Hey everyone! This guide is going to give a brief run down of differentiation, specifically, what you’ll need to know to answer questions in an exam scenario. I’ll give brief explanations of concepts, then we’ll look at how they are applied in past HSC questions. Reading this guide is an awesome way to revise for class tests or your trials. However, it is by no means an in depth coverage. If at any point you need some extra detail or explanation, there are awesome notes available
here. And of course, if anything is unclear,
pop a question below! You can register
here Either myself or one of your fellow HSC-ers will be sure to help you out.

So to begin, you should be familiar with the process of differentiation itself. Derivatives are without a doubt the most useful aspect of math and science; they allow us to take a function and measure the rate of change of one variable with respect to another. Geometrically, you would know this as a gradient, and in science, a rate of change. Be sure to
understand the process and how it works, you could be asked to explain it.
Stemming from this, you should understand the process of
differentiating from first principles. This is annoying, since you learn the quick way almost straight away, but it gets asked in the HSC a lot.
Like, a lot. Check out the example below, taken from a 2009 HSC Exam. It’s as simple as it gets, but be prepared for more complex examples. Rather than being difficult, these questions are more a test of notation and algebraic skills. Remember the formula!
EG: Use differentiation from first principles to show that
=1\\ )
=\lim_{h\rightarrow0}{\frac{f(x+h)-f(x)}{h}}=\lim _{h\rightarrow 0}{\frac { x+h-x }{ h }}\\\\ =\lim _{ h\rightarrow 0}{\frac{ x+h-x}{h}}\\\\ =1 = RHS<br />)
In this example, we apply the first principles formula and some basic algebra skills. These questions are more of a test of notation than anything else, make sure you are clear!
Of course, you don’t need to do this for every derivative. Unless you
REALLY want to. Instead, you should remember the rules for differentiating different functions (there is an awesome list available
here). You should also remember the
product rule, the chain rule, and the quotient rule, which allow us to differentiate functions in various combinations:
=u'v+v'u )
THE PRODUCT RULE
=\frac { u'v-v'u }{ { v }^{ 2 } })
THE QUOTIENT RULE

THE CHAIN RULE
Note: This may not be how you have learnt these rules, there are many appropriate notations.
You must know how to apply these rules. You will not be told which to use. See this example from the
2013 Paper:
EG: Differentiate

=u'v+v'u\\\\ \quad \quad \quad \quad \quad \quad \quad \quad =2x{ e }^{ x }+{ x }^{ 2 }{ e }^{ x }\\ \\ \quad \quad \quad \quad \quad \quad \quad \quad ={ e }^{ x }{ (x }^{ 2 }+2x)\\ )
In this example, we apply the product rule to differentiate and simplify. If it helps,
I highly recommend doing each derivative separately as shown, before combining them.Be sure to clear with your working in these sort of questions. It’s easy marks if you remember your formulae. And remember, you can differentiate as many times as you like! Though anything beyond the second derivative is useless for the purposes of this course.
More often, though, are questions which apply the derivative in various ways.
Pay extra attention here guys! This question is almost always asked. The most common is the good old, find the equation of the tangent/normal question. To do these, differentiate your function (remember, the derivative can give you the gradient at a specific point!). Combine this with some simple geometry to get your equation. See the example below:
EG: Find the equation of the tangent to the curve
}^{4}} )
at the point where

For this example, the first thing we do is differentiate using the chain rule. Use more working if you need it!
 }^{ 4 }\\ \\ y'=8(2x+1{ ) }^{ 3 } )
Then, we substitute

into the derivative to find the gradient at that point. At the same time, we substitute into the original equation to get the y value at that point.
READ THAT TWICE! I've gotten them backwards many a time.
When

When

Now we can apply the point gradient formula to get our equation:
\\ \\ y-1=-8(x+1)\\ \\ y-1=-8x+8\\ \\ 8x+y+7=0 )
Remember that general form is the
most correct way of giving an answer in the form of an equation. As I said, this question is 90% going to be in your exam! Could you do it blindfolded?

The second most common is a sketching question. These are giveaway marks as long as you know the process. For every question, you should:
1 - Check if the function is odd or even, it saves you lots of time!
2 - Check for asymptotes (vertical and horizontal!)
3 - Get intercepts with the axes
4 - Use derivatives to get turning points
5- Consider what happens as x gets very big and very small
6 - Sketch!Remember that the first derivative gives us the steepness of a curve. The second derivative gives us the concavity (think happy vs frowny face!). These combined can tell us a LOT about the shape of a function.
Note: It has been brought to my attention that I should DEFINITELY mention this. For people doing higher levels of math, be careful of your inflexion points. The second derivative being zero does not guarantee that an inflexion exists. 2U People, you can rest easy, they won't give you a case like this, though you must consider the possibility anyway! This is done by considering the sign of the second derivative either side of the point.
Check out this example from the
2011 HSC:
Let
={ x }^{ 3 }-3x+2 )
.
a) Find the coordinates of the stationary points of
 )
and determine their nature.
b) Hence, sketch a graph of
 )
, showing all stationary points and the y intercept.
For part (a), we get all our derivatives first. Remember, it is almost always easier to use the second derivative as a test for the turning point!

Next, we check for points when the first derivative is zero. Remember, this indicates when the graph is flat, so it is turning around!

Then use the second derivative to test their nature. Negative second derivative means a frowny face, so a maximum turning point, and the opposite for minimum. Note that though it is not shown, you substitute into the original equation to get your y coordinates.
x=1: y''>0, therefore (1,0) is a minimum turning point
x=-1: y''<0, therefore (-1,4) is a maximum turning point
For part (b), we use this data and go through the steps (don't write Step 1, Step 2 etc in an exam, otherwise the markers will be really confused as to what the hell is going on!).
1 - Neither odd nor even
2 - No asymptotes
3 - When x=0, y=2 so (0,2) is the y intercept.
When y=0, x=-2 so (-2, 0) is the x intercept
There is no easy way to find this x intercept, which is why it isn't compulsory to show it in this question.
4 - Done in previous part.
5 -

6- Sketch away!

There are other questions you could be asked too. Velocity and acceleration is common, as are rate of change considerations. Be sure to practice lots of different questions to be prepared for everything. But if you can do these sorts of questions, that’s an awesome set up for your exams!
Finally, a few hot exam tips to keep in mind for Trials and HSC (
post yours below!):
1 - Remember that if you aren't confident with the chain rule, you can expand the brackets! Don't attempt this with anything above a power of 3 though.
2 - With curve sketch questions, check your y value for minimum is less than that for your maximum, to make sure you haven't mixed them up somewhere.
3 - Be EXTREMELY clear with your logic, especially when dealing with multiple derivatives. Marks will be deducted for poor notation or explanation. Don't assume the markers know what you mean, be clear and methodical!
4 - Remember that if the first derivative is constant, then the curve is a straight line. If it is zero, it's a flat line.
5- Study hard!
Be sure to
shoot a message if anything here is unclear, or if there are other questions you would like explained or broken down! Remember to check out the notes for a more comprehensive breakdown of the concepts. And for people doing Extension 1, soon there will be a guide on your extra stuff in the calculus section, like simple harmonic motion, projectiles, and all the other cool stuff
