What is the gradient of a curve at a point?

It is the gradient of the tangent to the curve at that point. Unlike a straight line, a curve's gradient is constantly changing, so it has to be quoted at a particular point.

True or False? A tangent to a curve can never cross that curve anywhere.

False. At its own point the tangent only touches the curve without cutting through it. Elsewhere it may well cross the curve, which does not stop it being the tangent at the original point.

True or False? The trigonometric derivatives only work when the angle is measured in radians.

True. Every calculus result for trigonometric functions assumes radians. The formulas look identical whichever unit you use, so working in degrees produces answers that are wrong without looking wrong.

Define the normal to a curve.

The normal at a point is the straight line through that point which is perpendicular to the tangent there. It therefore cuts across the curve rather than running alongside it.

Define a stationary point .

A stationary point is a point on a curve where the gradient is zero . The tangent there is horizontal, which is why the curve is momentarily neither rising nor falling.

True or False? If the second derivative is zero at a stationary point, the point is a point of inflection.

False. A zero second derivative settles nothing: the point could be a minimum, a maximum or a point of inflection. When that happens you have to fall back on checking the sign of the first derivative on each side.

How does the first derivative reveal the nature of a stationary point?

Work out its sign just either side of the point. Negative then positive gives a local minimum , and positive then negative gives a local maximum .

Define optimisation .

Optimisation is finding the maximum or minimum output of a function. In a modelling question that means the largest volume, the smallest cost, the least fuel used, and so on.

How do you solve an optimisation problem?

Form a formula for the quantity in terms of a single variable, then differentiate it. Set that derivative to zero and solve, since the optimum is a stationary point, and substitute back to get the optimal value itself.

What are connected rates of change?

Problems with more than two variables , where the chain rule links several rates into one equation. Typically you know two of the rates and want the third.

What is the clue that a question is about rates of change?

The word rate itself, together with a unit "per" something. That tells you a derivative is involved before you have worked out which one.

Why must you be clear which letters are variables and which are constants?

Because they behave completely differently under differentiation: a constant differentiates to zero, while a variable does not. These questions carry a lot of letters, so treating a fixed radius as a variable, or the reverse, changes the answer entirely.

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Differentiation (Cambridge (CIE) IGCSE Additional Maths): Flashcards

Exam code: 0606

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FrontIntroduction to Differentiation
What is the gradient of a curve at a point?

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Cards in this collection (67)

What is the gradient of a curve at a point?
It is the gradient of the tangent to the curve at that point.
Unlike a straight line, a curve's gradient is constantly changing, so it has to be quoted at a particular point.
True or False?
A tangent to a curve can never cross that curve anywhere.
False.
At its own point the tangent only touches the curve without cutting through it.
Elsewhere it may well cross the curve, which does not stop it being the tangent at the original point.
What does "rate of change" mean in a calculus question?
It means the gradient.
So "the rate of change of $y$ with respect to $x$ " is asking for $\frac{d y}{d x}$ .
Define the derivative of a function.
The derivative is a function giving the gradient of the original curve at any value of $x$ .
It is also called the gradient function, so the derivative of $y = x^{2}$ is $2 x$ , meaning the gradient at $x = 5$ is $10$ .
Complete the rule for differentiating a power of $x$ :
$y = a x^{n} \Rightarrow \frac{d y}{d x} =______$
The completed rule is:
$y = a x^{n} \Rightarrow \frac{d y}{d x} = a n x^{n - 1}$
In words: bring down the power, then subtract one from the power.
Complete the two special cases:
$y = a x \Rightarrow \frac{d y}{d x} =______$
$y = a \Rightarrow \frac{d y}{d x} =______$
The completed cases are:
$y = a x \Rightarrow \frac{d y}{d x} = a$
$y = a \Rightarrow \frac{d y}{d x} = 0$
A constant has zero gradient because its graph is a horizontal line.
How do you differentiate $y = \frac{4}{x}$ ?
Rewrite it as a negative power first, so $y = 4 x^{- 1}$ .
The power rule then gives $\frac{d y}{d x} = - 4 x^{- 2}$ .
Why can you not differentiate $y = (2 x - 3) (x^{2} - 4)$ term by term as it stands?
The power rule works on a sum or difference of powers of $x$ , not on a product.
Expand the brackets first to get $y = 2 x^{3} - 3 x^{2} - 8 x + 12$ , which can then be differentiated term by term.
How do you differentiate $y = \frac{8 x^{6} - x^{3}}{2 x^{4}}$ using only the power rule?
Split the fraction over the single denominator, then simplify each part with the laws of indices:
$y = 4 x^{2} - \frac{1}{2} x^{- 1}$
Each term is now a power of $x$ , so differentiating gives $8 x + \frac{1}{2} x^{- 2}$ .
Complete the three standard trigonometric derivatives:
$y = \sin x \Rightarrow \frac{d y}{d x} =______$
$y = \cos x \Rightarrow \frac{d y}{d x} =______$
$y = \tan x \Rightarrow \frac{d y}{d x} =______$
The completed derivatives are:
$y = \sin x \Rightarrow \frac{d y}{d x} = \cos x$
$y = \cos x \Rightarrow \frac{d y}{d x} = - \sin x$
$y = \tan x \Rightarrow \frac{d y}{d x} = \sec^{2} x$
Only the cosine picks up a minus sign.
True or False?
The trigonometric derivatives only work when the angle is measured in radians.
True.
Every calculus result for trigonometric functions assumes radians.
The formulas look identical whichever unit you use, so working in degrees produces answers that are wrong without looking wrong.
What is the derivative of $y = \sin (a x + b)$ ?
It is $\frac{d y}{d x} = a \cos (a x + b)$ .
The bracket is copied across unchanged and the whole thing is multiplied by $a$ , the derivative of the inside.
Differentiate $y = 3 \sin 4 x - \cos (2 x - 3)$ .
Differentiate term by term, multiplying each by the derivative of its bracket:
$\frac{d y}{d x} = 12 \cos 4 x + 2 \sin (2 x - 3)$
The second term becomes positive, because the minus from the cosine rule meets the minus already in front of it.
Complete the two standard derivatives:
$y = e^{x} \Rightarrow \frac{d y}{d x} =______$
$y = \ln x \Rightarrow \frac{d y}{d x} =______$
The completed derivatives are:
$y = e^{x} \Rightarrow \frac{d y}{d x} = e^{x}$
$y = \ln x \Rightarrow \frac{d y}{d x} = \frac{1}{x}$
The second holds only for $x > 0$ , since that is where $\ln x$ exists.
What are the derivatives of $e^{a x + b}$ and $\ln (a x + b)$ ?
They are $a e^{a x + b}$ and $\frac{a}{a x + b}$ .
In both cases you multiply by $a$ , the derivative of the linear bracket inside.
True or False?
The derivative of $\ln (k x)$ is $\frac{k}{x}$ .
False.
The rule gives $\frac{k}{k x}$ , and the $k$ cancels, so the derivative is simply $\frac{1}{x}$ .
Stretching the inside of a logarithm makes no difference to its gradient function.
A curve is $y = e^{- 3 x} + 2 \ln x$ . Find its gradient at $x = 2$ , in exact form.
Differentiating term by term gives $\frac{d y}{d x} = - 3 e^{- 3 x} + \frac{2}{x}$ .
Substituting $x = 2$ gives $- 3 e^{- 6} + 1$ , which is written $1 - 3 e^{- 6}$ and left unevaluated because an exact value is wanted.
Complete the chain rule, where $y$ is a function of $u$ and $u$ is a function of $x$ :
$\frac{d y}{d x} =______\times______$
The completed rule is:
$\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x}$
The $d u$ terms behave as though they cancel, which is a useful way to remember it.
How do you recognise that you need the chain rule?
You have a composite function, a function of a function, which shows up as the variable not appearing alone.
So $\sin x$ needs no chain rule, but $\sin (3 x + 2)$ does, because $x$ is tripled and shifted before the sine is applied.
How do you use the chain rule by substitution?
Let $u$ be the inner function, so that $y$ becomes a function of $u$ alone.
Differentiate each part separately to get $\frac{d y}{d u}$ and $\frac{d u}{d x}$ , multiply them, then substitute $u$ back in terms of $x$ .
What is the derivative of $y = (a x + b)^{n}$ ?
It is $\frac{d y}{d x} = a n (a x + b)^{n - 1}$ .
The bracket keeps its contents, the power drops by one, and the whole thing is multiplied by $a$ , the derivative of the bracket.
How do you differentiate $y = \sqrt{a x + b}$ ?
Rewrite the root as a fractional power, $y = (a x + b)^{\frac{1}{2}}$ , and apply the same rule.
That gives $\frac{a}{2} (a x + b)^{- \frac{1}{2}}$ , usually written $\frac{a}{2 \sqrt{a x + b}}$ .
Differentiate $y = (x^{2} - 5 x + 7)^{7}$ .
The inner function is $x^{2} - 5 x + 7$ , whose derivative is $2 x - 5$ .
Bringing the power down and reducing it, then multiplying by that derivative, gives:
$\frac{d y}{d x} = 7 (2 x - 5) (x^{2} - 5 x + 7)^{6}$
What is the chain rule in words, once you are fluent with it?
Differentiate the outside function, leaving the inside alone, then multiply by the derivative of the inside.
So differentiating $\sin (e^{2 x})$ gives $\cos (e^{2 x})$ multiplied by $2 e^{2 x}$ .
Complete the product rule, where $u$ and $v$ are functions of $x$ :
$y = u v \Rightarrow \frac{d y}{d x} =______+______$
The completed rule is:
$\frac{d y}{d x} = u \frac{d v}{d x} + v \frac{d u}{d x}$
In dash notation this is $y' = u v' + v u'$ : each function in turn is differentiated while the other is left alone.
When do you need the product rule?
When you are differentiating two functions multiplied together, and the product cannot simply be expanded.
A product of two brackets in powers of $x$ can just be expanded instead, so the rule earns its keep when the factors are things like $x^{2} \sin x$ .
True or False?
$\sin x \cos x$ and $\sin (\cos x)$ both need the product rule.
False.
$\sin x \cos x$ is a product, "sin $x$ times cos $x$ ", so it needs the product rule.
$\sin (\cos x)$ is a composite, "sin of cos of $x$ ", so it needs the chain rule instead.
What is the first step in applying the product rule?
Write down clearly which function is $u$ and which is $v$ , then differentiate each to get $u'$ and $v'$ .
Setting the four out before substituting is what stops the terms getting mixed up.
Differentiate $y = x^{2} \sin x$ .
Take $u = x^{2}$ and $v = \sin x$ , so $u' = 2 x$ and $v' = \cos x$ .
Substituting into $y' = u v' + v u'$ gives:
$\frac{d y}{d x} = x^{2} \cos x + 2 x \sin x$
Why is the derivative of $u v$ not simply $u' v'$ ?
Because both functions are changing at once, so the product changes through each of them in turn.
A quick check settles it: with $u = v = x$ , the product $x^{2}$ differentiates to $2 x$ , whereas $u' v'$ would give only $1 \times 1 = 1$ .
Complete the quotient rule, where $u$ and $v$ are functions of $x$ :
$y = \frac{u}{v} \Rightarrow \frac{d y}{d x} = \frac{______-______}{______}$
The completed rule is:
$\frac{d y}{d x} = \frac{v \frac{d u}{d x} - u \frac{d v}{d x}}{v^{2}}$
In dash notation, $y' = \frac{v u' - u v'}{v^{2}}$ , where $v$ is always the denominator.
True or False?
The two terms in the numerator of the quotient rule can be written in either order.
False.
The numerator is a subtraction, so swapping the terms changes the sign of the whole answer.
It must be $v u' - u v'$ , with the denominator's own derivative in the second term.
When do you need the quotient rule?
When both the numerator and the denominator are functions of $x$ .
If only one of them involves $x$ , there is a quicker route than the formula.
What is the first step in applying the quotient rule?
Identify $u$ as the numerator and $v$ as the denominator, then differentiate each to get $u'$ and $v'$ .
Getting $u$ and $v$ the right way round matters more here than in the product rule, because of the minus sign.
How would you differentiate $y = \frac{5}{x^{3}}$ and $y = \frac{x^{2} + 1}{4}$ without the quotient rule?
The first has a constant numerator, so rewrite it as $y = 5 x^{- 3}$ and use the power rule.
The second has a constant denominator, so treat $\frac{1}{4}$ as a factor and differentiate $x^{2} + 1$ as usual.
What can you do if you cannot recall the quotient rule?
Rewrite $y = \frac{u}{v}$ as $y = u \times v^{- 1}$ and use the product rule together with the chain rule.
That is really where the quotient rule comes from, so the two routes must agree.
How do you find the gradient of a curve at a given point?
Differentiate to get $\frac{d y}{d x}$ , then substitute the $x$ -coordinate of the point into it.
The number you get is the gradient of the curve there, and equally the gradient of the tangent at that point.
A curve is $y = \frac{4}{3} x^{3} + 3 x - 8$ . Find its gradient at $x = 2$ .
Differentiating gives $\frac{d y}{d x} = 4 x^{2} + 3$ .
Substituting $x = 2$ gives $4 \times 4 + 3 = 19$ .
For that same curve, find the values of $x$ where the rate of change is 4.
Set the derivative equal to 4 and solve, so $4 x^{2} + 3 = 4$ gives $x^{2} = \frac{1}{4}$ .
That has two solutions, $x = \frac{1}{2}$ and $x = - \frac{1}{2}$ , which is why the question asks for values rather than a value.
Fill in the gaps about the sign of the gradient:
A function is increasing when $\frac{d y}{d x}______0$ , and decreasing when $\frac{d y}{d x}______0$ .
The completed sentence is:
A function is increasing when $\frac{d y}{d x} > 0$ , and decreasing when $\frac{d y}{d x} < 0$ .
Increasing means the graph goes up as $x$ increases, which is exactly a positive gradient.
How do you find the intervals where a function is increasing?
Differentiate, then solve the inequality $\frac{d y}{d x} > 0$ .
The answer is a range of $x$ values rather than a single number, since most functions increase over some stretches and decrease over others.
For what values of $x$ is $y = 2 x^{3} - 3 x^{2} + 5$ decreasing?
Solve $\frac{d y}{d x} < 0$ , which is $6 x (x - 1) < 0$ .
The gradient function is a positive quadratic with roots 0 and 1, so it lies below the axis between them, giving $0 < x < 1$ .
How do you find the equation of a tangent to a curve?
Differentiate and substitute the $x$ -coordinate to get the gradient, and substitute it into the curve to get the $y$ -coordinate.
Then put the point and the gradient into $y - y_{1} = m (x - x_{1})$ .
Define the normal to a curve.
The normal at a point is the straight line through that point which is perpendicular to the tangent there.
It therefore cuts across the curve rather than running alongside it.
What is the gradient of the normal at a point?
It is $\frac{- 1}{f' (x_{1})}$ , the negative reciprocal of the tangent's gradient.
So find the tangent gradient first, then flip it and change its sign before using the same straight-line equation.
What is the second derivative, and how do you find it?
It is the derivative of the derivative, so you simply differentiate twice.
For $y = 5 x^{3} + 10 x^{2}$ , the first derivative is $15 x^{2} + 20 x$ and the second is $30 x + 20$ .
What does the second derivative measure?
The rate of change of the gradient.
A positive second derivative means the gradient is increasing, as in a $\cup$ -shape, and a negative one means it is decreasing, as in a $\cap$ -shape.
In $\frac{d^{2} y}{d x^{2}}$ , why are the two squares in different places?
The top records that you have differentiated twice, so the 2 sits on the $d$ .
The bottom records that both differentiations were with respect to $x$ , so the 2 sits on the $x$ .
Define a stationary point.
A stationary point is a point on a curve where the gradient is zero.
The tangent there is horizontal, which is why the curve is momentarily neither rising nor falling.
How do you find the stationary points of a curve?
Differentiate, then solve $\frac{d y}{d x} = 0$ to get the $x$ -coordinates.
Substitute each of those back into the equation of the curve, not into the derivative, to get the matching $y$ -coordinates.
What three kinds of stationary point are there?
A local minimum, a local maximum, or a point of inflection.
A quadratic has only one stationary point, which is the overall minimum or maximum depending on the sign of $a$ .
At a stationary point, fill in what the second derivative tells you:
If $\frac{d^{2} y}{d x^{2}}$ is positive the point is a $______$ , and if it is negative the point is a $______$ .
The completed sentence is:
If $\frac{d^{2} y}{d x^{2}}$ is positive the point is a local minimum, and if it is negative the point is a local maximum.
Positive means the gradient is increasing through the point, which is the $\cup$ -shape of a minimum.
True or False?
If the second derivative is zero at a stationary point, the point is a point of inflection.
False.
A zero second derivative settles nothing: the point could be a minimum, a maximum or a point of inflection.
When that happens you have to fall back on checking the sign of the first derivative on each side.
How does the first derivative reveal the nature of a stationary point?
Work out its sign just either side of the point.
Negative then positive gives a local minimum, and positive then negative gives a local maximum.
Can a derivative involve variables other than $y$ and $x$ ?
Yes: a derivative relates any two variables.
So if $A = 4 π r^{2}$ then $\frac{d A}{d r} = 8 π r$ , which is the rate of change of area with respect to radius.
What does the phrase "increasing at a rate of" tell you?
That the rate is with respect to time, so the derivative is $\frac{d}{d t}$ .
A quantity decreasing at a rate gives a negative derivative.
A sphere has volume $V = \frac{4}{3} π r^{3}$ . Find the rate of change of volume with respect to radius.
Differentiate with respect to $r$ , treating $π$ as the constant it is:
$\frac{d V}{d r} = 4 π r^{2}$
That result is the surface area of the sphere, which is not a coincidence.
Define optimisation.
Optimisation is finding the maximum or minimum output of a function.
In a modelling question that means the largest volume, the smallest cost, the least fuel used, and so on.
How do you solve an optimisation problem?
Form a formula for the quantity in terms of a single variable, then differentiate it.
Set that derivative to zero and solve, since the optimum is a stationary point, and substitute back to get the optimal value itself.
A cuboid has volume $V = 12 x - 20 x^{2}$ . Find the value of $x$ giving the maximum volume.
Differentiate and set to zero:
$\frac{d V}{d x} = 12 - 40 x = 0$
So $x = 0.3$ , and substituting back gives a volume of $1.8$ .
How do you prove that your optimised value really is a maximum?
Find the second derivative and show it is negative.
For $V = 12 x - 20 x^{2}$ it is $- 40$ , which is negative everywhere, so the stationary point must be a maximum.
What are connected rates of change?
Problems with more than two variables, where the chain rule links several rates into one equation.
Typically you know two of the rates and want the third.
The height of a liquid is increasing at 4 cm per second. Write this as a derivative.
It is $\frac{d h}{d t} = 4$ .
The word rate signals a derivative, and "per second" tells you it is with respect to time.
True or False?
A quantity decreasing at 4 cm per second still gives a positive derivative.
False.
A decreasing quantity has a negative rate, so $\frac{d h}{d t} = - 4$ .
Missing the sign reverses the direction of the whole answer while leaving the arithmetic looking fine.
How do you connect two known rates to find a third?
Multiply them with the chain rule, choosing the parts so that the shared variable cancels.
For example $\frac{d V}{d t} = \frac{d V}{d r} \times \frac{d r}{d t}$ , where the $d r$ terms behave as though they cancel.
What is the clue that a question is about rates of change?
The word rate itself, together with a unit "per" something.
That tells you a derivative is involved before you have worked out which one.
Why must you be clear which letters are variables and which are constants?
Because they behave completely differently under differentiation: a constant differentiates to zero, while a variable does not.
These questions carry a lot of letters, so treating a fixed radius as a variable, or the reverse, changes the answer entirely.

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