Differentiation – Year 12 A level maths worksheet answers to download as PDF

← Back to the worksheet

1. Answer to Question 1

   1. Differentiate term by term using \(\dfrac{d}{dx}(x^n)=nx^{n-1}\); the constant \(-4\) differentiates to 0. \(\dfrac{dy}{dx}=\) \(15x^2-4x+7\).
   2. First rewrite the roots and fractions as powers of \(x\). \(y=3x^{1/2}-2x^{-2}\), so \(\dfrac{dy}{dx}=\tfrac32x^{-1/2}+4x^{-3}=\) \(\dfrac3{2\sqrt x}+\dfrac4{x^3}\). Take care with the sign: \(-2\times(-2)=+4\).
   3. A product of brackets cannot be differentiated term by term, so expand it first. Expand first: \(y=x^2-x-6\), so \(\dfrac{dy}{dx}=\) \(2x-1\). So the gradient function is \(2x-1\).
   4. Divide term by term: \(y=2x-3x^{-1}\), so \(\dfrac{dy}{dx}=2+3x^{-2}=\) \(2+\dfrac3{x^2}\). So the derivative is \(2+\dfrac3{x^2}\), which is always positive.

2. Answer to Question 2

   1. Use the definition \(f'(x)=\displaystyle\lim_{h\to0}\dfrac{f(x+h)-f(x)}h\) and expand \((x+h)^2=x^2+2xh+h^2\). \(\dfrac{f(x+h)-f(x)}h=\dfrac{3(x+h)^2-3x^2}h=\dfrac{6xh+3h^2}h=6x+3h\). As \(h\to0\), \(f'(x)=\) \(6x\). The term \(3h\) tends to 0, which leaves the derivative \(6x\), in agreement with the rule \(nx^{n-1}\).
   2. Expand \((x+h)^3=x^3+3x^2h+3xh^2+h^3\); the \(x^3\) terms cancel and every remaining term has a factor \(h\). \(\dfrac{(x+h)^3-x^3}h=\dfrac{3x^2h+3xh^2+h^3}h=3x^2+3xh+h^2\). As \(h\to0\), \(f'(x)=\) \(3x^2\). So \(f'(x)=3x^2\).

3. Answer to Question 3

   1. Find the point, then the gradient of the tangent from \(\dfrac{dy}{dx}\); the normal is perpendicular, so its gradient is \(-\dfrac1m\). \(y(2)=-1\). \(\dfrac{dy}{dx}=2x-3=1\) at \(x=2\). Tangent: \(y+1=x-2\), i.e. \(y=x-3\). Normal gradient \(-1\): \(y+1=-(x-2)\), i.e. \(y=-x+1\). The tangent is \(y=x-3\) and the normal is \(y=-x+1\).
   2. Use the same method: point, gradient, then \(y-y_1=m(x-x_1)\). \(y(1)=1\). \(\dfrac{dy}{dx}=6x^2-1=5\). Tangent: \(y-1=5(x-1)\), i.e. \(y=5x-4\). Normal gradient \(-\tfrac15\): \(y-1=-\tfrac15(x-1)\), i.e. \(x+5y=6\). The tangent is \(y=5x-4\) and the normal is \(x+5y=6\).

4. Answer to Question 4

   1. Differentiate, then take out the common factor 3 and factorise the quadratic. \(f'(x)=3x^2-12x+9=3(x^2-4x+3)=3(x-1)(x-3)\). So \(f'(x)=3(x-1)(x-3)\).
   2. \(f\) is increasing where \(f'(x)>0\); the roots of \(f'\) are 1 and 3. \(f'(x)>0\) when the U-shaped quadratic is above the axis: \(x<1\) or \(x>3\). So \(f\) is increasing for \(x<1\) and for \(x>3\).
   3. Between the roots the quadratic \(3(x-1)(x-3)\) is negative. Decreasing for \(1\lt x<3\). So \(f\) is decreasing for \(1\lt x<3\).

5. Answer to Question 5

   1. Stationary points occur where the gradient is zero. \(\dfrac{dy}{dx}=3x^2-6x-9=3(x-3)(x+1)=0\), so \(x=3\) or \(x=-1\). The stationary points have \(x\)-coordinates 3 and \(-1\).
   2. Substitute each \(x\)-value into the original equation for \(y\), not into the derivative. \(x=3\): \(27-27-27+5=-22\). \(x=-1\): \(-1-3+9+5=10\). Points \((3,-22)\) and \((-1,10)\).
   3. If \(\dfrac{d^2y}{dx^2}>0\) the point is a minimum; if it is negative, a maximum. \(\dfrac{d^2y}{dx^2}=6x-6\). At \(x=3\): \(12>0\), a minimum. At \(x=-1\): \(-12<0\), a maximum. So \((-1,10)\) is a maximum and \((3,-22)\) is a minimum.

6. Answer to Question 6

   1. Differentiate twice using the power rule. \(f'(x)=4x^3-16x\) and \(f''(x)=12x^2-16\).
   2. Solve \(f'(x)=0\) by taking out the factor \(4x\), then use \(f''\) to classify each point. \(4x(x^2-4)=0\) gives \(x=0,\ \pm2\). \(f(0)=0\); \(f(\pm2)=16-32=-16\). \(f''(0)=-16<0\): maximum at \((0,0)\). \(f''(\pm2)=32>0\): minima at \((2,-16)\) and \((-2,-16)\). The graph is a W shape, symmetrical about the \(y\)-axis.
   3. Use a sign diagram for \(4x(x-2)(x+2)\) with critical values \(-2\), 0 and 2. \(f'(x)=4x(x-2)(x+2)>0\) for \(-2\lt x<0\) and \(x>2\). So \(f\) is increasing for \(-2\lt x<0\) and for \(x>2\).

7. Answer to Question 7

   1. Differentiate the height with respect to time. \(\dfrac{dh}{dt}=14-9.8t\): the vertical velocity in m/s (positive upward). Its value at any time gives how fast the ball is rising.
   2. At the highest point the ball is momentarily at rest, so \(\dfrac{dh}{dt}=0\). Highest point when \(14-9.8t=0\), \(t=\dfrac{14}{9.8}=\) 1.43 s. \(h=1.2+14\times1.4286-4.9\times1.4286^2=\) 11.2 m. The ball reaches its maximum height of about 11.2 m after 1.43 s.
   3. The ball lands when \(h=0\); solve the quadratic and take the positive root. \(4.9t^2-14t-1.2=0\): \(t=\dfrac{14+\sqrt{196+23.52}}{9.8}=\dfrac{14+14.816}{9.8}=\) 2.94 s. The ball lands after 2.94 s.
   4. Substitute the landing time into \(\dfrac{dh}{dt}\). \(14-9.8\times2.9404=-14.8\): 14.8 m/s downwards. The negative sign shows that the ball is moving downwards.

8. Answer to Question 8

   1. Use the volume to express \(h\) in terms of \(x\); the open box has a base and four sides but no lid. \(x^2h=32\) so \(h=\dfrac{32}{x^2}\). Surface \(=x^2+4xh=x^2+\dfrac{128}x\). This is the required expression.
   2. At a minimum the derivative is zero. \(\dfrac{dS}{dx}=2x-\dfrac{128}{x^2}=0\), so \(x^3=64\) and \(x=\) 4 m (then \(h=2\) m). So the base should be 4 m by 4 m.
   3. A positive second derivative confirms a minimum. \(\dfrac{d^2S}{dx^2}=2+\dfrac{256}{x^3}=6>0\), a minimum. \(S=16+32=\) 48 m\(^2\). The minimum surface area is 48 m\(^2\).
   4. Multiply the minimum area by the cost per square metre. \(48\times15=\) £720. The minimum cost is £720.

9. Answer to Question 9

   1. Differentiate term by term. \(\dfrac{dP}{dx}=-4x+60\). This is the rate at which profit changes with output.
   2. Set \(\dfrac{dP}{dx}=0\) and remember the units: \(x\) is in hundreds of items and \(P\) is in hundreds of pounds. \(-4x+60=0\) gives \(x=15\) (1 500 items). \(P=-2(225)+900-300=150\), i.e. £15 000. The greatest profit is £15 000, made with 1 500 items a week.
   3. Use the second derivative test. \(\dfrac{d^2P}{dx^2}=-4<0\), so a maximum. The parabola opens downwards, so its only stationary point is the maximum.
   4. Solve \(-2x^2+60x-300>0\); dividing by \(-2\) reverses the inequality. \(P>0\): \(x^2-30x+150<0\), roots \(x=15\pm\sqrt{75}=15\pm8.66\). So \(6.34\lt x<23.66\). The firm makes a profit when it produces between about 634 and 2 366 items a week.

10. Answer to Question 10

    1. Differentiate displacement to get velocity, and velocity to get acceleration. \(v=3t^2-18t+24\) and \(a=6t-18\).
    2. The particle is at rest when \(v=0\). \(3(t^2-6t+8)=3(t-2)(t-4)=0\), so \(t=\) 2 s and 4 s. So it is instantaneously at rest at 2 s and at 4 s.
    3. Substitute these times into \(s\). \(s(2)=8-36+48=\) 20 m. \(s(4)=64-144+96=\) 16 m.
    4. Substitute the times into \(a=6t-18\). \(a(2)=-6\) m/s\(^2\) and \(a(4)=+6\) m/s\(^2\). At \(t=2\) the particle is at its furthest from the origin (a local max of \(s\)) and at \(t=4\) it is at a local minimum before moving away again.

11. Answer to Question 11

    1. Differentiate and substitute \(x=0.5\). \(\dfrac{dy}{dx}=3x^2-12x+9\); at \(x=0.5\): \(0.75-6+9=3.75\). A positive gradient means the track is rising at this point.
    2. Solve \(\dfrac{dy}{dx}=0\) and find the \(y\)-values from the original equation. \(3(x-1)(x-3)=0\): \(x=1\) (\(y=4\)) and \(x=3\) (\(y=0\)). So \((1,4)\) is a peak and \((3,0)\) is the bottom of the dip.
    3. To find where the gradient is smallest, differentiate the gradient function itself. The gradient function \(g=3x^2-12x+9\) has \(g'=6x-12=0\) at \(x=2\); \(g''=6>0\), so \(g\) is smallest there: \(g(2)=12-24+9=-3\).
    4. The steepest downhill section is at \(x=2\) where the gradient is \(-3\) (i.e. a drop of 3 units for every 1 unit across). This is the steepest descent of the ride.

12. Answer to Question 12

    1. Use the volume to eliminate \(h\); the can has two circular ends and a curved surface \(2\pi rh\). \(\pi r^2h=330\), so \(h=\dfrac{330}{\pi r^2}\). \(S=2\pi r^2+2\pi rh=2\pi r^2+\dfrac{660}r\). This is the required expression.
    2. Differentiate with respect to \(r\) and set the derivative equal to zero. \(\dfrac{dS}{dr}=4\pi r-\dfrac{660}{r^2}=0\Rightarrow r^3=\dfrac{660}{4\pi}=52.52\Rightarrow r=\) 3.74 cm. (\(\dfrac{d^2S}{dr^2}=4\pi+\dfrac{1320}{r^3}>0\) so a minimum.) The surface area is smallest when the radius is 3.74 cm.
    3. Substitute the unrounded value \(r=3.745\) to avoid rounding errors. \(h=\dfrac{330}{\pi\times3.745^2}=\) 7.49 cm. \(S=2\pi(14.02)+\dfrac{660}{3.745}=88.1+176.2=\) 264.4 cm\(^2\). The height is 7.49 cm and the minimum surface area is about 264 cm\(^2\).
    4. \(h\approx2r\): the best can has height equal to its diameter. Real cans are taller than this because of manufacturing and design considerations.

13. Answer to Question 13

    1. Substitute \(x=3\) into the curve, then differentiate. \(9-12+5=2\) ✓. \(\dfrac{dy}{dx}=2x-4=2\). So \(P\) is on the curve and the gradient there is 2.
    2. The normal is perpendicular to the tangent, so its gradient is \(-\tfrac12\). Normal gradient \(-\tfrac12\): \(y-2=-\tfrac12(x-3)\), so \(x+2y=7\). The normal at \(P\) is \(x+2y=7\).
    3. Solve the equations of the curve and the normal simultaneously; \(x=3\) must be one root because \(P\) lies on both. \(x^2-4x+5=\dfrac{7-x}2\Rightarrow2x^2-7x+3=0\Rightarrow(2x-1)(x-3)=0\). The other root is \(x=\tfrac12\), \(y=\dfrac{7-0.5}{2}=3.25\). So \(Q=(0.5,\,3.25)\).
    4. Use Pythagoras with \(\Delta x=2.5\) and \(\Delta y=1.25\). \(PQ=\sqrt{2.5^2+1.25^2}=\sqrt{7.8125}=2.80\) units \(=\) 28 m. So the lamp at \(Q\) is about 28 m from \(P\).

14. Answer to Question 14

    1. Differentiate \(P\) with respect to \(t\) and substitute \(t=10\). \(\dfrac{dP}{dt}=150-6t\); at \(t=10\): 90 hundred per year (9 000 people per year). In 2010 the population is increasing at about 9 000 people per year.
    2. The population stops growing when \(\dfrac{dP}{dt}=0\). \(150-6t=0\) gives \(t=25\), i.e. in 2025. The population stops growing in 2025.
    3. Substitute \(t=25\) into \(P\); \(\dfrac{d^2P}{dt^2}=-6<0\) confirms a maximum. \(P(25)=2000+3750-1875=3875\) hundred, i.e. 387 500 people. The maximum population is 387 500.
    4. Solve \(P=2000\), the value at \(t=0\), and reject the root \(t=0\). \(150t-3t^2=0\) gives \(t=50\): 2050. The model predicts the population is back to its 2000 value in 2050.

15. Answer to Question 15

    1. After cutting out the squares, the base measures \((24-2x)\) by \((15-2x)\) and the height is \(x\). \(V=x(24-2x)(15-2x)=x(360-78x+4x^2)=4x^3-78x^2+360x\). This is the required formula.
    2. Substitute \(x=3\): the base is 18 cm by 9 cm. \(V(3)=3\times18\times9=486\) ✓. The table value is correct.
    3. Differentiate, set \(\dfrac{dV}{dx}=0\) and factorise; \(x\) must be less than half of 15 cm. \(\dfrac{dV}{dx}=12x^2-156x+360=12(x^2-13x+30)=12(x-3)(x-10)\). \(x=3\) or \(10\); but \(x<7.5\) so \(x=\) 3. So squares of side 3 cm give the largest tray.
    4. Substitute \(x=3\) into the second derivative. \(\dfrac{d^2V}{dx^2}=24x-156=-84<0\) at \(x=3\): a maximum. \(V_{max}=\) 486 cm\(^3\). The largest possible tray holds 486 cm\(^3\).

16. Answer to Question 16

    1. Find the gradient and the point at \(x=2\), then use \(y-y_1=m(x-x_1)\). \(\dfrac{dy}{dx}=3x^2-3=9\) at \(x=2\); \(y=8-6+2=4\). Tangent: \(y-4=9(x-2)\), i.e. \(y=9x-14\). The tangent at \(A(2,4)\) is \(y=9x-14\).
    2. The tangent meets the curve where the two expressions for \(y\) are equal. \(x^3-3x+2=9x-14\Rightarrow x^3-12x+16=0\). This is the required cubic equation.
    3. Since the line is tangent at \(x=2\), \((x-2)^2\) is a factor: \(x^3-12x+16=(x-2)^2(x+4)\). The other root is \(x=-4\), \(y=9(-4)-14=-50\). So \(B=(-4,-50)\). (Check on \(C\): \(-64+12+2=-50\) ✓.) So the tangent meets the curve again at \(B(-4,-50)\).

17. Answer to Question 17

    1. Stationary points where \(\dfrac{dy}{dx}=3x^2+2ax+b=0\) at \(x=-2\) and \(x=1\), so \(\dfrac{dy}{dx}=3(x+2)(x-1)=3x^2+3x-6\). Compare: \(2a=3\), \(a=\) 1.5; \(b=\) \(-6\). So \(a=1.5\) and \(b=-6\).
    2. Substitute the point into the curve. At \((1,-2.5)\): \(1+1.5-6+c=-2.5\), so \(c=\) 1.
    3. Substitute the \(x\)-values into \(y\) and use the second derivative test. \(y=x^3+1.5x^2-6x+1\). \(x=-2\): \(-8+6+12+1=11\); \(\dfrac{d^2y}{dx^2}=6x+3=-9<0\): maximum \((-2,11)\). \(x=1\): \(-2.5\); second derivative \(9>0\): minimum \((1,-2.5)\).
    4. For a positive cubic, \(y\) decreases between the maximum and the minimum. Decreasing for \(-2\lt x<1\).

18. Answer to Question 18

    1. If the width is \(x\), the length is the area divided by the width. Length \(=\dfrac{100}x\). \(P=2\left(x+\dfrac{100}x\right)=2x+\dfrac{200}x\). This is the required formula.
    2. Substitute two values of \(x\) into the formula. \(x=8\): \(16+25=41\) ✓. \(x=20\): \(40+10=50\) ✓. Both entries agree with the table.
    3. Differentiate, set the derivative equal to zero and keep the positive root. \(\dfrac{dP}{dx}=2-\dfrac{200}{x^2}=0\Rightarrow x^2=100\Rightarrow x=10\). \(\dfrac{d^2P}{dx^2}=\dfrac{400}{x^3}>0\): minimum. The shape is a \(10\times10\) square and \(P=\) 40 m. The minimum perimeter is 40 m.
    4. Multiply the minimum perimeter by the cost per metre. \(40\times8.40=\) £336. The minimum cost is £336.

19. Answer to Question 19

    1. Differentiate and take out the common factor \(4x^2\). \(f'(x)=4x^3-12x^2=4x^2(x-3)\). This is the required result.
    2. Solve \(f'(x)=0\); the factor \(x^2\) gives a repeated root at 0. \(x=0\) (\(f=0\)) and \(x=3\) (\(f=81-108=-27\)). Points \((0,0)\) and \((3,-27)\).
    3. Since \(4x^2\ge0\), the sign of \(f'\) is the sign of \((x-3)\) on both sides of \(x=0\): negative on both sides. So \((0,0)\) is a point of inflection (stationary, not a turning point). At \(x=3\) the sign changes from negative to positive: minimum. So \((3,-27)\) is a minimum and \((0,0)\) is a stationary point of inflection.
    4. Differentiate \(f'\) again and factorise. \(f''(x)=12x^2-24x=12x(x-2)=0\) at \(x=0\) and \(x=2\); the curve changes concavity there.

20. Answer to Question 20

    1. Use Pythagoras with the horizontal distance \(x\) and the vertical distance \(x^2-3\). \(AP^2=x^2+(x^2-3)^2=x^2+x^4-6x^2+9=x^4-5x^2+9\). This is the required expression.
    2. Minimising \(AP\) is the same as minimising \(AP^2\), which avoids differentiating a square root. Let \(f=x^4-5x^2+9\). \(f'=4x^3-10x=2x(2x^2-5)=0\) gives \(x=0\) or \(x=\pm\sqrt{2.5}=\pm1.58\). \(f''=12x^2-10\). At \(x=0\): \(-10<0\) (maximum of \(AP^2\), i.e. a local farthest point). At \(x=\pm1.58\): \(f''=20>0\) (minima). So there are minima at \(x=\pm\sqrt{2.5}\) and a local maximum at \(x=0\).
    3. Substitute \(x^2=2.5\) into \(AP^2\) and take the square root. \(f(\pm\sqrt{2.5})=6.25-12.5+9=2.75=\tfrac{11}4\). Least distance \(=\sqrt{\tfrac{11}4}=\) \(\dfrac{\sqrt{11}}2\approx1.66\). So the least distance from \(A\) to the parabola is \(\dfrac{\sqrt{11}}2\).