P1.6 Differentiation
⤓ Presentation⤓ PowerPointPresentation: opens offline in any browser, simulations live. PowerPoint: editable.
This page covers what a derivative is, differentiation from first principles, the power rule for any rational index, second derivatives, rates of change, and the equations of tangents and normals (Pure Mathematics 1, AS). Stationary points and increasing and decreasing functions are in P2; the chain, product and quotient rules and the derivatives of \(e^x\), \(\ln x\) and trigonometric functions are in P3. Here every expression must be turned into a sum of powers of \(x\) first.
What this page covers
7 points
- The derivative as the gradient of the tangent, as the limit of chord gradients, and as a rate of change
- Differentiation from first principles for simple polynomials
- Differentiating \(x^n\) for any rational \(n\), constant multiples and sums
- Rewriting products, roots and simple fractions as sums of powers before differentiating
- Second derivatives, \(\dfrac{d^2y}{dx^2}\) and \(f''(x)\)
- Equations of tangents and normals
- Points where the gradient takes a given value
In one minute
6 points
- The derivative \(\dfrac{dy}{dx}\) (or \(f'(x)\)) is the gradient of the tangent to the curve at a point: the limit of the gradients of chords as the second point slides towards the first.
- First principles: \(f'(x)=\displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\).
- Power rule: if \(y=ax^n\) then \(\dfrac{dy}{dx}=anx^{n-1}\), for any rational \(n\). Constants differentiate to 0; sums are differentiated term by term.
- Prepare first: write roots as fractional powers, \(\dfrac{1}{x^k}\) as \(x^{-k}\), expand brackets and split fractions.
- The second derivative \(\dfrac{d^2y}{dx^2}\) is the rate of change of the gradient.
- Tangent at \((x_1, y_1)\): gradient \(m=\dfrac{dy}{dx}\) there. Normal: gradient \(-\dfrac1m\). Then use \(y-y_1=m(x-x_1)\).
Contents
14 sections
- What does a derivative measure?
- How do you differentiate from first principles?
- How do you differentiate xⁿ when n is negative or a fraction?
- What must you do to products and fractions before differentiating?
- What is the second derivative?
- How is a derivative a rate of change?
- How do you find the equation of a tangent or a normal?
- How do you find the points where the gradient has a given value?
- Exam tips for P1.6
- Common misconceptions
- How this topic connects
- Key formulae
- Check your understanding
- Specification checklist
What does a derivative measure?
A straight line has one gradient everywhere. A curve does not: its steepness changes from point to point. So we need to say what "the gradient of a curve at a point" means.
To find it, take a second point \(Q\) on the curve and work out the gradient of the chord \(PQ\). As \(Q\) moves towards \(P\), the chord turns towards the tangent, and its gradient approaches a limit. That limit is the gradient of the curve at \(P\).
The function that gives the gradient at every \(x\) is the derivative (or gradient function). If \(y=f(x)\), the derivative is written \(\dfrac{dy}{dx}\) or \(f'(x)\).
| \(Q\) has \(x\)-coordinate \(1+h\) | \(h\) | Gradient of \(PQ=\dfrac{(1+h)^2-1}{h}\) |
|---|---|---|
| \(Q_1\) | 2 | 4 |
| \(Q_2\) | 1 | 3 |
| \(Q_3\) | 0.5 | 2.5 |
| — | 0.1 | 2.10 |
| — | 0.01 | 2.010 |
Algebraically, \(\dfrac{(1+h)^2-1}{h}=\dfrac{2h+h^2}{h}=2+h\), which tends to 2 as \(h\to0\). The table is just this formula with numbers in. You cannot put \(h=0\) straight into the chord gradient: it would be \(\tfrac00\). You must simplify first, then let \(h\) tend to zero. That is the whole idea of the next section.
How do you differentiate from first principles?
\[f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\]Here \(f(x+h)-f(x)\) is the change in \(y\) between the two points, \(h\) is the change in \(x\), and the fraction is the gradient of the chord. This definition is not in the formulae list: learn it, because you need to use it, not just quote it. Some books write \(\delta x\) for \(h\), giving \(\dfrac{dy}{dx}=\displaystyle\lim_{\delta x\to0}\frac{\delta y}{\delta x}\). Either notation is fine if you use it consistently.
The method for a polynomial always has the same four steps:
- Substitute \(x+h\) into \(f\) and expand.
- Subtract \(f(x)\): every term without an \(h\) cancels.
- Divide by \(h\): every remaining term had at least one factor of \(h\).
- Let \(h\to0\): terms still containing \(h\) vanish, and what is left is \(f'(x)\).
Given that \(f(x)=3x^2-4x\), prove from first principles that \(f'(x)=6x-4\).
Substitute and expand. \(f(x+h)=3(x+h)^2-4(x+h)=3x^2+6xh+3h^2-4x-4h\).
Subtract \(f(x)\). \(f(x+h)-f(x)=6xh+3h^2-4h\).
Divide by \(h\) (\(h\ne0\)). \(\dfrac{f(x+h)-f(x)}{h}=6x+3h-4\).
Take the limit and conclude. \(f'(x)=\displaystyle\lim_{h\to0}(6x+3h-4)=\) \(6x-4\), as required
As a check, at \(x=2\) the gradient is \(6\times2-4=8\). The power rule in the next section gives the same answer in one line, but a "first principles" question asks for this method, not the rule.
Writing "\(=6x+3h-4=6x-4\)" with no limit. The two expressions are not equal; the second is the limit of the first. Write \(\displaystyle\lim_{h\to0}\) (or "as \(h\to0\)") on the line where \(h\) disappears.
How do you differentiate xⁿ when n is negative or a fraction?
\[y=ax^n\quad\Longrightarrow\quad\frac{dy}{dx}=anx^{n-1}\]Here \(a\) is any constant and \(n\) is any rational number (positive, negative or a fraction): "multiply by the power, then reduce the power by one". The rule is not in the formulae list, so learn it. Also, the derivative of a constant is 0 (its graph is a horizontal line), and a sum or difference is differentiated term by term.
The rule only works on terms written as \(ax^n\). Use the laws of indices from P1.1 to rewrite first:
| Term as given | Rewritten as \(ax^n\) | Derivative |
|---|---|---|
| \(\sqrt{x}\) | \(x^{\frac12}\) | \(\tfrac12x^{-\frac12}=\dfrac{1}{2\sqrt x}\) |
| \(\dfrac{5}{x}\) | \(5x^{-1}\) | \(-5x^{-2}\) |
| \(\dfrac{3}{x^4}\) | \(3x^{-4}\) | \(-12x^{-5}\) |
| \(\dfrac{2}{\sqrt x}\) | \(2x^{-\frac12}\) | \(-x^{-\frac32}\) |
| \(x\sqrt x\) | \(x^{\frac32}\) | \(\tfrac32x^{\frac12}\) |
| \(\sqrt[3]{x^2}\) | \(x^{\frac23}\) | \(\tfrac23x^{-\frac13}\) |
| \(\dfrac{1}{4x^2}\) | \(\tfrac14x^{-2}\) | \(-\tfrac12x^{-3}\) |
Given that \(y=4x^3-6\sqrt x+\dfrac{8}{x^2}-7\), \(x>0\), (a) find \(\dfrac{dy}{dx}\); (b) find the gradient of the curve at the point where \(x=4\).
(a) Rewrite every term as a power of \(x\): \(y=4x^3-6x^{\frac12}+8x^{-2}-7\).
Differentiate term by term (the constant \(-7\) gives 0): \(\dfrac{dy}{dx}=12x^2-3x^{-\frac12}-16x^{-3}\)
(b) Substitute \(x=4\), using \(4^{-\frac12}=\tfrac12\) and \(4^{-3}=\tfrac{1}{64}\): \(12\times16-3\times\tfrac12-16\times\tfrac{1}{64}=192-1.5-0.25=\) \(190.25\) \(\left(=\tfrac{761}{4}\right)\)
Reducing a negative power the wrong way. The power of \(8x^{-2}\) goes down by one, to \(-3\), not up to \(-1\). And \(-6x^{\frac12}\) becomes \(-6\times\tfrac12x^{-\frac12}=-3x^{-\frac12}\): the new power is \(\tfrac12-1=-\tfrac12\).
What must you do to products and fractions before differentiating?
In P1 there is no product rule and no quotient rule (those come in P3). So a product or quotient must be turned into a sum of separate powers of \(x\) first.
- Products: multiply out the brackets completely.
- Fractions with a single term underneath (such as \(x^2\), \(\sqrt x\) or \(3x\)): divide every term on top by the denominator, then simplify each power with the index laws.
- Never differentiate the top and bottom separately, and never differentiate the two brackets separately and multiply: the derivative of a product is not the product of the derivatives.
Find \(\dfrac{dy}{dx}\) when (a) \(y=(2x-1)(x^2+3)\); (b) \(y=\dfrac{2x^3+x-6}{\sqrt x}\), \(x>0\), and evaluate it when \(x=4\).
(a) Expand, then differentiate: \(y=2x^3-x^2+6x-3\), so \(\dfrac{dy}{dx}=6x^2-2x+6\)
(b) Divide each term by \(x^{\frac12}\) (subtract \(\tfrac12\) from each power): \(y=2x^{\frac52}+x^{\frac12}-6x^{-\frac12}\).
Differentiate: \(\dfrac{dy}{dx}=5x^{\frac32}+\tfrac12x^{-\frac12}+3x^{-\frac32}\). Note that \(-6\times\left(-\tfrac12\right)=+3\): the sign of the last term changes because the power is negative.
At \(x=4\): \(4^{\frac32}=8\), \(4^{-\frac12}=\tfrac12\), \(4^{-\frac32}=\tfrac18\), so \(5\times8+\tfrac12\times\tfrac12+3\times\tfrac18=40+0.25+0.375=\) \(40.625\) \(\left(=\tfrac{325}{8}\right)\)
Many questions split this into two parts: "Write \(y\) in the form \(ax^p+bx^q\), where \(a\), \(b\), \(p\) and \(q\) are constants" and then "Hence find \(\dfrac{dy}{dx}\)". State the values of \(a\), \(b\), \(p\) and \(q\) clearly in the first part; the rewritten form is a step in its own right.
What is the second derivative?
The derivative \(\dfrac{dy}{dx}\) is itself a function of \(x\), so it can be differentiated again. The result is the second derivative, written \(\dfrac{d^2y}{dx^2}\) or \(f''(x)\).
- It measures the rate of change of the gradient: positive where the gradient is increasing, negative where it is decreasing.
- Same rules, applied twice. Simplify the first derivative before you differentiate again.
In the figures below, the gradient function \(f'(x)\) of a cubic is a parabola. Its own gradient, \(f''(x)=6x-12\), is zero at \(x=2\), where the gradient of the cubic is least.
\(f(x)=2x^3+\dfrac{24}{x}\), \(x\ne0\). (a) Find \(f'(x)\) and \(f''(x)\). (b) Find \(f'(2)\) and \(f''(2)\). (c) Find the exact coordinates of the points on the curve \(y=f(x)\) where the gradient is zero.
(a) Write \(\dfrac{24}{x}\) as \(24x^{-1}\) and differentiate twice: \(f'(x)=6x^2-24x^{-2}\), \(f''(x)=12x+48x^{-3}\)
(b) Substitute \(x=2\): \(f'(2)=24-6=18\) and \(f''(2)=24+6=30\). \(f'(2)=18\), \(f''(2)=30\)
(c) Solve \(f'(x)=0\), multiplying through by \(x^2\): \(6x^2=\dfrac{24}{x^2}\Rightarrow x^4=4\Rightarrow x^2=2\Rightarrow x=\pm\sqrt2\).
Find \(y\) for each, using \(\dfrac{1}{\sqrt2}=\dfrac{\sqrt2}{2}\): \(f(\sqrt2)=2\times2\sqrt2+\dfrac{24}{\sqrt2}=4\sqrt2+12\sqrt2=16\sqrt2\). \((\sqrt2, 16\sqrt2)\) and \((-\sqrt2, -16\sqrt2)\)
\(x^4=4\) also gives \(x^2=-2\), which has no real solutions, so there are exactly two points. Deciding which is a maximum and which a minimum (using the sign of \(f''\)) is a P2 skill.
How is a derivative a rate of change?
\(\dfrac{dy}{dx}\) tells you how fast \(y\) changes as \(x\) changes. With other letters the meaning is the same: if \(A\) is an area and \(t\) is time, \(\dfrac{dA}{dt}\) is the rate of change of area with respect to time.
- Units: (units of the top) per (unit of the bottom), for example m² per hour, cm³ s⁻¹, £ per item.
- Sign: a positive derivative means the quantity is increasing; a negative one means it is decreasing.
- The second derivative tells you whether the rate itself is going up or down: a growing quantity can be growing more slowly.
- In mechanics the same idea gives velocity \(v=\dfrac{ds}{dt}\) and acceleration \(a=\dfrac{dv}{dt}\).
The area, \(A\) m², of an oil slick \(t\) hours after a leak starts is modelled by \(A=40t^{\frac32}\), \(0\le t\le16\). (a) Find the rate at which the area is increasing when \(t=9\). (b) Find the time at which the area is increasing at 120 m² per hour. (c) Find \(\dfrac{d^2A}{dt^2}\) and explain what its sign tells you.
(a) Differentiate with respect to \(t\): \(\dfrac{dA}{dt}=40\times\tfrac32t^{\frac12}=60t^{\frac12}\). At \(t=9\): \(60\times3=\) 180 m² per hour
(b) \(60t^{\frac12}=120\Rightarrow t^{\frac12}=2\Rightarrow\) \(t=4\) hours
(c) Differentiate again: \(\dfrac{d^2A}{dt^2}=30t^{-\frac12}=\dfrac{30}{\sqrt t}\)
For \(t>0\) this is positive, so the rate of increase of the area is itself increasing: the slick spreads faster and faster during the model's 16 hours (at \(t=9\) the rate is rising by 10 m² per hour, every hour).
When a question says "find the rate of change", give units, and when it asks you to interpret a sign, write a sentence in context: "the volume of water is decreasing at 120 m³ per day", not just "it is negative".
How do you find the equation of a tangent or a normal?
Perpendicular gradients multiply to \(-1\), so if the tangent has gradient \(m\), the normal has gradient \(-\dfrac1m\). The method:
- Find the coordinates of \(P\) (if you are only given \(x\), substitute into the equation of the curve to get \(y\)).
- Differentiate, and substitute the \(x\)-coordinate of \(P\) to get the gradient \(m\) of the tangent.
- For a normal, take \(-\dfrac1m\).
- Use \(y-y_1=m(x-x_1)\), then rearrange into the form asked for.
Special case: if the tangent is horizontal (\(m=0\)), the normal is the vertical line \(x=x_1\).
The curve \(C\) has equation \(y=x+\dfrac{6}{x}\), \(x>0\). The point \(P\) on \(C\) has \(x\)-coordinate 2. (a) Find an equation of the tangent to \(C\) at \(P\), in the form \(ax+by+c=0\) where \(a\), \(b\) and \(c\) are integers. (b) Find an equation of the normal to \(C\) at \(P\). (c) The tangent and the normal meet the \(y\)-axis at \(A\) and \(B\) respectively. Find the area of triangle \(APB\).
(a) \(y=2+3=5\), so \(P(2, 5)\). \(\dfrac{dy}{dx}=1-6x^{-2}=1-\tfrac64=-\tfrac12\) at \(x=2\).
Tangent: \(y-5=-\tfrac12(x-2)\), so \(2y-10=-x+2\), giving \(x+2y-12=0\)
(b) Normal gradient \(=-1\div\left(-\tfrac12\right)=2\): \(y-5=2(x-2)\), so \(y=2x+1\)
(c) Put \(x=0\) in each line: \(A(0, 6)\) and \(B(0, 1)\), so \(AB=5\); the height of the triangle from \(P\) to the \(y\)-axis is 2. Area \(=\tfrac12\times5\times2=\) 5
Using \(AB\), which lies along the \(y\)-axis, as the base makes the height simply the \(x\)-coordinate of \(P\).
Using the derivative itself as the gradient. \(\dfrac{dy}{dx}=1-6x^{-2}\) is a function; the gradient of the tangent is the number you get after substituting \(x=2\). An "equation of the tangent" containing \(x^{-2}\) is not a straight line.
How do you find the points where the gradient has a given value?
This is the previous section in reverse: you know the gradient and want the point. Set \(\dfrac{dy}{dx}\) equal to the given value and solve for \(x\); then substitute each \(x\) into the equation of the curve to get \(y\).
- "The tangent is parallel to the line \(y=4x-1\)" means the gradient is 4. If the line is given as \(ax+by+c=0\), rearrange it to read off its gradient \(-\dfrac{a}{b}\).
- "The normal is parallel to …" means \(-1\div\dfrac{dy}{dx}\) equals that gradient: find the tangent gradient first.
- "The tangent is horizontal" or "parallel to the \(x\)-axis" means \(\dfrac{dy}{dx}=0\).
- The equation in \(x\) is often a quadratic, so expect two points, or none if the discriminant is negative.
The curve \(C\) has equation \(y=x^3-3x^2-5x+4\). Find the coordinates of the points on \(C\) at which the tangent is parallel to the line \(y=4x-1\), and find the equation of each of these tangents.
The gradient must equal 4: \(\dfrac{dy}{dx}=3x^2-6x-5=4\Rightarrow3x^2-6x-9=0\Rightarrow x^2-2x-3=0\).
Solve, then find \(y\) from the curve: \((x-3)(x+1)=0\), and \(y(3)=27-27-15+4=-11\), \(y(-1)=-1-3+5+4=5\). \((3, -11)\) and \((-1, 5)\)
Tangents with gradient 4: \(y+11=4(x-3)\Rightarrow y=4x-23\) and \(y-5=4(x+1)\Rightarrow y=4x+9\). \(y=4x-23\) and \(y=4x+9\)
The last example puts the whole page together, at the length of a full P1 question.
The curve \(C\) has equation \(y=\dfrac{x^2+16}{2\sqrt x}\), \(x>0\).
(a) Write \(y\) in the form \(ax^p+bx^q\), where \(a\), \(b\), \(p\) and \(q\) are constants.
(b) Hence find \(\dfrac{dy}{dx}\).
(c) Show that the point \(P(4, 8)\) lies on \(C\), and find an equation of the normal to \(C\) at \(P\).
(d) The normal meets the \(x\)-axis at \(Q\). Find the exact length of \(PQ\).
(e) Find the exact \(x\)-coordinate of the point on \(C\) at which the tangent is parallel to the \(x\)-axis.
(a) Divide each term by \(2x^{\frac12}\): \(y=\dfrac{x^2}{2x^{\frac12}}+\dfrac{16}{2x^{\frac12}}=\) \(\tfrac12x^{\frac32}+8x^{-\frac12}\) (\(a=\tfrac12\), \(p=\tfrac32\), \(b=8\), \(q=-\tfrac12\))
(b) \(\dfrac{dy}{dx}=\tfrac34x^{\frac12}-4x^{-\frac32}\)
(c) \(y=(16+16)\div(2\times2)=8\) ✓, so \(P\) lies on \(C\). Gradient at \(x=4\): \(\tfrac34\times2-4\times\tfrac18=1.5-0.5=1\). Normal gradient \(=-1\): \(y-8=-(x-4)\), so \(x+y-12=0\)
(d) \(Q\) has \(y=0\), so \(Q(12, 0)\). \(PQ=\sqrt{(12-4)^2+(0-8)^2}=\sqrt{128}=\) \(8\sqrt2\)
(e) Tangent parallel to the \(x\)-axis means \(\dfrac{dy}{dx}=0\); multiply by \(x^{\frac32}\): \(\tfrac34x^{\frac12}=4x^{-\frac32}\Rightarrow\tfrac34x^2=4\Rightarrow x^2=\tfrac{16}{3}\Rightarrow x=\dfrac{4}{\sqrt3}=\) \(\dfrac{4\sqrt3}{3}\)
\(x=-\dfrac{4\sqrt3}{3}\) is rejected because \(x>0\) (and \(\sqrt x\) would not be defined).
Exam tips for P1.6
Your first line should show every term as \(ax^n\). It is a step worth showing on its own, and it is where the most costly slips (\(\sqrt x\) written as \(x^2\) instead of \(x^{\frac12}\), or \(\dfrac{1}{x^2}\) as \(x^2\) instead of \(x^{-2}\)) become visible.
If the question says "from first principles", show \(f(x+h)\) expanded, the subtraction, the division by \(h\) and the limit, and finish with a concluding statement. Quoting the power rule does not answer the question.
Substitute the \(x\)-coordinate into the equation of the curve for \(y_1\), and into \(\dfrac{dy}{dx}\) for \(m\). Mixing them up is the most common way to get a wrong tangent from correct calculus.
"\(ax+by+c=0\) where \(a\), \(b\) and \(c\) are integers" needs all terms on one side and no fractions. Multiply through by the denominator of the gradient at the end, and check your line passes through the point.
Many scientific calculators can evaluate a derivative at a point. Use it to check a gradient you have found by hand; the algebra that produces \(\dfrac{dy}{dx}\) still has to be written out.
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| The derivative of \(\dfrac{1}{x^3}\) is \(\dfrac{1}{3x^2}\). | The power rule only applies to \(x^n\); \(\dfrac{1}{x^3}\) must first be written as \(x^{-3}\). | \(\dfrac{d}{dx}(x^{-3})=-3x^{-4}=-\dfrac{3}{x^4}\). |
| To differentiate \((x+2)(x-5)\), differentiate each bracket and multiply: \(1\times1=1\). | The derivative of a product is not the product of the derivatives. | Expand to \(x^2-3x-10\), then differentiate: \(2x-3\). |
| The derivative of a constant such as 7 is 7. | The graph of \(y=7\) is horizontal, so its gradient is 0 everywhere. | Constants differentiate to 0. A constant multiple stays: \(\dfrac{d}{dx}(7x)=7\). |
| The gradient of the normal is \(\dfrac1m\) (or \(-m\)). | Perpendicular gradients must multiply to \(-1\); neither \(\dfrac1m\) nor \(-m\) does in general. | Normal gradient \(=-\dfrac1m\): take the reciprocal and change the sign. |
| In first principles you can put \(h=0\) as soon as you have the fraction. | At \(h=0\) the fraction is \(\tfrac00\), which is undefined. | Cancel the factor \(h\) first; only then let \(h\to0\). |
| A negative derivative means the quantity is negative. | The derivative is a rate, not a value: a large positive quantity can be falling. | A negative rate of change means the quantity is decreasing. |
How this topic connects
P1.1 Indices, P1.2 Simultaneous equations and P1.4 Straight-line graphs
Rewriting roots and fractions uses the index laws from P1.1; tangents and normals use \(y-y_1=m(x-x_1)\) and \(m_1m_2=-1\) from P1.4. A line that touches a curve can also be found with the discriminant method of P1.2.
P1.7 Integration and stationary points (P2); P3 and P4
Integration reverses the power rule. In P2, \(\dfrac{dy}{dx}=0\) locates maxima and minima, the second derivative classifies them, and the sign of \(\dfrac{dy}{dx}\) tells you where a function is increasing. The chain, product and quotient rules in P3 remove the need to expand or divide first; P3 and P4 add exponentials, logarithms, trigonometric functions, and implicit and parametric differentiation.
Mechanics (M1, M2) and IAL Physics
Velocity is the rate of change of displacement and acceleration the rate of change of velocity; M2 differentiates expressions in \(t\) exactly as in the rates-of-change section. The gradient of any physics graph is the same idea as the tangent gradient here.
Key formulae
| Formula | Use | In the formulae list? |
|---|---|---|
| \(f'(x)=\displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\) | differentiation from first principles | No – learn it |
| \(y=ax^n\Rightarrow\dfrac{dy}{dx}=anx^{n-1}\) | every derivative in P1, for rational \(n\) | No – learn it |
| \(\dfrac{d}{dx}(\text{constant})=0\); \(\dfrac{d}{dx}(f\pm g)=f'\pm g'\) | constants and sums of terms | No – learn it |
| \(\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left(\dfrac{dy}{dx}\right)\) | second derivative; rate of change of the gradient | No – learn it |
| \(y-y_1=m(x-x_1)\) | equation of a tangent or normal through \((x_1, y_1)\) | No – learn it |
| \(m_{\text{normal}}=-\dfrac{1}{m_{\text{tangent}}}\) | gradient of a normal | No – derive it from \(m_1m_2=-1\) |
Check your understanding
1. Differentiate \(y=\dfrac{5}{2x^3}\).
Write it as \(y=\tfrac52x^{-3}\). Then \(\dfrac{dy}{dx}=\tfrac52\times(-3)x^{-4}=-\tfrac{15}{2}x^{-4}=\) \(-\dfrac{15}{2x^4}\).
2. Find \(f''(x)\) when \(f(x)=x^4-8\sqrt x\).
\(f'(x)=4x^3-4x^{-\frac12}\), so \(f''(x)=12x^2+2x^{-\frac32}\) (since \(-4\times-\tfrac12=+2\)).
3. Find the gradient of the normal to \(y=x^2-3x\) at the point where \(x=4\).
\(\dfrac{dy}{dx}=2x-3=5\) at \(x=4\), so the normal has gradient \(-\tfrac15\).
4. Explain, using first principles, why the derivative of a constant function \(f(x)=c\) is zero.
\(f(x+h)-f(x)=c-c=0\) for every \(h\), so the chord gradient is \(\tfrac{0}{h}=0\) and its limit is 0. Geometrically, \(y=c\) is a horizontal line.
5. Find the point on \(y=x^2-6x+1\) at which the gradient is 4.
\(\dfrac{dy}{dx}=2x-6=4\) gives \(x=5\); then \(y=25-30+1=-4\). The point is \((5, -4)\).
6. Find the equation of the tangent to \(y=\sqrt x\) at the point where \(x=9\).
\(y=3\), so the point is \((9, 3)\). \(\dfrac{dy}{dx}=\tfrac12x^{-\frac12}=\tfrac12\times\tfrac13=\tfrac16\). Tangent: \(y-3=\tfrac16(x-9)\), so \(y=\tfrac16x+\tfrac32\) (or \(x-6y+9=0\)).
Specification checklist
Tick every box before the exam. The wording is ours; the numbers refer to the P1 specification sections.
- (P1 4.1) I can explain the derivative as the gradient of the tangent to a curve, found as the limit of the gradients of chords.
- (P1 4.1) I can differentiate simple polynomials from first principles, writing the limit correctly and concluding.
- (P1 4.1) I can interpret a derivative as a rate of change in context, with units and the meaning of its sign.
- (P1 4.1) I can find and use second derivatives, written \(\dfrac{d^2y}{dx^2}\) or \(f''(x)\).
- (P1 4.2) I can differentiate \(ax^n\) for any rational \(n\), and sums, differences and constant multiples of such terms.
- (P1 4.2) I can rewrite roots, reciprocals, products of brackets and simple fractions as sums of powers of \(x\) before differentiating.
- (P1 4.3) I can find equations of the tangent and the normal to a curve at a given point, in any required form.
- (P1 4.3) I can find the points on a curve where the gradient takes a given value, including tangents or normals parallel to a given line.