(f''(x)=(4x - 4)e^{-2x}\)

4. (d) (d) Use your answer to part (c) to show that the stationary point is a maximum, and find the \(x\)-coordinate of the point of inflexion. [2]

   • A1 for \(f''(\frac{1}{2})=- \frac{2}{e}<0\), so a maximum
   • A1 for \(f''=0\) at \(x=1\) (and \(f''\) changes sign)

The concentration, \(C\) ppm, of a pollutant at a distance of \(x\) km from a factory is modelled by \(C=\frac{160}{x^{2}}\). A drone flies directly away from the factory at a constant speed of 0.4 km per minute. [5]

1. (a) (a) Find \(\frac{dC}{dx}\). [2]

   • M1 for writing \(C=160x^{-2}\)
   • A1 for \(-\frac{320}{x^{3}}\)

2. (b) (b) Find the rate of change of \(C\) when \(x=2\). [2]

   • M1 for \(\frac{dC}{dt}=\frac{dC}{dx}\times\frac{dx}{dt}\)
   • A1 for \(-\frac{320}{2^{3}}\times 0.4=-16\)

3. (c) (c) Explain what the sign of your answer to part (b) tells you. [1]

   • R1 for the negative sign: the pollutant concentration is decreasing, by 16 ppm each minute

A particle moves in a straight line. The displacement from a fixed point O is \(s\) metres after \(t\) seconds, where \(s=2 t^{3} - 15 t^{2} + 24 t\) for \(t\ge0\). [7]

1. (a) (a) Find expressions for the velocity and the acceleration. [2]

   • A1 for \(v=6 t^{2} - 30 t + 24\)
   • A1 for \(a=12 t - 30\)

2. (b) (b) Find the times when the particle is at rest. [2]

   • M1 for \(6 t^{2} - 30 t + 24=0\)
   • A1 for \(t=1\) and \(t=4\)

3. (c) (c) Determine whether the particle is speeding up or slowing down when \(t=3\). Justify your answer. [3]

   • M1 for substituting \(t=3\): \(v=-12\), \(a=6\)
   • A1 for \(v\) negative and \(a\) positive
   • R1 for \(v\) and \(a\) having the opposite sign, so the particle is slowing down

A bead slides along a straight wire. The displacement from a fixed point O is \(s\) metres after \(t\) seconds, where \(s=8\left(1-e^{-0.5t}\right)\) for \(t\ge0\). [6]

1. (a) (a) Find an expression for the velocity. [2]

   • M1 for the chain rule: factor \(-1/2\) from the exponent
   • A1 for \(v=4e^{-0.5t}\)

2. (b) (b) Find the time when the velocity is \(1.5\) m s\(^{-1}\). [2]

   • M1 for \(4e^{-0.5t}=1.5\)
   • A1 for \(t=1.96\)

3. (c) (c) Find the distance from O that the particle approaches as \(t\) becomes very large. [2]

   • M1 for \(e^{-0.5t}\to0\) as \(t\to\infty\)
   • A1 for \(8\) m

A funnel is in the shape of an inverted cone of height 12 cm and radius 4 cm at the top. Liquid flows into the funnel at a constant rate of 5 cm³ per second. At time \(t\) seconds the depth of the liquid is \(h\) cm and the volume of liquid is \(V\) cm³. [12]

1. (a) (a) Show that \(V=\frac{\pi h^{3}}{27}\). [2]

   • M1 for similar triangles: \(\frac{r}{h}=\frac{4}{12}\), so \(r=\frac{h}{3}\)
   • A1 for \(V=\frac{1}{3}\pi\left(\frac{h}{3}\right)^{2}h=\frac{\pi h^{3}}{27}\) (AG)

2. (b) (b) Find the rate at which the depth is increasing at the instant when \(h=6\). [4]

   • M1 for \(\frac{dV}{dh}=\frac{\pi h^{2}}{9}\)
   • A1 for \(\frac{dV}{dh}=4 \pi\ \text{at}\ h=6\)
   • M1 for \(\frac{dh}{dt}=\frac{dV}{dt}\div\frac{dV}{dh}=5\div 4 \pi\)
   • A1 for \(\frac{5}{4 \pi}\) = \(0.398\)

3. (c) (c) Find the rate at which the area of the liquid surface is increasing at the instant when \(h=6\). [4]

   • M1 for \(A=\pi r^{2}=\frac{\pi h^{2}}{9}\)
   • A1 for \(\frac{dA}{dh}=\frac{2 \pi h}{9}\)
   • M1 for \(\frac{dA}{dt}=\frac{dA}{dh}\times\frac{dh}{dt}\)
   • A1 for \(\frac{5}{3}\) = \(1.67\)

4. (d) (d) Explain why the depth rises more and more slowly while the liquid flows in at a constant rate. [2]

   • R1 for \(\frac{dh}{dt}=\frac{45}{\pi h^{2}}\) decreasing as \(h\) increases
   • R1 for the reason: the surface area grows with \(h\), so the same volume spreads over a larger area

The function \(f\) is defined by \(f(x)=x^{3}+ax^{2}+bx+5\), where \(a\) and \(b\) are constants. The graph of \(y=f(x)\) has a point of inflexion at \(x=1\) and a stationary point at \(x=3\). [13]

1. (a) (a) Find the value of \(a\) and the value of \(b\). [4]

   • M1 for \(f''(x)=6x+2a\)
   • A1 for \(6(1)+2a=0\), so \(a=-3\)
   • M1 for \(f'(x)=3x^{2}+2ax+b\) and \(f'(3)=0\)
   • A1 for \(b=-9\)

2. (b) (b) Find the coordinates of the stationary points of the graph of \(y=f(x)\). Use the second derivative to decide the nature of each. [5]

   • M1 for \(f'(x)=3x^{2} - 6x - 9=0\) solved
   • A1 for \(x=-1\) and \(x=3\)
   • A1 for \((-1,10)\) and \((3,-22)\)
   • M1 for \(f''(x)=6 \left(x - 1\right)\) evaluated at both points
   • A1 for \(f''(-1)=-12<0\) (maximum) and \(f''(3)=12>0\) (minimum)

3. (c) (c) Write down the set of values of \(x\) for which the graph is concave down. Explain why there is a point of inflexion at \(x=1\). [2]

   • A1 for \(x<1\) (\(f''(x)<0\))
   • R1 for \(f''\) changes sign at \(x=1\)

4. (d) (d) Find the set of values of \(k\) for which \(f(x)=k\) has exactly three solutions. [2]

   • A1 for the values lie between the local minimum and the local maximum: \(-22<k<10\)
   • A1 for strict inequalities (at \(k=10\) or \(k=-22\) there are only two solutions)

A drinks company makes closed cylindrical cans of volume 330 cm³. A can has base radius \(r\) cm and height \(h\) cm. The company wants to use as little metal as possible. [12]

1. (a) (a) Write down an expression for \(h\) in terms of \(r\). [1]

   • A1 for \(V=\pi r^{2}h=330\) rearranged: \(h=\frac{330}{\pi r^{2}}\)

2. (b) (b) Show that the total surface area is \(A=2\pi r^{2}+\frac{660}{r}\). [2]

   • M1 for \(A=2\pi r^{2}+2\pi rh\)
   • A1 for \(2\pi r\times\frac{330}{\pi r^{2}}=\frac{660}{r}\) (AG)

3. (c) (c) Find \(\frac{dA}{dr}\). [2]

   • M1 for rewriting \(\frac{660}{r}=660r^{-1}\)
   • A1 for \(4 \pi r - \frac{660}{r^{2}}\)

4. (d) (d) Find the value of \(r\) that minimizes the surface area. Give your answer to 3 significant figures. [3]

   • M1 for \(4 \pi r - \frac{660}{r^{2}}=0\)
   • A1 for \(r^{3}=\frac{165}{\pi}\)
   • A1 for \(r=3.74\)

5. (e) (e) Use the second derivative to justify that this value of \(r\) gives a minimum. [2]

   • M1 for \(\frac{d^{2}A}{dr^{2}}=4 \left(\pi + \frac{330}{r^{3}}\right)\)
   • R1 for positive for all \(r>0\), so a minimum (value 37.7 at \(r=3.74\))

6. (f) (f) Find the minimum surface area. [2]

   • M1 for substituting \(r=3.74\) into \(A\)
   • A1 for \(264\)

The number of deer, \(P\), on an island \(t\) years after they are first counted is modelled by \(P(t)=\frac{1200}{1+11e^{-0.4t}}\), \(t\geq 0\). [13]

1. (a) (a) Find the initial population at \(t=0\). [2]

   • M1 for substituting \(t=0\): \(\frac{1200}{1+11}\)
   • A1 for \(100\)

2. (b) (b) Write down the value that \(P\) approaches as \(t\to\infty\). [1]

   • A1 for \(1200\)

3. (c) (c) Find \(\frac{dP}{dt}\). [3]

   • M1 for writing \(1200\left(1+11e^{-0.4t}\right)^{-1}\) and using the chain rule
   • A1 for \(1200\times(-1)\left(1+11e^{-0.4t}\right)^{-2}\) (outer function)
   • A1 for \(\frac{dP}{dt}=\frac{5280e^{-0.4t}}{\left(1+11e^{-0.4t}\right)^{2}}\)

4. (d) (d) Find the rate at which the population is changing when \(t=2\). [2]

   • M1 for substituting \(t=2\) into \(\frac{dP}{dt}\)
   • A1 for \(67.2\)

5. (e) (e) Use your calculator to find the greatest rate of change of \(P\) and the value of \(t\) at which it occurs. [3]

   • M1 for a graph or table of \(\frac{dP}{dt}\) and finding its maximum
   • A1 for \(t=5.99\)
   • A1 for greatest rate \(120\)

6. (f) (f) Find the value of \(P\) at this time and describe what happens to the graph of \(P\) at this point. [2]

   • A1 for \(P=600\) (half of the limiting value \(1200\))
   • R1 for a point of inflexion: the gradient is greatest here so \(P''=0\) and the concavity changes from up to down

A robot arm slides a component along a straight rail. Its velocity at time \(t\) seconds is \(v\) m s\(^{-1}\), where \(v=3\cos(2t)\) for \(t\ge0\). The displacement of the component from a fixed point O is \(s\) metres, and \(s=1\) when \(t=0\). [12]

1. (a) (a) Find an expression for the acceleration. [2]

   • M1 for differentiating with the chain rule, factor \(2\)
   • A1 for \(a=-6\sin(2t)\)

2. (b) (b) Find the first time \(t>0\) when the component is momentarily at rest. [2]

   • M1 for \(v=0\), i.e. \(\cos(2t)=0\) (radian mode)
   • A1 for \(t=0.785\)

3. (c) (c) Write down the greatest speed of the component. [1]

   • A1 for \(3\)

4. (d) (d) Find an expression for \(s\) in terms of \(t\). [2]

   • M1 for integrating \(v\) with respect to \(t\)
   • A1 for \(s=1.5\sin(2t)+1\) using \(s(0)=1\)

5. (e) (e) Find the displacement of the component from its starting position when \(t=4\). [2]

   • M1 for \(s(4)-s(0)\) or \(\int_0^{4}v\,dt\)
   • A1 for \(1.48\)

6. (f) (f) Find the total distance travelled by the component in the first 4 seconds. [3]

   • M1 for \(\int_0^{4}|v|\,dt\)
   • (M1) for the use of the absolute value, so that backward motion counts as positive
   • A1 for \(7.52\)