Lesson 01 · Unit 12 · Differentiation

What a derivative is

A straight line has one gradient. A curve has a different gradient at every point. This lesson builds the tool that measures it, then gives you the six derivatives you must know by heart.

You already know how to find the gradient of a straight line. Two points, rise over run, and the answer is the same wherever you measure it. That is what makes a line straight.

A curve is not like that. On \(y=x^{2}\) the curve is almost flat near the origin and climbing steeply by \(x=3\). So asking “what is the gradient of this curve” is not a fair question. You have to say where.

Measuring something we cannot measure yet

Fix a point \(P(1,1)\) on \(y=x^{2}\). We want the gradient of the curve exactly at P. We cannot do that directly — gradient needs two points, and we only have one.

So take a second point Q further along the curve and join them. The straight line PQ is called a chord, and its gradient we can work out, because now we have two points. Then slide Q toward P and watch what happens.

Interactive The chord closing onto the tangent

the curve \(y=x^{2}\)
the chord PQ (dashed) and the tangent at P (solid)
the value the chord gradient is heading for

Here are the numbers, written out. Each row moves Q closer to P.

Q at \(x=\)	\(\delta x\)	\(\delta y\)	gradient of chord \(=\dfrac{\delta y}{\delta x}\)
2	1	3	3
1.5	0.5	1.25	2.5
1.1	0.1	0.21	2.1
1.01	0.01	0.0201	2.01
1.001	0.001	0.002001	2.001

The gradients are not settling near 2 by luck. Here is why they must.

Worked example The chord gradient, in algebra

Let Q be the point on \(y=x^{2}\) with \(x = 1+h\), where \(h\) is any small number.

\(P = (1,\ 1) \qquad Q = \big(1+h,\ (1+h)^{2}\big)\)both points are on the curve
\(\delta x = (1+h) - 1 = h\)the run
\(\delta y = (1+h)^{2} - 1 = 1 + 2h + h^{2} - 1 = 2h + h^{2}\)the rise, expanded
\(\dfrac{\delta y}{\delta x} = \dfrac{2h + h^{2}}{h} = \dfrac{h(2+h)}{h} = 2 + h\)\(h \neq 0\), so we may cancel

So the gradient of the chord is \(2 + h\) — exactly, for every \(h\). Put \(h = 1\) and you get 3. Put \(h=0.01\) and you get 2.01. The table was never going to say anything else.

Now make \(h\) smaller and smaller. The chord gradient \(2+h\) gets as close to 2 as you like. We write this

\(\dfrac{\delta y}{\delta x} \to 2 \quad \text{as} \quad \delta x \to 0\)

and we say the gradient of the curve at P is 2.

Marks note

You will never be asked to do this in the exam. Differentiation from first principles is not on the 0606 syllabus. It is here so that you know what the number means — because in two lessons’ time you will be asked what a gradient of \(-3\) tells you about a curve, and a student who only learned the recipe cannot answer that.

Marks note

“The gradient of the curve at the point” and “the gradient of the tangent at the point” are the same number. Questions use both wordings and expect you to know they are the same.

Guided The same argument at a different point

Take the same curve \(y=x^{2}\), but this time fix P at \(x=3\), so \(P(3,9)\).

Let Q be at \(x=3.5\), so \(\delta x = 0.5\). Find \(\delta y\).

Now find the gradient of that chord.

Repeat with Q at \(x=3.1\). What is the chord gradient now?

In general the chord gradient here is \(6+h\). So what is the gradient of the curve at \(x=3\)?

At \(x=1\) the gradient was 2. At \(x=3\) it is 6. Both are double the \(x\)-value. Hold on to that — it is the whole of the next tab but one.

Fluency Chord gradients on \(y=x^{2}\) 12 questions

On \(y=x^{2}\), the chord from \(x=2\) to \(x=3\). What is its gradient?
On \(y=x^{2}\), the chord from \(x=2\) to \(x=2.5\).
On \(y=x^{2}\), the chord from \(x=2\) to \(x=2.1\).
On \(y=x^{2}\), the chord from \(x=2\) to \(x=2.01\).
Those four are closing in on something. What is the gradient of the tangent at \(x=2\)?
On \(y=x^{2}\), the chord from \(x=5\) to \(x=7\).
On \(y=x^{2}\), the chord from \(x=-1\) to \(x=3\).
On \(y=x^{2}\), the chord from \(x=-2\) to \(x=-1\).
On \(y=x^{2}\), \(x\) goes from \(3\) to \(3.2\). What is \(\delta y\)?
And what is the gradient of that chord?
On \(y=x^{2}\), the chord from \(x=4\) to \(x=4+h\) has gradient \(8+h\). What is it when \(h=0.5\)?
And as \(h\) shrinks towards zero?

Every chord on this curve has gradient equal to the sum of its two \(x\)-values — which is why the last one in each run is so easy to see coming.

Give the gradient of the chord as a decimal. Each one is the difference in \(y\) divided by the difference in \(x\), and nothing more than that.

We now have a number: the gradient of a curve at a point. It needs a name and a symbol, and the exam uses several. All of them mean the same thing.

The two main notations

If the curve is written \(y = \ldots\), the gradient function is written \(\dfrac{dy}{dx}\), said “dee y by dee x”.

If the same thing is written as a function, \(f(x) = \ldots\), the gradient function is written \(f'(x)\), said “f dashed of x”.

They are interchangeable. Which one appears depends only on how the question was phrased.

Symbol	Said aloud	What it means
\(\delta x\)	“delta x”	A small change in \(x\). One symbol, not \(\delta\) times \(x\).
\(\delta y\)	“delta y”	The change in \(y\) that \(\delta x\) causes.
\(\dfrac{\delta y}{\delta x}\)	“delta y over delta x”	Gradient of the chord. A real fraction.
\(\delta x \to 0\)	“as delta x tends to zero”	Shrink the chord. It never becomes zero.
\(\dfrac{dy}{dx}\)	“dee y by dee x”	Gradient of the curve: the value \(\dfrac{\delta y}{\delta x}\) heads for.
\(f'(x)\)	“f dashed of x”	The same thing, function notation.
\(\dfrac{d^{2}y}{dx^{2}}\)	“dee two y by dee x squared”	Differentiate, then differentiate the answer.
\(f''(x)\)	“f double dashed of x”	The same thing, function notation.

Marks note

\(\dfrac{d^{2}y}{dx^{2}}\) means differentiate twice. It does not mean \(\left(\dfrac{dy}{dx}\right)^{2}\). Squaring the first derivative instead of differentiating it again is a mistake that appears every session, and it destroys every mark that follows it.

The word “derivative”

\(\dfrac{dy}{dx}\) is called the derivative, or the derived function. Finding it is called differentiating. All three words appear in questions.

It is a function, not a number. Feed it an \(x\) and it gives you the gradient there. On \(y=x^{2}\) we found the derivative is \(2x\): at \(x=1\) it gives 2, at \(x=3\) it gives 6.

Four ways of asking for exactly the same thing

Find \(\dfrac{dy}{dx}\) when \(x=4\).
Find \(f'(4)\).
Find the gradient of the curve at the point where \(x=4\).
Find the gradient of the tangent to the curve at \(x=4\).

Recognising this is worth real marks, because a question that looks unfamiliar is usually a familiar one in different clothes.

Check yourself Which one is different?

A curve has equation \(y = f(x)\). Which of these is not asking for the same thing as the other three?

\(f'(2)\)
The value of \(\dfrac{dy}{dx}\) at \(x=2\)
\(f(2)\)
The gradient of the tangent at \(x=2\)

Marks note

Write what you mean. If your working line says \(y = 6x\) when you have just differentiated, you have claimed the curve is a straight line. Examiners follow your written statements, not your intentions.

Doing that chord argument for every curve would be unbearable. It has been done once, for each standard function, and the results are what you learn. Here is the first and most important.

If \(\quad y = x^{n} \quad\) then \(\quad \dfrac{dy}{dx} = n x^{\,n-1}\)

for any rational \(n\)

Said out loud: multiply by the power, then knock one off the power. Check it against what we already found: \(y=x^{2}\) gives \(\dfrac{dy}{dx} = 2x^{1} = 2x\). At \(x=1\) that is 2, at \(x=3\) it is 6. Both agree with the last tab.

Worked example Whole-number powers

\(y = x^{5} \ \Rightarrow\ \dfrac{dy}{dx} = 5x^{4}\)\(5\) down, power \(5-1=4\)
\(y = x^{7} \ \Rightarrow\ \dfrac{dy}{dx} = 7x^{6}\)the same two moves, every time
\(y = x^{1} = x \ \Rightarrow\ \dfrac{dy}{dx} = 1x^{0} = 1\)\(x^{0}=1\); the line \(y=x\) has gradient 1

One power at a time

Everything on this tab is a single power of \(x\) with nothing in front of it. Expressions like \(4x^{3}\), or \(x^{2} + 5x\), need two more rules — a number multiplying a function, and a sum of terms — and those are the next tab. Get this one rule solid first.

Fluency \(y = x^{n}\), find the gradient at a point 14 questions

\(y=x^{5}\). Find \(\dfrac{dy}{dx}\).
\(y=x^{9}\). Find \(\dfrac{dy}{dx}\).
\(y=x^{2}\). Find \(\dfrac{dy}{dx}\).
\(y=x\). Find \(\dfrac{dy}{dx}\).
\(y=x^{8}\). Find \(\dfrac{dy}{dx}\).
\(y=x^{7}\). Find the gradient at \(x=1\).
\(y=x^{4}\). Find the gradient at \(x=2\).
\(y=x^{3}\). Find the gradient at \(x=-2\).
\(y=x^{5}\). Find the gradient at \(x=-1\).
\(y=x^{6}\). Find the gradient at \(x=-1\).
\(y=x^{3}\). Find the gradient at \(x=3\).
\(y=x^{n}\) has \(\dfrac{dy}{dx} = 6x^{5}\). Find \(n\).
\(y=x^{4}\). At what value of \(x\) is the gradient \(32\)?
\(y=x^{12}\). At what value of \(x\) is the gradient \(12\)?

Nothing here has a number in front of it and nothing is a sum — those are two separate rules, and they are the next tab.

Watch the negative values of \(x\). A negative number to an even power is positive; to an odd power it stays negative.

Negative and fractional powers

The rule says any rational \(n\). That includes \(-3\) and \(\tfrac12\). The difficulty is never the differentiating — it is that the question hides the power inside a fraction or a root.

Rewrite first. Always.

As written	As a power
\(\dfrac{1}{x}\)	\(x^{-1}\)
\(\dfrac{1}{x^{2}}\)	\(x^{-2}\)
\(\dfrac{1}{x^{3}}\)	\(x^{-3}\)
\(\sqrt{x}\)	\(x^{1/2}\)
\(\dfrac{1}{\sqrt{x}}\)	\(x^{-1/2}\)
\(x\sqrt{x}\)	\(x^{3/2}\)
\(x^{2}\sqrt{x}\)	\(x^{5/2}\)

Worked example \(y = \dfrac{1}{x^{3}}\), find \(\dfrac{dy}{dx}\) at \(x=2\)

\(y = x^{-3}\)rewrite before differentiating
\(\dfrac{dy}{dx} = -3x^{-4}\)\(-3-1 = -4\), not \(-2\)
\(\dfrac{dy}{dx} = -\dfrac{3}{x^{4}}\)back to the form the question used
At \(x=2:\ -\dfrac{3}{16}\)\(2^{4}=16\)

Marks note

Subtracting one from a negative power moves it down: \(-3\) becomes \(-4\). Candidates who rush write \(-3x^{-2}\) and lose every mark after that line.

Check yourself Differentiate \(y=\sqrt{x}\)

\(\tfrac{1}{2}\sqrt{x}\)
\(\dfrac{1}{2\sqrt{x}}\)
\(\dfrac{1}{\sqrt{x}}\)
\(\tfrac{2}{3}x^{3/2}\)

Guided \(y = \dfrac{1}{x^{2}}\), at \(x = 2\)

One curve, four steps. The first two turn the fraction into something the power rule can touch, the third puts a number to it, and the fourth asks what that number actually tells you about the curve.

Write \(y\) as \(x^{p}\). What is \(p\)?

Differentiating gives \(\dfrac{dy}{dx} = q\,x^{-3}\). What is \(q\)?

Now find the value of \(\dfrac{dy}{dx}\) at \(x=2\).

Is the curve rising or falling at \(x = 2\)? Enter 1 for rising, 2 for falling.

Fluency Roots and reciprocals 12 questions

\(y=\sqrt{x}\). Find \(\dfrac{dy}{dx}\).
\(y=\dfrac{1}{x}\). Find \(\dfrac{dy}{dx}\).
\(y=\dfrac{1}{x^{3}}\). Find \(\dfrac{dy}{dx}\).
\(y=x^{-5}\). Find \(\dfrac{dy}{dx}\).
\(y=x\sqrt{x}\). Rewrite it as a power first, then find \(\dfrac{dy}{dx}\).
\(y=\dfrac{1}{\sqrt{x}}\). Find \(\dfrac{dy}{dx}\).
\(y=\sqrt{x}\). Find the gradient at \(x=9\).
\(y=\dfrac{1}{x}\). Find the gradient at \(x=2\).
\(y=\dfrac{1}{x^{2}}\). Find the gradient at \(x=1\).
Written as \(x^{n}\), what is \(n\) for \(\dfrac{1}{x^{4}}\)?
Written as \(x^{n}\), what is \(n\) for \(x^{2}\sqrt{x}\)?
\(y=\sqrt{x}\). At what value of \(x\) is the gradient \(\tfrac14\)?

Rewrite first, every time. A root is a fractional power and a reciprocal is a negative one; once they are written that way there is only the one rule to apply.

Answers may be fractions. Type them as decimals or as \(a/b\) — both are accepted. Rewrite as a power before you differentiate anything.

Error hunt One line here is wrong. Which?

Question. Given \(y = x\sqrt{x}\), find the value of \(\dfrac{dy}{dx}\) when \(x=4\).

\(y = x\sqrt{x}\)
\(y = x^{1/2}\)
\(\dfrac{dy}{dx} = \tfrac12 x^{-1/2}\)
At \(x=4:\quad \dfrac{dy}{dx} = \tfrac12 \times \dfrac{1}{2}\)
\(\dfrac{dy}{dx} = \dfrac{1}{4}\)

Click the line you think is wrong. Every line after the mistake is done correctly — the differentiating was never the problem. That is why the rewrite deserves as much care as the rule itself.

Real questions are not single powers. They are things like \(4x^{3}-7x^{2}+2x-9\). Two short rules let you handle all of them.

Constant multiple

A number multiplying the function just comes along for the ride: \(\dfrac{d}{dx}\big(c\,f(x)\big) = c\,f'(x)\). So \(y = 6x^{4}\) gives \(\dfrac{dy}{dx} = 6 \times 4x^{3} = 24x^{3}\).

Sums and differences

Differentiate a sum term by term: \(\dfrac{d}{dx}\big(u + v\big) = \dfrac{du}{dx} + \dfrac{dv}{dx}\). Subtraction works the same way.

Marks note

Term by term works for sums only. It does not work for products or quotients: \(\dfrac{d}{dx}(x^{2}\sin x)\) is not \(2x\cos x\). Those need the product and quotient rules, in lesson 02. Differentiating a product term by term is the single most common way of losing every mark on a question you understood.

Worked example \(y = 4x^{3} - 7x^{2} + 2x - 9\)

\(\dfrac{d}{dx}\big(4x^{3}\big) = 12x^{2}\)\(4 \times 3 = 12\), power drops to 2
\(\dfrac{d}{dx}\big(-7x^{2}\big) = -14x\)the sign travels with the term
\(\dfrac{d}{dx}\big(2x\big) = 2\)gradient of \(y=2x\) is 2
\(\dfrac{d}{dx}\big(-9\big) = 0\)a constant has no gradient
\(\dfrac{dy}{dx} = 12x^{2} - 14x + 2\)put the pieces back together
At \(x=2:\ 48 - 28 + 2 = 22\)substitute only at the end

Marks note

The \(-9\) disappears completely. Sliding a curve up or down does not change its steepness anywhere, so the constant cannot survive. Every year candidates carry it through as \(-9\).

Guided \(y = 5x^{4} - 3x^{2} + 6\)

Differentiating \(5x^{4}\) gives \(a\,x^{3}\). What is \(a\)?

What does the \(+6\) become?

Find the value of \(\dfrac{dy}{dx}\) at \(x=1\).

And at \(x=-1\)?

Fluency Cubics — gradient at a point 16 questions

\(y=3x^{2}+5x\). Find \(\dfrac{dy}{dx}\).
\(y=x^{3}-4x^{2}+7\). Find \(\dfrac{dy}{dx}\).
\(y=2x^{4}-3x\). Find \(\dfrac{dy}{dx}\).
\(y=5x^{3}+2x^{2}-x+9\). Find \(\dfrac{dy}{dx}\).
\(y=7\). Find \(\dfrac{dy}{dx}\).
\(y=6x-4\). Find \(\dfrac{dy}{dx}\).
\(y=x(x-4)\). Multiply out first.
\(y=(2x+1)(x-3)\). Multiply out first.
\(y=4x^{3}\). Find the gradient at \(x=2\).
\(y=x^{3}-6x\). Find the gradient at \(x=1\).
\(y=2x^{3}+3x^{2}-12x\). Find the gradient at \(x=0\).
\(y=x^{3}+x^{2}+x+1\). Find the gradient at \(x=-1\).
\(y=2x^{3}-9x^{2}+12x\). Find the gradient at \(x=1\).
\(y=ax^{3}\) has \(\dfrac{dy}{dx} = 15x^{2}\). Find \(a\).
\(y=x^{3}-3x^{2}\). At what positive value of \(x\) is the gradient zero?
\(y=x^{2}-6x+4\). At what value of \(x\) is the gradient zero?

Differentiating twice

\(\dfrac{dy}{dx}\) is itself a function of \(x\), so it can be differentiated too. The result is the second derivative, \(\dfrac{d^{2}y}{dx^{2}}\) or \(f''(x)\). You will need it in lesson 04 to decide whether a stationary point is a maximum or a minimum, so learn the mechanics now while there is nothing else to think about.

Worked example \(y = 2x^{3} - 5x^{2} + x\)

\(\dfrac{dy}{dx} = 6x^{2} - 10x + 1\)differentiate once
\(\dfrac{d^{2}y}{dx^{2}} = 12x - 10\)differentiate that answer
At \(x=3:\ 36 - 10 = 26\)substitute into the second derivative

Notice what happens to the \(+x\): it becomes \(+1\) on the first differentiation, and that constant then vanishes on the second. Each differentiation strips one more term from the bottom.

Fluency Second derivatives 10 questions

\(y=x^{5}\). Find \(\dfrac{dy}{dx}\) first, then \(\dfrac{d^{2}y}{dx^{2}}\).
\(y=3x^{4}-2x^{2}\). Find \(\dfrac{d^{2}y}{dx^{2}}\).
\(y=x^{3}-6x^{2}+5x\). Find \(\dfrac{d^{2}y}{dx^{2}}\).
\(y=5x^{2}+3x+1\). Find \(\dfrac{d^{2}y}{dx^{2}}\).
\(\dfrac{dy}{dx} = 4x^{3}-6x\). Find \(\dfrac{d^{2}y}{dx^{2}}\).
\(y=2x^{5}\). Find \(\dfrac{d^{2}y}{dx^{2}}\) at \(x=1\).
\(y=x^{3}\). Find \(\dfrac{d^{2}y}{dx^{2}}\) at \(x=2\).
\(y=x^{4}-4x^{3}\). Find \(\dfrac{d^{2}y}{dx^{2}}\) at \(x=1\).
\(y=x^{3}-3x\). At what value of \(x\) is \(\dfrac{d^{2}y}{dx^{2}} = 0\)?
Differentiating \(y=x^{n}\) twice gives \(6x\). Find \(n\).

Differentiating twice is not a new rule — it is the same rule, used again on whatever the first one produced.

Error hunt Now that multiples and sums are allowed

Question. Given \(y = 2\sqrt{x} + \dfrac{5}{x}\), find the value of \(\dfrac{dy}{dx}\) when \(x=4\).

\(y = 2\sqrt{x} + \dfrac{5}{x}\)
\(y = 2x^{1/2} + 5x^{-1}\)
\(\dfrac{dy}{dx} = 2 \times \tfrac12 x^{1/2} - 5x^{-2}\)
\(\dfrac{dy}{dx} = x^{1/2} - 5x^{-2}\)
At \(x=4:\quad \dfrac{dy}{dx} = 2 - \dfrac{5}{16}\)
\(\dfrac{dy}{dx} = \dfrac{27}{16}\)

Click the line you think is wrong. Notice that every line after the mistake is arithmetically correct — that is exactly why this kind of error survives a re-read.

Five more derivatives complete the list. With \(x^{n}\) that makes six, and those six are the whole of what you are required to know by heart for Unit 12.

Marks note · read this one twice

Every derivative below assumes \(x\) is in radians. A calculator left in degrees gives wrong answers that look perfectly reasonable, so nothing warns you. In this topic that single setting has cost more marks than any other mistake.

\(y\)	\(\dfrac{dy}{dx}\)	Watch for
\(x^{n}\)	\(n x^{\,n-1}\)	Any rational \(n\). Rewrite roots and fractions first.
\(\sin x\)	\(\cos x\)	No sign change.
\(\cos x\)	\(-\sin x\)	The minus sign. See below.
\(\tan x\)	\(\sec^{2} x\)	\(\sec x = \dfrac{1}{\cos x}\), so \(\sec^{2}x = \dfrac{1}{\cos^{2}x}\).
\(e^{x}\)	\(e^{x}\)	Unchanged. The only function that does this.
\(\ln x\)	\(\dfrac{1}{x}\)	The \(\ln\) disappears entirely.

Versions with something inside, such as \(\sin(3x+1)\) or \(e^{2x}\), need the chain rule. That is lesson 02. For now, the inside is always just \(x\).

Marks note

\(\cos\) differentiates to minus \(\sin\). Dropping that minus is the most frequently made error in the whole of Unit 12. It is worth saying the pair out loud every time: “sin gives cos, cos gives minus sin.”

Marks note

\(e^{x}\) is its own derivative, so \(5e^{x}\) differentiates to \(5e^{x}\) — the 5 stays. But \(e^{2x}\) does not differentiate to \(e^{2x}\), and \(2^{x}\) is not \(e^{x}\) at all. Only \(e^{x}\) itself has this property.

Worked example Three of them together

(a) \(y = 3\sin x + 2\cos x\). Find \(\dfrac{dy}{dx}\), and its value at \(x=0\).

\(\dfrac{dy}{dx} = 3\cos x - 2\sin x\)the minus comes from \(\cos\)
At \(x=0:\ 3(1) - 2(0) = 3\)\(\cos 0 = 1\), \(\sin 0 = 0\)

(b) \(y = 5e^{x} - 4\ln x\). Find \(\dfrac{dy}{dx}\), and its value at \(x=1\), to 3 significant figures.

\(\dfrac{dy}{dx} = 5e^{x} - \dfrac{4}{x}\)\(\ln x \to \frac1x\), and the 4 stays
At \(x=1:\ 5e - 4\)\(e^{1}=e\)
\(= 9.59\)\(5(2.71828) - 4 = 9.5914\)

(c) \(y = 2\tan x\). Find \(\dfrac{dy}{dx}\) at \(x = \tfrac{\pi}{4}\).

\(\dfrac{dy}{dx} = 2\sec^{2} x\)the 2 is a constant multiple
\(\cos\tfrac{\pi}{4} = \tfrac{1}{\sqrt2}\ \Rightarrow\ \sec^{2}\tfrac{\pi}{4} = 2\)\(\sec^{2} = 1/\cos^{2}\)
\(\dfrac{dy}{dx} = 4\)\(2 \times 2\)

Check yourself Differentiate \(y = \ln x + e^{x}\)

\(\dfrac{1}{x} + x e^{x-1}\)
\(x + e^{x}\)
\(\dfrac{1}{x} + e^{x}\)
\(\dfrac{1}{x} + 1\)

Guided \(y = 6\sin x - 2e^{x}\) at \(x = 0\)

Differentiate the \(6\sin x\) term and put \(x=0\). What value do you get?

Now the \(-2e^{x}\) term at \(x=0\). What value?

So what is \(\dfrac{dy}{dx}\) at \(x=0\)?

Fluency The five standard derivatives 16 questions

\(y=\sin x\). Find \(\dfrac{dy}{dx}\).
\(y=\cos x\). Find \(\dfrac{dy}{dx}\).
\(y=\tan x\). Find \(\dfrac{dy}{dx}\).
\(y=e^{x}\). Find \(\dfrac{dy}{dx}\).
\(y=\ln x\). Find \(\dfrac{dy}{dx}\).
\(y=3\sin x\). Find \(\dfrac{dy}{dx}\).
\(y=5e^{x}-2\ln x\). Find \(\dfrac{dy}{dx}\).
\(y=4\cos x + x^{2}\). Find \(\dfrac{dy}{dx}\).
\(y=\sin x\). Find the gradient at \(x=0\).
\(y=\cos x\). Find the gradient at \(x=0\).
\(y=e^{x}\). Find the gradient at \(x=0\).
\(y=\ln x\). Find the gradient at \(x=2\).
\(y=\tan x\). Find the gradient at \(x=0\).
\(y=2\sin x + 3\cos x\). Find the gradient at \(x=0\).
\(y=k\ln x\) has gradient \(4\) at \(x=1\). Find \(k\).
\(y=e^{x}-x\). At what value of \(x\) is the gradient zero?

Radians throughout. Type \(\sec^{2}x\) as sec(x)^2, \(e^{x}\) as e^x and \(\ln x\) as ln(x).

Every angle is in radians. \(\cos 0 = 1\), \(\sin 0 = 0\), \(\cos\pi = -1\), \(\sin\tfrac{\pi}{2} = 1\), \(e^{0} = 1\).

Error hunt Find the lost mark

Question. Given \(y = 3\cos x + 4\ln x\), find \(\dfrac{dy}{dx}\) when \(x = 2\), correct to 3 significant figures.

\(y = 3\cos x + 4\ln x\)
\(\dfrac{dy}{dx} = 3\sin x + \dfrac{4}{x}\)
At \(x = 2:\quad \dfrac{dy}{dx} = 3\sin 2 + \dfrac{4}{2}\)
\(= 3(0.9093) + 2\)
\(= 4.7279\)
\(= 4.73\) (3 s.f.)

Four questions in the style of the papers. Write out full solutions before you reveal mine — reading a solution is not the same as producing one, and only one of those two earns marks in November.

Exam-style · Paper 1 No calculator

A curve has equation \(y = 2x^{3} - 5x^{2} + 4\).

(a)Find \(\dfrac{dy}{dx}\).[2]
(b)Find the gradient of the curve at the point where \(x=2\).[2]
(c)Find \(\dfrac{d^{2}y}{dx^{2}}\).[1]

A curve has equation \(y = \dfrac{3x+2}{\sqrt{x}}\) for \(x > 0\).

(a)Show that \(y = 3x^{1/2} + 2x^{-1/2}\).[1]
(b)Find \(\dfrac{dy}{dx}\).[3]
(c)Hence find the value of \(\dfrac{dy}{dx}\) when \(x=4\).[2]

Exam-style · Paper 2 Scientific calculator, radians

A curve has equation \(y = 4\sin x + e^{x}\).

(a)Find \(\dfrac{dy}{dx}\).[2]
(b)Find the value of \(\dfrac{dy}{dx}\) when \(x = 1.2\), giving your answer correct to 3 significant figures.[2]

A function is defined by \(f(x) = x^{3} - 6x^{2} + 9x + 1\).

(a)Find \(f'(x)\).[2]
(b)Solve the equation \(f'(x) = 0\).[3]
(c)Find \(f''(x)\), and its value at each solution of part (b).[3]

Error hunt A harder one — the notation itself is misread

Question. Given \(y = x^{4} - 2x^{3}\), find the value of \(\dfrac{d^{2}y}{dx^{2}}\) when \(x=1\).

\(\dfrac{dy}{dx} = 4x^{3} - 6x^{2}\)
\(\dfrac{d^{2}y}{dx^{2}} = \left(4x^{3} - 6x^{2}\right)^{2}\)
\(= 16x^{6} - 48x^{5} + 36x^{4}\)
At \(x=1:\quad 16 - 48 + 36\)
\(= 4\)

Before you move on to lesson 02

You should now be able to

Explain what \(\dfrac{dy}{dx}\) means, using a chord and a tangent
Read and write \(f'(x)\), \(\dfrac{dy}{dx}\), \(f''(x)\), \(\dfrac{d^{2}y}{dx^{2}}\), \(\delta x\), \(\delta x \to 0\)
Differentiate \(x^{n}\) for any rational \(n\), rewriting roots and fractions first
Differentiate sums, differences and constant multiples
Write down the derivatives of \(\sin x\), \(\cos x\), \(\tan x\), \(e^{x}\) and \(\ln x\) without looking
Find a second derivative
Substitute a value and evaluate a gradient, in radians

If any line there is shaky, go back to that tab and work its questions again until it is not. Lesson 02 assumes every one of these is automatic.

The marks notes from this lesson, in one place

First principles is not examined — but the meaning of the derivative is.
Gradient of the curve at a point = gradient of the tangent there.
\(\dfrac{d^{2}y}{dx^{2}}\) means differentiate twice, not square the first derivative.
Rewrite roots and fractions as powers before differentiating.
Subtracting one from a negative power moves it down: \(-3 \to -4\).
Term by term works for sums only — never for products or quotients.
A constant term differentiates to zero and disappears.
\(\cos x\) gives minus \(\sin x\).
\(e^{x}\) is its own derivative, but \(e^{2x}\) is not.
Radians, always. Check the calculator mode before you start.