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.

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

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

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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.
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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