What is meant by the gradient of a curve at a particular point?

It is the gradient of the tangent to the curve at that point. The gradient of a curve is not fixed: it changes from point to point, which is why a gradient can only be given at a particular point .

Define tangent to a curve.

A straight line that touches a curve at a given point, running in the same direction as the curve at that point. At any given point of a smooth curve there is exactly one tangent.

True or False? A line that is a tangent to a curve at one point cannot meet the curve anywhere else.

False. A tangent only has to touch the curve at the point where it is a tangent. Further along it may well cut the curve at another point.

Why can the gradient of a chord be found exactly, when the gradient of a tangent cannot be read off directly?

Because a chord joins two known points on the curve, so both of their coordinates can be worked out from the equation. A tangent touches at only one point, so there is no second point to work with and drawing one gives only an estimate.

Define the normal to a curve.

The normal at a point is the line through that point which is perpendicular to the tangent there. Where there is no tangent, as at a sharp corner, there is no normal either.

What is the second derivative mainly used for?

Determining the nature of a stationary point , that is whether it is a maximum or a minimum. Its sign at that point is what distinguishes the two.

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Differentiation (Edexcel International AS Maths: Pure 1): Flashcards

Exam code: XMA01

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What is meant by the gradient of a curve at a particular point?

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Cards in this collection (31)

What is meant by the gradient of a curve at a particular point?
It is the gradient of the tangent to the curve at that point.
The gradient of a curve is not fixed: it changes from point to point, which is why a gradient can only be given at a particular point.
Define tangent to a curve.
A straight line that touches a curve at a given point, running in the same direction as the curve at that point.
At any given point of a smooth curve there is exactly one tangent.
True or False?
A line that is a tangent to a curve at one point cannot meet the curve anywhere else.
False.
A tangent only has to touch the curve at the point where it is a tangent. Further along it may well cut the curve at another point.
A tangent has been drawn to a curve at the point $A$ on an accurately drawn graph. Which two points should you use to work out the gradient of the curve at $A$ ?
Any two points that lie on the tangent line, chosen where it passes through easily read coordinates; points on the curve itself must not be used.
The gradient is then the change in $y$ divided by the change in $x$ between those two points. So, for example, a tangent through $(1, - 2)$ and $(3, 4)$ has gradient:
$\frac{4 - (- 2)}{3 - 1} = \frac{6}{2} = 3$
Are there points on a graph where a curve has no gradient?
Yes. A tangent can only be drawn where the curve is smooth.
At a sharp corner, such as the vertex of the graph of $y = | f (x) |$ , no tangent can be drawn, so the gradient there is undefined.
Define derivative.
The derivative of a function is its rate of change with respect to a variable.
It is also the gradient function of the graph $y = f (x)$ , giving the gradient of the curve at any value of $x$ .
True or False?
The derivative of a function is itself a function, rather than a single number.
True.
The gradient of a curve changes from one value of $x$ to the next, so the derivative has to give a rule rather than one answer.
For $y = x^{2}$ it is close to zero near the origin and grows as $x$ increases, while for a straight line it happens to be the same everywhere.
What does the notation $\frac{d}{d x} (f (x))$ tell you to do?
Differentiate $f (x)$ with respect to $x$ .
Written this way it is an instruction, with the function it acts on placed in the brackets after it, and what it produces is the derivative.
Why can the gradient of a chord be found exactly, when the gradient of a tangent cannot be read off directly?
Because a chord joins two known points on the curve, so both of their coordinates can be worked out from the equation.
A tangent touches at only one point, so there is no second point to work with and drawing one gives only an estimate.
Complete the gradient of the chord joining the points on $y = f (x)$ at $x$ and at $x + h$ :
$gradient = \frac{f (x + h) -______}{______}$
The completed expression is:
$gradient = \frac{f (x + h) - f (x)}{h}$
The denominator is $h$ because the horizontal separation of the two points is $(x + h) - x$ .
In the chord construction for a derivative, why do you let $h$ tend to zero?
Because as the second point slides towards the first, the chord turns until it lies along the tangent.
The gradient of the tangent is therefore the limit the chord gradients approach, and that limit is what the derivative is defined to be.
On $y = x^{2}$ , the chord joining the points at $x = 5$ and $x = 5 + h$ , where $h > 0$ , has gradient $10 + h$ . What is the gradient of the curve at $x = 5$ ?
$10$ , the value that $10 + h$ approaches as $h$ gets closer to zero.
Every one of these chords is slightly steeper than the tangent, and the extra $h$ is exactly that difference.
Complete the rule for differentiating a power of $x$ , filling in the missing coefficient and index:
$\frac{d}{d x} (x^{n}) =______x^{______}$
The completed rule is:
$\frac{d}{d x} (x^{n}) = n x^{n - 1}$
Multiply by the index, then reduce the index by 1.
This works for any constant $n$ , including negative and fractional values.
How do you differentiate a term such as $- 4 x^{5}$ , where a power of $x$ has a constant multiplier?
Differentiate the power of $x$ and keep the multiplier:
$\frac{d}{d x} (a x^{n}) = a n x^{n - 1}$
So, for example, $- 4 x^{5}$ differentiates to $- 4 \times 5 x^{4} = - 20 x^{4}$ .
True or False?
Differentiating $x^{- 4}$ gives $- 4 x^{- 3}$ .
False.
The index is always reduced by 1, even when it is already negative: $- 4 - 1 = - 5$ , so the derivative is $- 4 x^{- 5}$ .
Negative and fractional indices are a common place to lose marks for exactly this reason.
What must you do to terms such as $\sqrt{x}$ and $\frac{4}{x}$ before you can differentiate them?
Rewrite each one as a power of $x$ , so that the rule for differentiating $x^{n}$ can be applied:
$\sqrt{x} = x^{\frac{1}{2}}$
$\frac{4}{x} = 4 x^{- 1}$
The derivative is often converted back into root or fraction form at the end.
How do you differentiate an expression that is a sum or difference of several terms?
Differentiate the terms one at a time, keeping them in the same order.
So, for example, $y = x^{4} - \sqrt{x} + \frac{4}{x}$ gives:
$\frac{d y}{d x} = 4 x^{3} - \frac{1}{2 \sqrt{x}} - \frac{4}{x^{2}}$
Notice that the last term changes sign, because differentiating $4 x^{- 1}$ gives $- 4 x^{- 2}$ .
True or False?
The derivative of $y = \sqrt{7}$ is $\frac{1}{2 \sqrt{7}}$ .
False.
$\sqrt{7}$ is a constant: there is no $x$ in it. The derivative of any constant is zero, so $\frac{d y}{d x} = 0$ .
The graph of a constant is a horizontal line, which has gradient zero everywhere.
What is the derivative of a term such as $- 5 x$ , where $x$ has no visible index?
The coefficient on its own, so $- 5$ . In general:
$\frac{d}{d x} (a x) = a$
This makes sense because $y = a x$ is a straight line with gradient $a$ , the same at every point.
What is the difference between $f' (x)$ and $\frac{d y}{d x}$ ?
There is no difference in meaning: both stand for the derivative.
$f' (x)$ is used when a function has been defined as $f (x)$ , and $\frac{d y}{d x}$ when the curve is written as $y$ in terms of $x$ .
How do you find the gradient of a curve at a particular point?
Differentiate to get $f' (x)$ , then substitute the $x$ -coordinate of the point into it.
The derivative is a formula for the gradient, so it has to be evaluated at the point you actually want.
Define the normal to a curve.
The normal at a point is the line through that point which is perpendicular to the tangent there.
Where there is no tangent, as at a sharp corner, there is no normal either.
The tangent to $y = f (x)$ at the point $(a, f (a))$ is:
$y - f (a) =______(x - a)$
$y - f (a) = f' (a) (x - a)$
It is simply $y - y_{1} = m (x - x_{1})$ , with the gradient supplied by the derivative.
If the tangent at a point has gradient $f' (a)$ , what is the gradient of the normal there?
$- \frac{1}{f' (a)}$ .
The normal is perpendicular to the tangent, so the two gradients must multiply to $- 1$ .
You are told only the $x$ -coordinate of the point where a tangent touches. What else do you need, and how do you get it?
The $y$ -coordinate, found by substituting that $x$ into the original function.
Two different substitutions are needed: into $f' (x)$ for the gradient, and into $f (x)$ for the point.
True or False?
Where the tangent to a curve is horizontal, the normal is vertical.
True.
If $f' (a) = 0$ the tangent is horizontal, so the normal must be vertical.
Its equation is then $x = a$ , because $- \frac{1}{f' (a)}$ has no value when the derivative is zero.
Define second derivative.
The result of differentiating a function twice, written $f'' (x)$ or $\frac{d^{2} y}{d x^{2}}$ .
It does not mean squaring the first derivative: the superscripts are part of the notation, not powers.
Complete the second derivative notation:
$\frac{d^{2} y}{______}$
The completed notation is:
$\frac{d^{2} y}{d x^{2}}$
Note the positions: the $2$ sits on the $d$ on top and on the $x$ underneath, never on the $y$ .
What does the second derivative measure?
The rate of change of the gradient.
The first derivative says how fast $y$ is changing; the second says how fast that rate is itself changing.
What is the second derivative mainly used for?
Determining the nature of a stationary point, that is whether it is a maximum or a minimum.
Its sign at that point is what distinguishes the two.
True or False?
The second derivative of a straight line is zero.
True.
A straight line has a constant gradient, so the rate at which that gradient changes is zero.
Differentiating $y = m x + c$ gives $m$ , and differentiating again gives $0$ .

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