True or False? First principles is the quickest way to find derivatives.

False. First principles is not the quickest way to find derivatives; there are much quicker methods. Only differentiate by first principles if an exam question tells you to.

Define the term rate of change .

A rate of change is a measure of how a quantity is changing with respect to another quantity.

What is meant by related rates of change ?

Related rates of change are rates of change connected by a linking variable or parameter , often time. For example, if a container is being filled with water the rates of change for the volume and height of the water in the container are related rates of change connected by time.

True or False? A positive rate of change always indicates a decrease .

False. A positive rate of change indicates an increase .

What is implicit differentiation?

Implicit differentiation is a method of differentiating equations where y is not given explicitly as a function of x.

True or False? Implicit differentiation always uses the chain rule .

True. Implicit differentiation always uses the chain rule .

True or False? In optimisation questions using implicit differentiation, the locations of minimums and maximums will always occur at turning points .

False. In optimisation questions using implicit differentiation, the locations of minimums and maximums will not always occur at turning points . It is possible that there may not be a turning point. Also the minimum or maximum could be at the start or end of a given or appropriate interval.

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What does it mean to differentiate from first principles?

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What does it mean to differentiate from first principles?
Differentiating from first principles means using the definition of the derivative to show what the derivative of a function is.
The definition is $f' (x) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h}$ .
What is the definition of the derivative of a function $f (x)$ ?
The definition of the derivative of a function $f (x)$ is $f' (x) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h}$ .
This formula is in the exam formula booklet.
True or False?
When $h = 0$ , $\frac{f (x + h) - f (x)}{h}$ is always defined.
False.
When $h = 0$ , $\frac{f (x + h) - f (x)}{h}$ is always undefined as it is equal to $\frac{0}{0}$ .
True or False?
First principles is the quickest way to find derivatives.
False.
First principles is not the quickest way to find derivatives; there are much quicker methods.
Only differentiate by first principles if an exam question tells you to.
What are the four steps in differentiating from first principles?
The four steps in differentiating from first principles are:
1. Identify the function $f (x)$ and substitute this into the first principles formula.
2. Expand $f (x + h)$ in the numerator.
3. Simplify the numerator, factorise and cancel $h$ with the denominator.
4. Evaluate the remaining expression as $h$ tends to zero.
Define the term rate of change.
A rate of change is a measure of how a quantity is changing with respect to another quantity.
What is meant by related rates of change?
Related rates of change are rates of change connected by a linking variable or parameter, often time.
For example, if a container is being filled with water the rates of change for the volume and height of the water in the container are related rates of change connected by time.
True or False?
A positive rate of change always indicates a decrease.
False.
A positive rate of change indicates an increase.
How is the chain rule used in solving related rates of change problems?
In solving related rates of change problems, the chain rule is used to set up an equation linking rates of change.
E.g. $\frac{d V}{d t}$ and $\frac{d r}{d t}$ can be linked using the chain rule equation $\frac{d V}{d t} = \frac{d V}{d r} \times \frac{d r}{d t}$ .
State the chain rule equation for related rates of change, connecting $\frac{d y}{d x}$ , $\frac{d y}{d t}$ and $\frac{d x}{d t}$ .
The chain rule equation for related rates of change, connecting $\frac{d y}{d x}$ , $\frac{d y}{d t}$ and $\frac{d x}{d t}$ is $\frac{d y}{d t} = \frac{d y}{d x} \times \frac{d x}{d t}$ .
This is not in the exam formula booklet (but it is a variant of the general chain rule formula which is in the booklet).
What is a useful mathematical relationship between $\frac{d y}{d x}$ and $\frac{d x}{d y}$ ?
A useful mathematical relationship between $\frac{d y}{d x}$ and $\frac{d x}{d y}$ is $\frac{d y}{d x} = \frac{1}{(\frac{d x}{d y})}$ .
This allows $\frac{d y}{d x}$ to be found by finding $\frac{d x}{d y}$ instead and then inverting it.
True or False?
If you know $y = f (x)$ and want to find the derivative of the inverse function $y = f^{- 1} (x)$ , it might be easiest to start with $x = f (y)$ and find $\frac{d x}{d y}$ .
True.
If you know $y = f (x)$ and want to find the derivative of the inverse function $y = f^{- 1} (x)$ , it might be easiest to start with $x = f (y)$ and find $\frac{d x}{d y}$ .
E.g. for $y = \sqrt{{(5 x + 1)}^{3}}$ , you can start with $x = \sqrt{{(5 y + 1)}^{3}}$ , find $\frac{d x}{d y}$ , and then use $\frac{d y}{d x} = \frac{1}{(\frac{d x}{d y})}$ to find the derivative of the inverse of $y = \sqrt{{(5 x + 1)}^{3}}$ .
True or False?
When using $y = f^{- 1} (x) \Leftrightarrow x = f (y)$ to find the derivatives of inverse functions, the final result will always be in terms of x.
False.
When using $y = f^{- 1} (x) \Leftrightarrow x = f (y)$ to find the derivatives of inverse functions, the final result will not always be in terms of x.
The final result will usually be in terms of y, but it may be possible to use a substitution to get the answer in terms of x if required.
The equation for the volume, $V$ , of a right cone is $V = \frac{1}{3} π r^{2} h$ .
How can you find an expression for the rate of change $\frac{d V}{d t}$ expressed in terms of the rates of change $\frac{d r}{d t}$ and $\frac{d h}{d t}$ ?
The equation for the volume, $V$ , of a right cone is $V = \frac{1}{3} π r^{2} h$ .
To find an expression for the rate of change $\frac{d V}{d t}$ expressed in terms of the rates of change $\frac{d r}{d t}$ and $\frac{d h}{d t}$ , differentiate the formula for $V$ implicitly with respect to $t$ :
$\frac{d V}{d t} = \frac{d}{d t} (\frac{1}{3} π r^{2} h) = \frac{2}{3} π r h \frac{d r}{d t} + \frac{1}{3} π r^{2} \frac{d h}{d t}$
(Note that this involves using the product rule along with implicit differentiation.)
What is implicit differentiation?
Implicit differentiation is a method of differentiating equations where y is not given explicitly as a function of x.
True or False?
Implicit differentiation always uses the chain rule.
True.
Implicit differentiation always uses the chain rule.
What is $\frac{d}{d x} (f (y))$ equal to using implicit differentiation?
In implicit differentiation, $\frac{d}{d x} (f (y)) = f' (y) \frac{d y}{d x}$ .
E.g. $\frac{d}{d x} (y^{2}) = 2 y \frac{d y}{d x}$
True or False?
When using implicit differentiation, $\frac{d y}{d x}$ will always be found in terms of $x$ only.
False.
When using implicit differentiation, $\frac{d y}{d x}$ will not always be found in terms of $x$ only.
It will usually be found in terms of both $x$ and $y$ .
True or False?
The product rule is never used in implicit differentiation.
False.
The product rule is often used (along with the chain rule) in implicit differentiation.
E.g. $\frac{d}{d x} (x^{2} y^{2}) = x^{2} \frac{d}{d x} (y^{2}) + y^{2} \frac{d}{d x} (x^{2}) = 2 x^{2} y \frac{d y}{d x} + 2 x y^{2}$
True or False?
At points on a curve where the tangent is horizontal $\frac{d x}{d y} = 0$ , and at points where the tangent is vertical $\frac{d y}{d x} = 0$
False.
At points on a curve where the tangent is horizontal $\frac{d y}{d x} = 0$ , and at points where the tangent is vertical $\frac{d x}{d y} = 0$
True or False?
In optimisation questions using implicit differentiation, the locations of minimums and maximums will always occur at turning points.
False.
In optimisation questions using implicit differentiation, the locations of minimums and maximums will not always occur at turning points.
It is possible that there may not be a turning point. Also the minimum or maximum could be at the start or end of a given or appropriate interval.
What is the derivative of $f (x) = \sec x$ ?
The derivative of $f (x) = \sec x$ is $f^{'} (x) = \sec x \tan x$ .
This is in the exam formula booklet.
True or False?
The derivative of $f (x) = cosec x$ is $f^{'} (x) = cosec x \cot x$ .
False.
The derivative of $f (x) = cosec x$ is $f^{'} (x) = - cosec x \cot x$ .
This is in the exam formula booklet.
True or False?
The derivatives for $\sec x$ and $cosec x$ can be derived using the quotient rule.
True.
The derivatives for $\sec x$ and $cosec x$ can be derived using the quotient rule.
First write $\sec x = \frac{1}{\cos x}$ and $cosec x = \frac{1}{\sin x}$ , then the quotient rule can be used to work out the derivatives of the reciprocal fraction forms.
What is the derivative of $f (x) = \cot x$ ?
The derivative of $f (x) = \cot x$ is $f^{'} (x) = - {cosec}^{2} x$ .
This is in the exam formula booklet.
What is the derivative of $f (x) = \arcsin x$ ?
The derivative of $f (x) = \arcsin x$ is $f^{'} (x) = \frac{1}{\sqrt{1 - x^{2}}}$ .
This is in the exam formula booklet.
True or False?
The derivative of $\arccos x$ is positive for all values in its domain.
False.
The derivative of $\arccos x$ is negative for all values in its domain.
The derivative of $f (x) = \arccos x$ is $f^{'} (x) = - \frac{1}{\sqrt{1 - x^{2}}}$ .
This is in the exam formula booklet.
What is the derivative of $f (x) = \arctan x$ ?
The derivative of $f (x) = \arctan x$ is $f^{'} (x) = \frac{1}{1 + x^{2}}$ .
This is in the exam formula booklet.
What is the derivative of $f (x) = a^{x}$ ?
The derivative of $f (x) = a^{x}$ is $f^{'} (x) = a^{x} (\ln a)$ .
This is in the exam formula booklet.
What is the derivative of $f (x) = \log_{a} x$ ?
The derivative of $f (x) = \log_{a} x$ is $f^{'} (x) = \frac{1}{x \ln a}$ .
This is in the exam formula booklet.
True or False?
To derive the form of the derivative for $\arcsin x$ , start by rewriting $y = \arcsin x$ as $\sin y = x$ .
True.
To derive the form of the derivative for $\arcsin x$ , start by rewriting $y = \arcsin x$ as $\sin y = x$ .
From there, the derivative of $\arcsin x$ can be found by using implicit differentiation and then rearranging.

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