ASMathsCambridge (CIE)Pure 2FlashcardsDifferentiationFurther Differentiation

Further Differentiation (Cambridge (CIE) AS Maths: Pure 2): Flashcards

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FrontDifferentiating Other Functions
Complete the three standard trigonometric derivatives:
$\frac{d}{d x} (\sin x) =______$
$\frac{d}{d x} (\cos x) =______$
$\frac{d}{d x} (\tan x) =______$

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Cards in this collection (18)

Complete the three standard trigonometric derivatives:
$\frac{d}{d x} (\sin x) =______$
$\frac{d}{d x} (\cos x) =______$
$\frac{d}{d x} (\tan x) =______$
The completed derivatives are:
$\frac{d}{d x} (\sin x) = \cos x$
$\frac{d}{d x} (\cos x) = - \sin x$
$\frac{d}{d x} (\tan x) = \sec^{2} x$
All three are standard results given in the formulae booklet.
What is the derivative of $y = e^{k x}$ , where $k$ is a constant?
For $y = e^{k x}$ , $\frac{d y}{d x} = k e^{k x}$ : the exponential itself is unchanged, and the constant comes down as a multiplier.
So, for example, $y = e^{- 3 x}$ differentiates to $- 3 e^{- 3 x}$ .
True or False?
At every point on the curve $y = e^{x}$ , the gradient is equal to the $y$ -coordinate.
True.
Differentiating gives $\frac{d y}{d x} = e^{x}$ , which is the same expression as $y$ itself.
So at $(0, 1)$ the gradient is 1, and at $(2, e^{2})$ it is $e^{2}$ .
Differentiate $y = \ln 5 x$ , and explain why the 5 disappears.
The derivative is $\frac{1}{x}$ , exactly the same as for $y = \ln x$ .
The laws of logarithms split it into $\ln 5 x = \ln 5 + \ln x$ , and $\ln 5$ is a constant, whose derivative is zero.
Any positive constant in place of the 5 behaves the same way.
True or False?
Since $\frac{d}{d x} (\ln x) = \frac{1}{x}$ , it follows that $\frac{d}{d x} (\ln x^{2}) = \frac{1}{x^{2}}$ .
False.
Writing $\ln x^{2}$ as $2 \ln x$ for $x > 0$ shows that the derivative is $\frac{2}{x}$ .
Replacing $x$ by a new expression inside a standard derivative never means replacing it in the answer as well.
Differentiate $y = \cos (4 x - 1)$ .
For $y = \cos (4 x - 1)$ , $\frac{d y}{d x} = - 4 \sin (4 x - 1)$ .
Differentiating the cosine gives $- \sin (4 x - 1)$ , and the bracket is then differentiated separately to contribute the extra factor of 4.
The expression inside the bracket is never changed.
Why can the rule for differentiating $x^{n}$ not be used on $y = e^{x}$ ?
Because in $x^{n}$ the variable is in the base and the power is a constant, whereas in $e^{x}$ the variable is in the exponent.
They are different kinds of function, so $e^{x}$ has its own standard derivative, which is $e^{x}$ itself.
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:
$\frac{d y}{d x} = u \frac{d v}{d x} + v \frac{d u}{d x}$
In the shorter notation that is $y' = u v' + v u'$ : each function is paired with the other one's derivative.
What is the difference between a product of two functions and a composite function?
A product is two functions multiplied together, while a composite is one function applied to the output of another.
With $f (x) = 3 x^{2}$ and $g (x) = x - 3$ , the product is $3 x^{3} - 9 x^{2}$ but the composite $fg (x)$ is $3 x^{2} - 18 x + 27$ .
Products need this rule; composites need the chain rule.
True or False?
The derivative of a product is the product of the derivatives.
False.
Take $y = x \times x = x^{2}$ , which differentiates to $2 x$ .
Multiplying the separate derivatives would give $1 \times 1 = 1$ , which is not the same thing at all.
How do you apply the product rule?
Label the two factors $u$ and $v$ , differentiate each of them separately, then substitute all four expressions into the rule.
Setting the working out as four short labelled lines before you combine them is what keeps the terms from getting mixed up.
True or False?
It does not matter which factor you call $u$ and which you call $v$ .
True.
The rule adds its two terms together, and addition can be done in either order, so swapping the labels simply produces the same two terms the other way round.
That is not true of every rule of this kind, so it is worth knowing which ones tolerate it.
What do you do when one factor of a product needs the chain rule?
Differentiate that factor with the chain rule as a separate step, then feed the result into the product rule as usual.
Differentiating $(5 \sin x - 7 \cos x) e^{3 x + 2}$ needs the chain rule for $e^{3 x + 2}$ , giving $3 e^{3 x + 2}$ , before the product rule is applied at all.
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.

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