International A Level (IAL)MathsEdexcelPure 3FlashcardsDifferentiationFurther Differentiation

Further Differentiation (Edexcel International A Level (IAL) Maths: Pure 3): Flashcards

Exam code: YMA01

1/22

0Still learning

Know0

FrontChain Rule
If $y$ is a function of $u$ , and $u$ is a function of $x$ , complete the chain rule:
$\frac{d y}{d x} = \frac{d y}{d u} \times______$

Was this flashcard helpful?

Cards in this collection (22)

If $y$ is a function of $u$ , and $u$ is a function of $x$ , complete the chain rule:
$\frac{d y}{d x} = \frac{d y}{d u} \times______$
The completed rule is:
$\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x}$
The $d u$ terms look as though they cancel, which is a useful way to remember it, though it is not a proof.
What kind of function does the chain rule differentiate?
A composite function, meaning one function applied to the output of another.
So $\sin (x^{2})$ needs it: the squaring happens first, and the sine acts on the result.
How do you differentiate ${(f (x))}^{n}$ ?
Bring the power down, reduce it by one, and multiply by the derivative of the inside:
$n {(f (x))}^{n - 1} f' (x)$
The inside function is differentiated but not otherwise changed.
What does $\frac{d y}{d x} = \frac{1}{\frac{d x}{d y}}$ allow you to do?
Differentiate a relationship that is given as $x$ in terms of $y$ , rather than the other way round.
Differentiate as it stands to get $\frac{d x}{d y}$ , then take the reciprocal.
What is the derivative of $\ln (f (x))$ ?
$\frac{f' (x)}{f (x)}$ .
This shape, a derivative sitting over the original function, is worth recognising: it is the pattern that makes certain fractions integrate to a logarithm later in the course.
True or False?
$\frac{d}{d x} ({(3 x + 1)}^{5}) = 5 {(3 x + 1)}^{4}$
False.
The derivative of the inside, which is $3$ , has to be included as well, giving $15 {(3 x + 1)}^{4}$ .
Forgetting that final factor is the commonest chain rule error.
For $y = u v$ , where $u$ and $v$ are functions of $x$ , complete the product rule:
$\frac{d y}{d x} = u \frac{d v}{d x} +______$
The completed rule is:
$\frac{d y}{d x} = u \frac{d v}{d x} + v \frac{d u}{d x}$
It is often remembered in the shorter form $y' = u v' + v u'$ , and it is not in the formulae booklet.
What is the difference between a product of two functions and a composite function?
A product is two functions multiplied together, such as $x^{2} \sin x$ .
A composite is a function of a function, such as $\sin (x^{2})$ , and it needs the chain rule instead.
Why is the derivative of a product not the product of the derivatives?
Because each term of the product rule keeps one of the two functions unchanged.
Differentiating both at once, turning $x^{2} \sin x$ into $2 x \cos x$ , throws that structure away entirely and gives the wrong answer.
Differentiate $y = x^{2} \sin x$ .
Taking $u = x^{2}$ and $v = \sin x$ :
$\frac{d y}{d x} = x^{2} \cos x + 2 x \sin x$
Each of the two terms still contains one of the original functions untouched, which is the pattern to check your answer against.
True or False?
In the product rule it makes no difference which function you call $u$ and which you call $v$ .
True.
The two terms are added together, so swapping $u$ and $v$ simply writes the same answer in the other order.
The rule is symmetric in the two functions.
For $y = \frac{u}{v}$ , where $u$ and $v$ are functions of $x$ , complete the quotient rule:
$\frac{d y}{d x} = \frac{v \frac{d u}{d x} - u \frac{d v}{d x}}{______}$
The completed rule is:
$\frac{d y}{d x} = \frac{v \frac{d u}{d x} - u \frac{d v}{d x}}{v^{2}}$
The denominator is the bottom function squared, not the derivative of anything, and the whole formula is given in the formulae booklet.
Why does the order of the two terms matter in the quotient rule?
Because of the minus sign in the numerator: swapping the terms reverses the sign of the whole answer.
The term beginning with $v$ , the bottom function, is the one that comes first.
How can you recognise a quotient rule question written as $g (x) {(h (x))}^{- 1}$ ?
A negative power applied to a whole function is a division in disguise, since $g (x) {(h (x))}^{- 1} = \frac{g (x)}{h (x)}$ .
It can be done with the product and chain rules instead, but the quotient rule is usually quicker.
Differentiate $y = \frac{\sin x}{x}$ .
Taking $u = \sin x$ and $v = x$ :
$\frac{d y}{d x} = \frac{x \cos x - \sin x}{x^{2}}$
Answers from the quotient rule rarely simplify much, so leaving the result as a single fraction is normal.
True or False?
Every quotient has to be differentiated using the quotient rule.
False.
A quotient that simplifies should be simplified first: $\frac{x^{3} + x}{x}$ is just $x^{2} + 1$ , which differentiates in one line.
The rule is for quotients that cannot be reduced to a sum of simpler terms.
Complete the derivatives of these reciprocal trigonometric functions:
$\frac{d}{d x} (\sec x) =______$
$\frac{d}{d x} (\cot x) =______$
The completed derivatives are:
$\frac{d}{d x} (\sec x) = \sec x \tan x$
$\frac{d}{d x} (\cot x) = - {cosec}^{2} x$
The third of the set is $\frac{d}{d x} (cosec x) = - cosec x \cot x$ , and notice that the two functions beginning with "co" are the two carrying a minus sign.
Complete the derivatives of these inverse trigonometric functions:
$\frac{d}{d x} (\arcsin x) =______$
$\frac{d}{d x} (\arctan x) =______$
The completed derivatives are:
$\frac{d}{d x} (\arcsin x) = \frac{1}{\sqrt{1 - x^{2}}}$
$\frac{d}{d x} (\arctan x) = \frac{1}{1 + x^{2}}$
Neither of these, nor the $\arccos$ result, is given in the formulae booklet.
How do you derive the derivative of $\sec x$ ?
Write $\sec x = {(\cos x)}^{- 1}$ and apply the chain rule.
That gives $- {(\cos x)}^{- 2} \times (- \sin x) = \frac{\sin x}{\cos^{2} x}$ , which splits into $\sec x \tan x$ .
How do you derive the derivative of $\arcsin x$ ?
Write $y = \arcsin x$ , so that $x = \sin y$ , and differentiate to get $\frac{d x}{d y} = \cos y$ .
Taking the reciprocal gives $\frac{d y}{d x} = \frac{1}{\cos y}$ , and $\sin^{2} y + \cos^{2} y \equiv 1$ turns that into $\frac{1}{\sqrt{1 - x^{2}}}$ .
True or False?
The derivatives of $\arcsin x$ and $\arccos x$ differ only by a sign.
True.
$\arcsin x$ differentiates to $\frac{1}{\sqrt{1 - x^{2}}}$ , and $\arccos x$ to the negative of it.
The reason is that $\arcsin x + \arccos x$ is a constant, so the two derivatives have to cancel each other out.
For which values of $x$ is the derivative of $\arcsin x$ defined?
Only for $- 1 < x < 1$ .
At $x = \pm 1$ the denominator $\sqrt{1 - x^{2}}$ becomes zero, which matches the graph of $\arcsin x$ turning vertical at each end of its domain.

Sign up to unlock flashcards

By signing up you agree to our Terms and Privacy Policy

Previous:Modelling with Exponentials & LogarithmsNext:Further Integration