Hard A-Level First principles and derivatives Questions

Challenging, exam-style A-Level First principles and derivatives questions with worked solutions. Stretch yourself on the hardest first-principles, limit, power-rule, polynomial problems.

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A-Level34 questionsStep-by-step solutions
Question 1
6 markschallenging
Find the equation of the tangent to the curve y=x3−xy = x^3 - x at the point where x=2x = 2.
Show worked solution

Worked solution

  1. Write down the curve and the point

    y=x3−x,x=2y = x^{3} - x,\quad x = 2

    We need the equation of the tangent to the curve at the given x-value.

  2. Find the y-coordinate of the point

    y=6  ⇒  (2,6)y = 6\;\Rightarrow\;\left(2, 6\right)

    Substitute the x-value into the curve to get the exact point the line passes through.

  3. Differentiate to get the gradient function

    dydx=3x2−1\frac{dy}{dx} = 3x^{2} - 1

    The derivative gives the gradient of the curve (and of the tangent) at any point.

  4. Find the gradient of the curve at the point

    At x=2:dydx=11\text{At } x = 2:\quad \frac{dy}{dx} = 11

    Substitute the x-value into the gradient function to get the gradient of the tangent at this point.

  5. Use the straight-line equation

    y−y1=m(x−x1)y - y_1 = m\left(x - x_1\right)

    A line through a known point with a known gradient is found with this formula from coordinate geometry.

  6. Substitute the point and gradient

    y−6=11(x−2)y - 6 = 11\left(x - 2\right)

    Put the gradient and the coordinates of the point into the straight-line formula.

  7. Expand the bracket

    y−6=11x−22y - 6 = 11 x - 22

    Multiply out the bracket carefully, keeping track of signs.

  8. Rearrange into the form y = mx + c

    y=11x−16y = 11x - 16

    Make y the subject to write the final equation of the tangent neatly.

  9. Write the gradient function

    dydx=3x2−1\frac{dy}{dx} = 3x^{2} - 1

    The derivative is called the gradient function because putting any x-value into it gives the gradient of the curve at that point.

  10. Check with a sample value

    At x=1:dydx=2\text{At } x=1:\quad \frac{dy}{dx} = 2

    As a quick sanity check we substitute x = 1. This is exactly how the gradient function is used once we have it.

  11. Link the notation

    dydx=f′(x)\frac{dy}{dx} = f'(x)

    Remember from the introduction to calculus that dy/dx and f'(x) mean the same thing. Different exam boards use different notation for the derivative.

  12. Interpret the sign of the gradient

    dydx>0⇒increasing,dydx<0⇒decreasing\frac{dy}{dx} > 0 \Rightarrow \text{increasing}, \quad \frac{dy}{dx} < 0 \Rightarrow \text{decreasing}

    Where the gradient function is positive the curve goes uphill; where it is negative it goes downhill. This links the derivative back to the shape of the graph.

  13. Use the linearity of differentiation

    ddx(a f(x)+b g(x))=a f′(x)+b g′(x)\frac{d}{dx}\big(a\,f(x) + b\,g(x)\big) = a\,f'(x) + b\,g'(x)

    Differentiation can be done one term at a time, and constants stay multiplied on the front. That is why the working above splits so neatly.

  14. Try a second sample value

    At x=2:dydx=11\text{At } x=2:\quad \frac{dy}{dx} = 11

    Substituting x = 2 gives a different gradient, showing that the steepness of the curve changes as we move along it.

  15. Recap the method

    xn  →  n xn−1x^{n} \;\to\; n\,x^{n-1}

    In words: bring the power down to the front as a multiplier, then reduce the power by one. Repeat this for every term.

Answer
y=11x−16y = 11x - 16
Question 2
6 markschallenging
Find the coordinates of the stationary points on the curve y=x3−12xy = x^3 - 12x.
Show worked solution

Worked solution

  1. Write down the curve

    y=x3−12xy = x^{3} - 12x

    Stationary points are where the curve momentarily levels off, so the gradient there is zero.

  2. Differentiate to get the gradient function

    dydx=3x2−12\frac{dy}{dx} = 3x^{2} - 12

    Differentiate using the power rule; stationary points occur where this equals zero.

  3. Set the gradient to zero

    3x2−12=03x^{2} - 12 = 0

    At a stationary point the tangent is horizontal, so the gradient is 0. We solve this equation.

  4. Solve for the x-coordinates

    x=−2,  2x = -2 ,\; 2

    These x-values are where the curve has zero gradient.

  5. Find the y-coordinate for x = -2

    y=16y = 16

    Substitute this x-value into the original curve to find the height of the stationary point.

  6. Find the y-coordinate for x = 2

    y=−16y = -16

    Substitute this x-value into the original curve to find the height of the stationary point.

  7. Differentiate again for classification

    d2ydx2=6x\frac{d^2y}{dx^2} = 6x

    The second derivative tells us the type of each stationary point: positive means a minimum, negative means a maximum.

  8. Classify the point (-2, 16)

    d2ydx2=−12  ⇒  maximum\frac{d^2y}{dx^2} = -12\;\Rightarrow\;\text{maximum}

    Because the second derivative is negative here, this stationary point is a maximum.

  9. Classify the point (2, -16)

    d2ydx2=12  ⇒  minimum\frac{d^2y}{dx^2} = 12\;\Rightarrow\;\text{minimum}

    Because the second derivative is positive here, this stationary point is a minimum.

  10. State the stationary points

    (−2,16),  (2,−16)\left(-2, 16\right),\; \left(2, -16\right)

    These are the coordinates of the stationary points, together with their types.

  11. Write the gradient function

    dydx=3x2−12\frac{dy}{dx} = 3x^{2} - 12

    The derivative is called the gradient function because putting any x-value into it gives the gradient of the curve at that point.

  12. Check with a sample value

    At x=1:dydx=−9\text{At } x=1:\quad \frac{dy}{dx} = -9

    As a quick sanity check we substitute x = 1. This is exactly how the gradient function is used once we have it.

  13. Link the notation

    dydx=f′(x)\frac{dy}{dx} = f'(x)

    Remember from the introduction to calculus that dy/dx and f'(x) mean the same thing. Different exam boards use different notation for the derivative.

  14. Interpret the sign of the gradient

    dydx>0⇒increasing,dydx<0⇒decreasing\frac{dy}{dx} > 0 \Rightarrow \text{increasing}, \quad \frac{dy}{dx} < 0 \Rightarrow \text{decreasing}

    Where the gradient function is positive the curve goes uphill; where it is negative it goes downhill. This links the derivative back to the shape of the graph.

  15. Use the linearity of differentiation

    ddx(a f(x)+b g(x))=a f′(x)+b g′(x)\frac{d}{dx}\big(a\,f(x) + b\,g(x)\big) = a\,f'(x) + b\,g'(x)

    Differentiation can be done one term at a time, and constants stay multiplied on the front. That is why the working above splits so neatly.

Answer
(−2,16),  (2,−16)\left(-2, 16\right),\; \left(2, -16\right)
Question 3
6 markschallenging
Differentiate y=x4−xx2y = \dfrac{x^4 - x}{x^2}, giving your answer as powers of xx.
Show worked solution

Worked solution

  1. Identify what to differentiate

    y=x4−xx2y = \frac{x^{4}-x}{x^{2}}

    Write the function down clearly first, so we are certain exactly what we are differentiating.

  2. Recall the power rule

    ddx(xn)=n xn−1\frac{d}{dx}\left(x^{n}\right) = n\,x^{n-1}

    The power rule says: multiply by the old power, then subtract 1 from that power. It is the main tool for differentiating powers of x.

  3. Rewrite every term as a power of x

    x4−xx2=x2−x−1\frac{x^{4}-x}{x^{2}} = x^{2} - x^{-1}

    Using the index laws from earlier work, roots become fractional powers and fractions become negative powers. This lets us use the power rule on every term.

  4. Differentiate x^{2}

    ddx(x2)=2x\frac{d}{dx}\left(x^{2}\right) = 2x

    Multiply the term by its power 2 and then reduce the power by 1. Any number multiplying x is kept along for the ride.

  5. Differentiate -x^{-1}

    ddx(−x−1)=x−2\frac{d}{dx}\left(-x^{-1}\right) = x^{-2}

    Multiply the term by its power -1 and then reduce the power by 1. Any number multiplying x is kept along for the ride.

  6. Combine the differentiated terms

    dydx=2x+x−2\frac{dy}{dx} = 2x + x^{-2}

    Add the separate results back together. We can differentiate a sum term by term, which keeps the work simple.

  7. State the derivative

    dydx=2x+x−2\frac{dy}{dx} = 2x + x^{-2}

    This is the final gradient function for the curve.

  8. Write the gradient function

    dydx=2x+x−2\frac{dy}{dx} = 2x + x^{-2}

    The derivative is called the gradient function because putting any x-value into it gives the gradient of the curve at that point.

  9. Check with a sample value

    At x=1:dydx=3\text{At } x=1:\quad \frac{dy}{dx} = 3

    As a quick sanity check we substitute x = 1. This is exactly how the gradient function is used once we have it.

  10. Link the notation

    dydx=f′(x)\frac{dy}{dx} = f'(x)

    Remember from the introduction to calculus that dy/dx and f'(x) mean the same thing. Different exam boards use different notation for the derivative.

  11. Interpret the sign of the gradient

    dydx>0⇒increasing,dydx<0⇒decreasing\frac{dy}{dx} > 0 \Rightarrow \text{increasing}, \quad \frac{dy}{dx} < 0 \Rightarrow \text{decreasing}

    Where the gradient function is positive the curve goes uphill; where it is negative it goes downhill. This links the derivative back to the shape of the graph.

  12. Use the linearity of differentiation

    ddx(a f(x)+b g(x))=a f′(x)+b g′(x)\frac{d}{dx}\big(a\,f(x) + b\,g(x)\big) = a\,f'(x) + b\,g'(x)

    Differentiation can be done one term at a time, and constants stay multiplied on the front. That is why the working above splits so neatly.

  13. Try a second sample value

    At x=2:dydx=174\text{At } x=2:\quad \frac{dy}{dx} = \frac{17}{4}

    Substituting x = 2 gives a different gradient, showing that the steepness of the curve changes as we move along it.

  14. Recap the method

    xn  →  n xn−1x^{n} \;\to\; n\,x^{n-1}

    In words: bring the power down to the front as a multiplier, then reduce the power by one. Repeat this for every term.

  15. Relate the size to the steepness

    ∣dydx∣ large⇒steeper curve\left|\frac{dy}{dx}\right| \text{ large} \Rightarrow \text{steeper curve}

    The bigger the size of the gradient, the steeper the curve at that point. A gradient near zero means the curve is nearly flat there.

Answer
2x+x−22x + x^{-2}
Question 4
6 markschallenging
Differentiate y=x3+3xxy = \dfrac{x^3 + 3x}{\sqrt{x}}, giving your answer as powers of xx.
Show worked solution

Worked solution

  1. Identify what to differentiate

    y=x3+3xxy = \frac{x^{3}+3x}{\sqrt{x}}

    Write the function down clearly first, so we are certain exactly what we are differentiating.

  2. Recall the power rule

    ddx(xn)=n xn−1\frac{d}{dx}\left(x^{n}\right) = n\,x^{n-1}

    The power rule says: multiply by the old power, then subtract 1 from that power. It is the main tool for differentiating powers of x.

  3. Rewrite every term as a power of x

    x3+3xx=x52+3x12\frac{x^{3}+3x}{\sqrt{x}} = x^{\frac{5}{2}} + 3x^{\frac{1}{2}}

    Using the index laws from earlier work, roots become fractional powers and fractions become negative powers. This lets us use the power rule on every term.

  4. Differentiate x^{frac{5}{2}}

    ddx(x52)=52x32\frac{d}{dx}\left(x^{\frac{5}{2}}\right) = \frac{5}{2}x^{\frac{3}{2}}

    Multiply the term by its power \frac{5}{2} and then reduce the power by 1. Any number multiplying x is kept along for the ride.

  5. Differentiate 3x^{frac{1}{2}}

    ddx(3x12)=32x−12\frac{d}{dx}\left(3x^{\frac{1}{2}}\right) = \frac{3}{2}x^{-\frac{1}{2}}

    Multiply the term by its power \frac{1}{2} and then reduce the power by 1. Any number multiplying x is kept along for the ride.

  6. Combine the differentiated terms

    dydx=52x32+32x−12\frac{dy}{dx} = \frac{5}{2}x^{\frac{3}{2}} + \frac{3}{2}x^{-\frac{1}{2}}

    Add the separate results back together. We can differentiate a sum term by term, which keeps the work simple.

  7. State the derivative

    dydx=52x32+32x−12\frac{dy}{dx} = \frac{5}{2}x^{\frac{3}{2}} + \frac{3}{2}x^{-\frac{1}{2}}

    This is the final gradient function for the curve.

  8. Write the gradient function

    dydx=52x32+32x−12\frac{dy}{dx} = \frac{5}{2}x^{\frac{3}{2}} + \frac{3}{2}x^{-\frac{1}{2}}

    The derivative is called the gradient function because putting any x-value into it gives the gradient of the curve at that point.

  9. Check with a sample value

    At x=1:dydx=4\text{At } x=1:\quad \frac{dy}{dx} = 4

    As a quick sanity check we substitute x = 1. This is exactly how the gradient function is used once we have it.

  10. Link the notation

    dydx=f′(x)\frac{dy}{dx} = f'(x)

    Remember from the introduction to calculus that dy/dx and f'(x) mean the same thing. Different exam boards use different notation for the derivative.

  11. Interpret the sign of the gradient

    dydx>0⇒increasing,dydx<0⇒decreasing\frac{dy}{dx} > 0 \Rightarrow \text{increasing}, \quad \frac{dy}{dx} < 0 \Rightarrow \text{decreasing}

    Where the gradient function is positive the curve goes uphill; where it is negative it goes downhill. This links the derivative back to the shape of the graph.

  12. Use the linearity of differentiation

    ddx(a f(x)+b g(x))=a f′(x)+b g′(x)\frac{d}{dx}\big(a\,f(x) + b\,g(x)\big) = a\,f'(x) + b\,g'(x)

    Differentiation can be done one term at a time, and constants stay multiplied on the front. That is why the working above splits so neatly.

  13. Try a second sample value

    At x=2:dydx=2324\text{At } x=2:\quad \frac{dy}{dx} = \frac{23 \sqrt{2}}{4}

    Substituting x = 2 gives a different gradient, showing that the steepness of the curve changes as we move along it.

  14. Recap the method

    xn  →  n xn−1x^{n} \;\to\; n\,x^{n-1}

    In words: bring the power down to the front as a multiplier, then reduce the power by one. Repeat this for every term.

  15. Relate the size to the steepness

    ∣dydx∣ large⇒steeper curve\left|\frac{dy}{dx}\right| \text{ large} \Rightarrow \text{steeper curve}

    The bigger the size of the gradient, the steeper the curve at that point. A gradient near zero means the curve is nearly flat there.

Answer
52x32+32x−12\frac{5}{2}x^{\frac{3}{2}} + \frac{3}{2}x^{-\frac{1}{2}}
Question 5
6 markschallenging
Differentiate y=x (x+1)y = \sqrt{x}\,(x + 1), giving your answer as powers of xx.
Show worked solution

Worked solution

  1. Identify what to differentiate

    y=x (x+1)y = \sqrt{x}\,(x+1)

    Write the function down clearly first, so we are certain exactly what we are differentiating.

  2. Recall the power rule

    ddx(xn)=n xn−1\frac{d}{dx}\left(x^{n}\right) = n\,x^{n-1}

    The power rule says: multiply by the old power, then subtract 1 from that power. It is the main tool for differentiating powers of x.

  3. Rewrite every term as a power of x

    x (x+1)=x32+x12\sqrt{x}\,(x+1) = x^{\frac{3}{2}} + x^{\frac{1}{2}}

    Using the index laws from earlier work, roots become fractional powers and fractions become negative powers. This lets us use the power rule on every term.

  4. Differentiate x^{frac{3}{2}}

    ddx(x32)=32x12\frac{d}{dx}\left(x^{\frac{3}{2}}\right) = \frac{3}{2}x^{\frac{1}{2}}

    Multiply the term by its power \frac{3}{2} and then reduce the power by 1. Any number multiplying x is kept along for the ride.

  5. Differentiate x^{frac{1}{2}}

    ddx(x12)=12x−12\frac{d}{dx}\left(x^{\frac{1}{2}}\right) = \frac{1}{2}x^{-\frac{1}{2}}

    Multiply the term by its power \frac{1}{2} and then reduce the power by 1. Any number multiplying x is kept along for the ride.

  6. Combine the differentiated terms

    dydx=32x12+12x−12\frac{dy}{dx} = \frac{3}{2}x^{\frac{1}{2}} + \frac{1}{2}x^{-\frac{1}{2}}

    Add the separate results back together. We can differentiate a sum term by term, which keeps the work simple.

  7. State the derivative

    dydx=32x12+12x−12\frac{dy}{dx} = \frac{3}{2}x^{\frac{1}{2}} + \frac{1}{2}x^{-\frac{1}{2}}

    This is the final gradient function for the curve.

  8. Write the gradient function

    dydx=32x12+12x−12\frac{dy}{dx} = \frac{3}{2}x^{\frac{1}{2}} + \frac{1}{2}x^{-\frac{1}{2}}

    The derivative is called the gradient function because putting any x-value into it gives the gradient of the curve at that point.

  9. Check with a sample value

    At x=1:dydx=2\text{At } x=1:\quad \frac{dy}{dx} = 2

    As a quick sanity check we substitute x = 1. This is exactly how the gradient function is used once we have it.

  10. Link the notation

    dydx=f′(x)\frac{dy}{dx} = f'(x)

    Remember from the introduction to calculus that dy/dx and f'(x) mean the same thing. Different exam boards use different notation for the derivative.

  11. Interpret the sign of the gradient

    dydx>0⇒increasing,dydx<0⇒decreasing\frac{dy}{dx} > 0 \Rightarrow \text{increasing}, \quad \frac{dy}{dx} < 0 \Rightarrow \text{decreasing}

    Where the gradient function is positive the curve goes uphill; where it is negative it goes downhill. This links the derivative back to the shape of the graph.

  12. Use the linearity of differentiation

    ddx(a f(x)+b g(x))=a f′(x)+b g′(x)\frac{d}{dx}\big(a\,f(x) + b\,g(x)\big) = a\,f'(x) + b\,g'(x)

    Differentiation can be done one term at a time, and constants stay multiplied on the front. That is why the working above splits so neatly.

  13. Try a second sample value

    At x=2:dydx=724\text{At } x=2:\quad \frac{dy}{dx} = \frac{7 \sqrt{2}}{4}

    Substituting x = 2 gives a different gradient, showing that the steepness of the curve changes as we move along it.

  14. Recap the method

    xn  →  n xn−1x^{n} \;\to\; n\,x^{n-1}

    In words: bring the power down to the front as a multiplier, then reduce the power by one. Repeat this for every term.

  15. Relate the size to the steepness

    ∣dydx∣ large⇒steeper curve\left|\frac{dy}{dx}\right| \text{ large} \Rightarrow \text{steeper curve}

    The bigger the size of the gradient, the steeper the curve at that point. A gradient near zero means the curve is nearly flat there.

Answer
32x12+12x−12\frac{3}{2}x^{\frac{1}{2}} + \frac{1}{2}x^{-\frac{1}{2}}

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