A LevelFurther MathsEdexcelFurther Pure 1FlashcardsFurther CalculusMethods in Calculus

Methods in Calculus (Edexcel A Level Further Maths: Further Pure 1): Flashcards

Exam code: 9FM0

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FrontLeibnitz's Theorem
Leibnitz's theorem gives the $n$ th derivative of a product $y = f (x) g (x)$ . Complete the three missing derivative orders:
$\frac{d^{n} y}{d x^{n}} = (\binom{n}{0}) f^{(0)} (x) g^{(______)} (x) + (\binom{n}{1}) f^{(1)} (x) g^{(______)} (x) + \dots + (\binom{n}{n}) f^{(n)} (x) g^{(______)} (x)$

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Leibnitz's theorem gives the $n$ th derivative of a product $y = f (x) g (x)$ . Complete the three missing derivative orders:
$\frac{d^{n} y}{d x^{n}} = (\binom{n}{0}) f^{(0)} (x) g^{(______)} (x) + (\binom{n}{1}) f^{(1)} (x) g^{(______)} (x) + \dots + (\binom{n}{n}) f^{(n)} (x) g^{(______)} (x)$
The completed theorem is:
$\frac{d^{n} y}{d x^{n}} = (\binom{n}{0}) f^{(0)} (x) g^{(n)} (x) + (\binom{n}{1}) f^{(1)} (x) g^{(n - 1)} (x) + \dots + (\binom{n}{n}) f^{(n)} (x) g^{(0)} (x)$
Here $f^{(0)} (x)$ means $f (x)$ itself, so the first term differentiates only $g$ and the last differentiates only $f$ .
What does Leibnitz's theorem save you from having to do?
It saves you differentiating the product $n$ times over, applying the product rule again at every stage and simplifying as you go.
Instead it builds $\frac{d^{n} y}{d x^{n}}$ in one step out of the separate derivatives of the two factors, which are usually far easier to write down.
What does Leibnitz's theorem reduce to when $n = 1$ ?
It reduces to the product rule, $\frac{d y}{d x} = f (x) g' (x) + f' (x) g (x)$ .
Both of the coefficients $(\binom{1}{0})$ and $(\binom{1}{1})$ are equal to $1$ , so the two terms are left with nothing in front of them.
Leibnitz's theorem is usually worked from a table of the derivatives of $f$ and of $g$ . How are the two columns paired up?
From opposite ends: the top of the $f$ column goes with the bottom of the $g$ column, the second entry of one with the second from last of the other, and so on inwards.
The theorem requires it, because the two derivative orders in every term have to add up to $n$ .
True or False?
The coefficients in Leibnitz's theorem are the same numbers as the coefficients in the expansion of ${(a + b)}^{n}$ .
True.
Both are the binomial coefficients $(\binom{n}{r})$ , so row $n$ of Pascal's triangle supplies them directly.
For a fourth derivative they are $1$ , $4$ , $6$ , $4$ and $1$ , exactly as in the expansion of ${(a + b)}^{4}$ .
How many terms should $\frac{d^{n} y}{d x^{n}}$ have before it is simplified?
Exactly $n + 1$ of them, so a fourth derivative should show five terms.
The sum runs over every value of $r$ from $0$ to $n$ inclusive, which is one more than the value of $n$ , and counting the terms is a quick check that none has been missed.
To prove a general result for $\frac{d^{n} y}{d x^{n}}$ the binomial coefficients have to be written in terms of $n$ . What are $(\binom{n}{0})$ , $(\binom{n}{1})$ and $(\binom{n}{2})$ ?
They are $1$ , $n$ and $\frac{1}{2} n (n - 1)$ .
The last comes from $\frac{n!}{2! (n - 2)!}$ , in which the $(n - 2)!$ cancels and leaves $\frac{n (n - 1)}{2}$ .
For $y = x^{2} e^{2 x}$ , why do all but the first three terms of Leibnitz's theorem vanish?
Taking $f (x) = x^{2}$ , its third and every later derivative is zero, and every term beyond the third contains one of them as a factor.
Only the terms in $f^{(0)}$ , $f^{(1)}$ and $f^{(2)}$ survive, whatever the value of $n$ , which is what makes a general result for $\frac{d^{n} y}{d x^{n}}$ reachable at all.
When $\lim_{x \to a} (\frac{f (x)}{g (x)})$ takes one of the two forms the rule allows, complete what L'Hospital's rule says it is equal to:
$\lim_{x \to a} (\frac{f (x)}{g (x)}) = \lim_{x \to a} (\frac{______}{______})$
The completed rule is:
$\lim_{x \to a} (\frac{f (x)}{g (x)}) = \lim_{x \to a} (\frac{f' (x)}{g' (x)})$
The numerator and the denominator are differentiated separately, so the quotient rule plays no part at all and would give a different expression altogether.
What must you check before applying L'Hospital's rule to a quotient?
That substituting the value $x$ is tending to gives either $\frac{0}{0}$ or $\pm \frac{\infty}{\infty}$ .
Those are the only two forms the rule applies to, and the check is made by direct substitution rather than by any working.
Define the indeterminate form of a limit.
The indeterminate form of a limit is what you get by substituting the value $x$ is tending to, when the result has no value of its own: expressions such as $\frac{0}{0}$ , $\frac{\infty}{\infty}$ , $0 \times \infty$ , $\infty - \infty$ and $1^{\infty}$ .
It says what kind of limit you are dealing with, but it does not by itself tell you the limit's value.
True or False?
L'Hospital's rule can be used to find $\lim_{x \to 2} (\frac{x^{2} - 4}{x + 2})$ .
False.
Substituting $x = 2$ gives $\frac{0}{4}$ , which is simply $0$ , so there is nothing indeterminate here and the limit is $0$ .
Applying the rule anyway would give $\frac{2 x}{1} \to 4$ , which is the wrong answer.
In a limit involving trigonometric functions, which angle measure is assumed?
Radians, unless a question says otherwise.
The derivatives of $\sin x$ and $\cos x$ hold only when $x$ is measured in radians, so a limit reached by differentiating them would be wrong in degrees.
You apply L'Hospital's rule once and the new quotient is still $\frac{0}{0}$ . What now?
Apply it again, differentiating the new numerator and denominator in turn, and keep going until substitution gives something that is not indeterminate.
For $\lim_{x \to 0} (\frac{x - \sin x}{x^{2}})$ the first application gives $\frac{1 - \cos x}{2 x}$ , still $\frac{0}{0}$ , and a second gives $\frac{\sin x}{2}$ , whose limit is $0$ .
L'Hospital's rule needs a quotient. What do you do with a limit of the form $0 \times \infty$ or $\infty - \infty$ ?
Rearrange it into a single quotient before applying the rule.
For a product, write one factor as the reciprocal of its reciprocal, so $x \cot x$ becomes $\frac{x}{\tan x}$ , which is $\frac{0}{0}$ as $x \to 0$ .
For a difference of algebraic fractions, combine them over a common denominator, so $\frac{1}{x} - \cot x$ becomes $\frac{\sin x - x \cos x}{x \sin x}$ .
How do you find $\lim_{x \to 0} ({(1 + \sin x)}^{\frac{1}{x}})$ , which has the form $1^{\infty}$ ?
Take the logarithm first, which brings the index down and turns the expression into $\frac{\ln (1 + \sin x)}{x}$ , of the form $\frac{0}{0}$ .
L'Hospital's rule then gives $\frac{\cos x}{1 + \sin x} \to 1$ , and the answer to the original limit is $e$ raised to that, namely $e$ .
Define the Weierstrass substitution.
The Weierstrass substitution is the substitution $t = \tan (\frac{θ}{2})$ , used to carry out an integration in $θ$ .
It is also called the tangent half-angle substitution, and it is the same substitution that produces the t-formulae, here put to work inside an integral.
Under the Weierstrass substitution $t = \tan (\frac{θ}{2})$ , complete the conversion of $d θ$ :
$d θ = \frac{______}{______} d t$
The completed conversion is:
$d θ = \frac{2}{1 + t^{2}} d t$
Differentiating the substitution gives $\frac{d t}{d θ} = \frac{1}{2} \sec^{2} (\frac{θ}{2})$ , and $1 + \tan^{2} A \equiv \sec^{2} A$ turns that into $\frac{1}{2} (1 + t^{2})$ , which inverts to the result above.
What kind of expression does the Weierstrass substitution turn a trigonometric integrand into?
A rational function of $t$ , that is one polynomial divided by another.
That is the whole point of the substitution: a rational function is something the ordinary methods of integration can handle, whereas a trigonometric expression in $θ$ often is not.
Every t-formula has $1 + t^{2}$ in it, and so does $d θ$ . What happens to those factors?
They cancel.
Clearing the t-formulae out of the integrand means multiplying top and bottom by $1 + t^{2}$ , which leaves a factor of $1 + t^{2}$ in the numerator, and that is exactly what the $1 + t^{2}$ in the denominator of $d θ$ removes.
It is why the integral finally left to do is usually far simpler than the working leading up to it.
True or False?
After a Weierstrass substitution, the integral in $t$ always has to be done by partial fractions.
False.
Partial fractions is often the route, as it is for $\int \frac{2}{1 - t^{2}} d t$ , but it is only one possibility.
The substitution may equally leave something as simple as $\int \frac{1}{t + 2} d t$ , or a standard form such as $\int \frac{2}{t^{2} + 3} d t$ .
What happens to the limits of a definite integral under the Weierstrass substitution?
Each $θ$ limit is converted into a $t$ limit through $t = \tan (\frac{θ}{2})$ , so $θ = \frac{π}{3}$ becomes $t = \tan \frac{π}{6} = \frac{\sqrt{3}}{3}$ and $θ = \frac{π}{2}$ becomes $t = 1$ .
Once the limits are in $t$ the integral is finished entirely in $t$ , and there is no need to convert the answer back to $θ$ at all.
Before evaluating a definite integral in $θ$ by the Weierstrass substitution, what must you check about the range of integration?
Whether it contains a value at which $t = \tan (\frac{θ}{2})$ is undefined.
If it does, the integral has to be split there into two improper integrals, one with its limit approached from below and one from above.
An integral in $θ$ has been split at a value where $t = \tan (\frac{θ}{2})$ is undefined. Why does one part run up to $\infty$ while the other starts at $- \infty$ ?
Because $\frac{θ}{2}$ is passing through $\frac{π}{2}$ at that point, and the tangent graph jumps from $+ \infty$ on one side of $\frac{π}{2}$ to $- \infty$ on the other.
So the first part runs from its lower $t$ limit up to $\infty$ , and the second begins at $- \infty$ and runs to its upper $t$ limit.

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