A-Level First principles and derivatives Practice Questions

Free A-Level First principles and derivatives practice questions with full step-by-step worked solutions. Covers power-rule, differentiation, polynomial, constant. Practise exam-style problems and check your method.

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A-Level70 questionsStep-by-step solutions
Question 1
2 markseasy
Differentiate y=x5y = x^5 with respect to xx.
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Worked solution

  1. Identify what to differentiate

    y=x5y = x^{5}

    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. Differentiate x^{5}

    ddx(x5)=5x4\frac{d}{dx}\left(x^{5}\right) = 5x^{4}

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

  4. State the derivative

    dydx=5x4\frac{dy}{dx} = 5x^{4}

    This is the final gradient function for the curve.

  5. Write the gradient function

    dydx=5x4\frac{dy}{dx} = 5x^{4}

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

Answer
5x45x^{4}
Question 2
2 markseasy
Using the power rule, the derivative of xnx^n is:
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Worked solution

  1. State the power rule

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

    The power rule says to bring the power down as a multiplier, then reduce the power by one.

  2. Check the multiplier

    multiply by the old power n\text{multiply by the old power } n

    The old power n moves to the front and multiplies the term.

  3. Check the new power

    new power=n−1\text{new power} = n - 1

    We then subtract one from the power.

  4. State the result

    n xn−1n\,x^{n-1}

    Putting these together, the derivative of x^n is n x^(n-1).

Answer
nxn−1n x^{n-1}
Question 3
3 marksintermediate
For displacement ss as a function of time tt, what does dsdt\dfrac{ds}{dt} represent?
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Worked solution

  1. Recall that a derivative is a rate

    dsdt=rate of change of s\frac{ds}{dt} = \text{rate of change of } s

    A derivative measures how fast one quantity changes as another changes.

  2. Identify the quantities

    s=displacement,t=times = \text{displacement}, \quad t = \text{time}

    Here displacement s is changing as time t goes on.

  3. Interpret the rate

    dsdt=change in displacementchange in time\frac{ds}{dt} = \frac{\text{change in displacement}}{\text{change in time}}

    Change in displacement over change in time is exactly what we call velocity.

  4. Name it

    dsdt=velocity\frac{ds}{dt} = \text{velocity}

    So the rate of change of displacement with respect to time is the velocity.

  5. Contrast with acceleration

    dvdt=acceleration\frac{dv}{dt} = \text{acceleration}

    Differentiating again (velocity with respect to time) would give the acceleration, not this first step.

  6. Conclude

    dsdt=velocity\frac{ds}{dt} = \text{velocity}

    The derivative of displacement with respect to time is the velocity.

Answer
the velocity
Question 4
5 markshard
Differentiate f(x)=x3f(x) = x^3 from first principles.
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Worked solution

  1. Write the first principles definition

    f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

    First principles means finding the gradient straight from the definition, not the power rule. It measures the gradient of a chord as the two points squeeze together.

  2. Write down the function

    f(x)=x3f(x) = x^{3}

    This is the function we are differentiating from first principles.

  3. Work out f(x + h)

    f(x+h)=(x+h)3f(x+h) = (x+h)^{3}

    Replace every x in the function with (x + h). This is the value of the function a small step h further along.

  4. Expand the cube

    (x+h)3=x3+3x2h+3xh2+h3(x+h)^{3} = x^{3} + 3x^{2}h + 3xh^{2} + h^{3}

    Use the binomial expansion of a cube. Writing all four terms carefully now avoids slips later.

  5. Write f(x + h) in full

    f(x+h)=x3+3x2h+3xh2+h3f(x+h) = x^{3} + 3x^{2}h + 3xh^{2} + h^{3}

    Multiply the expansion by the coefficient so every term is written out ready to subtract.

  6. Subtract f(x)

    f(x+h)−f(x)=3x2h+3xh2+h3f(x+h) - f(x) = 3x^{2}h + 3xh^{2} + h^{3}

    Take away the original function. The x^{3} term cancels, leaving only the parts that still contain h.

  7. Factorise out h

    f(x+h)−f(x)=h(3x2+3xh+h2)f(x+h) - f(x) = h\left(3x^{2} + 3xh + h^{2}\right)

    Every remaining term has a factor of h, so we take it outside a bracket ready to cancel.

  8. Divide by h

    f(x+h)−f(x)h=3x2+3xh+h2\frac{f(x+h)-f(x)}{h} = 3x^{2} + 3xh + h^{2}

    Cancel the h on top with the h on the bottom. This is allowed because h is not yet zero.

  9. Take the limit as h tends to 0

    f′(x)=lim⁡h→0(3x2+3xh+h2)f'(x) = \lim_{h \to 0}\left(3x^{2} + 3xh + h^{2}\right)

    Now let h shrink to zero. Every term that still contains h disappears in the limit.

  10. State the derivative

    f′(x)=3x2f'(x) = 3x^{2}

    This is the gradient function found purely from first principles.

  11. Check with the power rule

    ddx(x3)=3x2\frac{d}{dx}\left(x^{3}\right) = 3x^{2}

    The quick power rule gives the same answer, confirming that our first-principles working is correct.

  12. Interpret the result

    f′(x)=3x2≥0 for all xf'(x) = 3x^{2} \ge 0 \text{ for all } x

    Because this gradient is never negative, the curve x^{3} never goes downhill. First principles has told us about the shape as well.

Answer
3x23x^{2}
Question 5
6 markschallenging
Find the equation of the tangent to the curve y=x3−xy = x^3 - x at the point where x=2x = 2.
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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

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