Define differentiation .

Differentiation is the process of finding the derivative , or gradient function, of a function. It turns the equation of a curve into a new function whose outputs are gradients.

How is the gradient of a curve at a point defined?

It is the gradient of the tangent to the curve at that point. A tangent is a straight line touching the curve there, and its steepness is what the derivative reports.

What is the advantage of having a gradient function?

It gives the gradient at any point of the curve, replacing the need to draw a tangent and measure it. That turns a drawing problem into an algebraic one, with no sketching required at all.

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Differentiation (Edexcel IGCSE Further Pure Maths): Flashcards

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FrontIntroduction to Derivatives
Define differentiation.

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Cards in this collection (25)

Define differentiation.
Differentiation is the process of finding the derivative, or gradient function, of a function.
It turns the equation of a curve into a new function whose outputs are gradients.
What does the gradient of a curve measure?
How fast $y$ changes as $x$ changes, the rate of change of $y$ with respect to $x$ .
In everyday terms it is steepness, and it is what makes things like the speed of a car measurable.
How is the gradient of a curve at a point defined?
It is the gradient of the tangent to the curve at that point.
A tangent is a straight line touching the curve there, and its steepness is what the derivative reports.
True or False?
A straight line has a different gradient at every point.
False.
A straight line has the same gradient everywhere, which is the $m$ in $y = m x + c$ .
It is a curve whose gradient changes from point to point, and that is exactly why calculus is needed for curves.
What are the two ways of writing the derivative?
$\frac{d y}{d x}$ , read as 'd $y$ by d $x$ ', when the curve is given as $y = \dots$
$f' (x)$ , read as 'f dash of $x$ ', when it is given in function notation instead.
What is the advantage of having a gradient function?
It gives the gradient at any point of the curve, replacing the need to draw a tangent and measure it.
That turns a drawing problem into an algebraic one, with no sketching required at all.
Complete the rule for differentiating a power of $x$ :
$if f (x) = x^{n} then f' (x) =______x^{______}$
The completed rule is:
$if f (x) = x^{n} then f' (x) = n x^{n - 1}$
Bring the power down as a multiplier, then subtract $1$ from it; any constant factor is simply carried along.
What are the derivatives of $a x$ and of a constant $a$ ?
The derivative of $a x$ is $a$ , and the derivative of a constant is $0$ .
Both follow from the power rule, since $a x$ is $a x^{1}$ and $a$ is $a x^{0}$ .
How do you differentiate $\frac{4}{\sqrt{x}}$ ?
Rewrite it as a power first, so that $\frac{4}{\sqrt{x}} = 4 x^{- \frac{1}{2}}$ .
The power rule then gives $4 \times (- \frac{1}{2}) x^{- \frac{3}{2}} = - 2 x^{- \frac{3}{2}}$ .
Why can you not differentiate $(2 x - 3) (x^{2} - 4)$ term by term?
Because the power rule applies to a sum of powers, not to a product of two brackets.
Expand it to $2 x^{3} - 3 x^{2} - 8 x + 12$ first, and then differentiate term by term.
Complete the derivatives of the two trigonometric functions:
$\frac{d}{d x} (\sin a x) =______\cos a x and \frac{d}{d x} (\cos a x) =______\sin a x$
The completed derivatives are:
$\frac{d}{d x} (\sin a x) = a \cos a x and \frac{d}{d x} (\cos a x) = - a \sin a x$
The minus sign belongs to $\cos$ alone, and the factor $a$ comes from the inside of the bracket.
Find the gradient of $y = 3 \sin \frac{x}{2}$ at $x = \frac{π}{3}$ .
Here $a = \frac{1}{2}$ , so $\frac{d y}{d x} = \frac{3}{2} \cos \frac{x}{2}$ .
At $x = \frac{π}{3}$ that gives $\frac{3}{2} \cos \frac{π}{6} = \frac{3}{2} \times \frac{\sqrt{3}}{2} = \frac{3 \sqrt{3}}{4}$ .
True or False?
Differentiating $\sin 3 x$ gives $\cos 3 x$ .
False.
The factor from inside the function has to come out as well, so the derivative is $3 \cos 3 x$ .
Losing that factor is the commonest slip whenever the angle is a multiple of $x$ .
Complete the derivative of the exponential function:
$\frac{d}{d x} (e^{a x}) =______e^{______}$
The completed derivative is:
$\frac{d}{d x} (e^{a x}) = a e^{a x}$
The exponential itself is left completely unchanged, and simply picks up a factor of $a$ from the index.
Find the gradient of $y = 2 e^{- 3 x}$ at $x = \frac{1}{2}$ .
Differentiating gives $\frac{d y}{d x} = - 6 e^{- 3 x}$ , since the factor $- 3$ comes out.
At $x = \frac{1}{2}$ that is $- 6 e^{- \frac{3}{2}} = - 1.34$ to three significant figures.
Complete the product rule for $y = u v$ :
$\frac{d y}{d x} =______v + u______$
The completed rule is:
$\frac{d y}{d x} = u' v + u v'$
Differentiate each factor in turn, leaving the other one alone, then add the two results together.
How do you tell a product of two functions from a function of a function?
$\sin x \cos x$ is a product, sine times cosine, and needs the product rule.
$\sin (\cos x)$ is a function of a function, sine of cosine, and needs the chain rule instead.
Differentiate $y = 5 x^{2} \cos 3 x$ .
Take $u = 5 x^{2}$ and $v = \cos 3 x$ , so that $u' = 10 x$ and $v' = - 3 \sin 3 x$ .
Then $u' v + u v'$ gives $10 x \cos 3 x - 15 x^{2} \sin 3 x$ .
What is the quotient rule for $y = \frac{u}{v}$ ?
It is $\frac{d y}{d x} = \frac{u' v - u v'}{v^{2}}$ .
The order matters: unlike the product rule this one subtracts, and the term carrying $u'$ must come first.
When is the quotient rule not the quickest method?
When the numerator is a constant, since $\frac{k}{g (x)}$ can be written as $k {[g (x)]}^{- 1}$ and done by the chain rule.
When the denominator is a constant, treat it as a factor instead: $\frac{f (x)}{k}$ is just $\frac{1}{k} f (x)$ .
Complete the chain rule, where $y$ is a function of $u$ and $u$ is a function of $x$ :
$\frac{d y}{d x} = \frac{d y}{d______} \times \frac{d______}{d x}$
The completed rule is:
$\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x}$
The two $d u$ parts behave as though they cancel, which is the easiest way to remember the shape of it.
How do you differentiate ${[f (x)]}^{n}$ quickly?
Use $\frac{d y}{d x} = n f' (x) {[f (x)]}^{n - 1}$ , the chain rule specialised to a power.
It saves expanding: ${(x^{2} - 5 x + 7)}^{7}$ differentiates to $7 (2 x - 5) {(x^{2} - 5 x + 7)}^{6}$ .
How do you differentiate $y = \frac{1}{\sqrt{2 x - 3}}$ ?
Rewrite it as a power first, $y = {(2 x - 3)}^{- \frac{1}{2}}$ , then use the power-of-a-function rule.
Square roots and denominators are hidden powers, and spotting them is what makes the chain rule usable.
Differentiate $y = e^{\cos x}$ .
Set $u = \cos x$ , so $y = e^{u}$ with $\frac{d y}{d u} = e^{u}$ and $\frac{d u}{d x} = - \sin x$ .
Multiplying the two and substituting back gives $- (\sin x) e^{\cos x}$ .
True or False?
The chain rule is needed to differentiate $\cos (3 x^{2} - 1)$ .
True.
The angle is itself a function of $x$ , so this is a function of a function rather than a simple $\cos a x$ .
Setting $u = 3 x^{2} - 1$ gives $\frac{d y}{d x} = - 6 x \sin (3 x^{2} - 1)$ .

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