    graemethin
bannerdive
graemethin2

EASIER THAN YOU THINK...

# Derivatives and the Gradient Function  Once the method of finding derivatives from first principles was discovered, mathematicians quickly attempted to learn the rules that would govern how they worked.  The applied these principles to as many functions as they could.

The illustration at right shows all six basic derivative patterns taught in schools here in NSW, Australia.  They are the derivatives for the six basic functions that you see … three trigonometric functions, the polynomial function, the exponential function, and the logarithmic function.

Mathematicians then had to learn what rules applied when functions were added, subtracted, multiplied, or divided.  It transpired that the derivative of a string of functions (that were added and subtracted) was simply the sum of, and difference between, their derivatives.  Multiplication and division were a different matter, however, and the Product Rule and Quotient Rule were devised for such compound functions.  The simple case of the product rule, where a function was multiplied by a constant, turned out to be remarkably simple.  The constant simply multiplied the derivative as well.

So, using the symbol D meaning derivative (i.e. d/dx), and the dash notation where D[a(x)] = a'(x), and D[g(x)] = g'(x), we now have the following rules:

D[a(x) + g(x)] = a'(x) + g'(x)

D[a(x) – g(x)] = a'(x) – g'(x)

D[a(x) ⋅ g(x)] = a'(x)⋅g(x) + a(x)⋅g'(x) ~ Product Rule

D[a(x) / g(x)] = [a'(x)⋅g(x) – a(x)⋅g'(x)]/[g(x)]² ~ Quotient Rule

D[k⋅a(x)] = k⋅D[a(x)] =k⋅a'(x) ~ Multiplication by a constant

It was also not long before mathematicians worked out how to find the derivatives when there were functions within functions (functions of functions):

D[a(g(x))] = a'(g(x))⋅g'(x), and

D[a(g(h(x)))] = a'(g(h(x)))⋅g'(h(x))⋅h'(x) ~ Chain Rule (or Function of a Function Rule)

Then the rules were worked out for finding the derivatives for all kinds of combinations of functions (compound functions)!  Of course, there was a lot of work being carried out concerning what all this meant and how to use these derivatives, but I am talking about the basic skills of how to calculate derivatives for the moment.

If you have had difficulty finding derivatives of compound functions, then please watch all four of these videos. They are long (15-25 minutes each), but I am sure you will find them worth-while! I start by reviewing the derivatives of the six basic functions and then show you, step-by-step, how to calculate the derivatives of most functions encountered at school.  With a little practice, you should be able to find the derivatives of most functions within 5-15 seconds and provide a ‘tidied up’ solution within 20-30 seconds.  Only the most complicated functions at school should take you more than a minute.  I am speaking of the normal advanced level courses, by the way.  In the very top courses, students may encounter some quite involved problems.

Using the Chain Rule With Speed and Finesse (20 seconds per question)

In this first video you will learn to use the Chain Rule to find derivatives of simple functions within about 20 seconds (per question).

Most of us are taught to find the derivatives of compound functions by substitution (in the case of the Chain Rule) or by a substitution pattern, for example, for the Product Rule (u'v + v'u) and the Quotient Rule [(u'v - v'u)/v²]. Many students find these confusing and, particularly in the case of chain rule substitution, end up with 'bits and pieces' all over the page.

The method that you will learn here (for using the Chain Rule) uses the substitution concept (or pattern) but avoids the technical process ... and allows you to calculate derivatives in one line of work instead of up to six lines of work. Any further work is algebraic simplification of the expression. If you learn and practise this technique, you will be able to find the derivatives of even the most complicated chains in one process, rapidly and without losing track of where you are up to.

Using the Product Rule With Pizzaz! (20 seconds per question)

In this second video you will learn to use the Product Rule to find derivatives of simple functions within about 20 seconds (per question).

You were probably taught (as I was) to find the derivatives of the products of functions by using a substitution pattern (u'v + v'u). Many students find the use of this pattern confusing and, if there are more than two functions multiplied together, the use of this pattern gets quite complicated! The method that you will learn here (for using the Product Rule) uses the same foundational theory (or pattern) but avoids the technical process ... and allows you to calculate derivatives in one line of work. Any further work is only algebraic simplification of the expression. If you learn and practise this technique, you will be able to find the derivatives of even the most extensive products in one process (or step), rapidly and without losing track of where you are up to.

Twelve Demonstrations of the Product Rule in Action!

This is the first of four videos in response to a YouTube subscriber's request.

Chelsea asked me to demonstrate how to use the product rule to find the derivatives of a set of 12 questions that she provided.

In this video I find the derivative of these functions:

• y = (x² + 6x +5)(x² -4x - 5)
• y = (x² - 3x +1)(x² + x + 5)
• y = (x + 1)(x² - 3)³

I show how you can use the product rule by following a pattern ... i.e. using no substitutions (and, therefore, no 'bits and pieces' all over the page).

To help you practise this skill, I have created a FREE PDF FILE containing a wide variety of exercises (and their solutions). You might like to work through them either on your own or with friends. You may download the file from here.

Thank you, Chelsea, for requesting these videos. I hope they help you and many others.

This is the second of four videos in response to a YouTube subscriber's request (thank you, Chelsea).

In this video I calculate the derivative of these functions:

• y = (x + 2)(1 - x²)^4
• y = (x² - 1)²(x + 5)²
• y = (x² + 1)²(x + 1)²

In this video I find the derivative of these functions:

• y = x²(1 - 6x)^-3
• y = x³(1 - 2x)^-4
• y = x³√(1 - 5x)

In this video I find the derivative of these functions:

• y = x²√(2x + 2)
• y = (x² + 1)√(x + 1)
• y = √(x - 1).(x² - 1)

Using the Quotient Rule With Aplomb! (20-30 seconds per question)

In this third video you will learn to use the Quotient Rule to find derivatives of simple functions within about 20-30 seconds (per question).

You were probably taught (as I was) to find the derivative of a ratio of functions (i.e. one function divided by another function) by using a substitution pattern (u'v - v'u)/v². Many students find the use of this pattern confusing!

The method that you will learn here (for using the Quotient Rule) uses the same foundational theory (or pattern) but avoids the technical process ... and allows you to calculate derivatives in one line of work. Any further work is only algebraic simplification of the expression. In the video you will be shown how to set your work out and what steps to follow (and in what order) to find a derivative. If you learn and practise this technique, you should be able to find the derivatives of even quite difficult functions in one process (or step), rapidly, and without losing track of what you are doing.

To help you practise this skill, I have created a FREE PDF FILE containing a wide variety of exercises (and their solutions). You might like to work through them either on your own or with friends. You may download the file from here.

Using all 3 Rules with Compound Functions

The beauty of recognising and using structures for the Chain, Product and Quotient Rules is that these same structures can help us find the derivatives of compound functions with relative ease ... and in just one line of work!

This method uses no substitutions! It uses a structure (or pattern) of setting your work out and it allows you to find the derivatives of quite complicated functions in just one line of work.  The rest of your work will be devoted to simplifying and tidying up your derivative.

Simple structures, less work and more ease means that you will work more rapidly and with less risk of making careless mistakes. So, the speed and accuracy with which you calculate derivatives should both improve!

If the steps appear too difficult for you, please watch my previous three videos so you can understand the patterns that are being used.

To help you practise this skill, I have created a FREE PDF FILE containing a wide variety of exercises (and their solutions). You might like to work through them either on your own or with friends. You may download the file from here.

One of my YouTube subscribers asked me to explain implicit differentiation (thank you, Janjira). This is the first of four videos that will do just that!

This one is a long video (27:54), giving a fairly full explanation, with examples. The following three videos are much shorter ... simply providing examples of implicit differentiation.

What you will have studied so far are explicit functions and their derivatives (using the chain, product and quotient rules). An explicit function is easy to spot! It simply has the y term on the LHS of the = sign and the function of x on the RHS. There are no x terms on the left, and no y terms on the right. In this way, y is explicitly defined as a function (or a relation) in terms of x. So far, so good.

An implicit function is basically any other equation involving x and y. The terms may be jumbled (moved from one side of the equation to the other) and even combined (expressions may involve both x and y terms). In particular, if all the terms are moved to the LHS, we get statements like x² -xy³ + 4sin(2x + y) = 0. In general, we say that any function A(x,y) = 0 is an impicit function ... that is, the relationship between x and y is implied in the equation, but not yet made explicit.

Of course, these equations/formulae are not always functions. Sometimes (often) they are relations. But finding their derivative implicitly in terms of x and y allows us to find the gradient of the tangent even to such relations.

Finding a derivative implicitly is actually fairly easy and normally only takes about five lines of working. The chain, product and quotient rules still apply. The only new thing to remember is that, when finding the derivative (with respect to x) of a term involving y, as the chain rule is applied, the last term multiplied will be dy/dx. For ease of use and simplcity in writing, we usually write this as y'. In this video I give examples of this process.

The result will always be a linear equation in y'. We simple collect all those terms on the LHS of the equation, all move all the other terms (that do not include a y') to the RHS, remove y' as a common factor from the LHS and then perform a simple division. This process is demonstrated using three simple examples. The following three videos demonstrate more typical examples.

This is a long video but I wanted to cover all the essential bases fairly thoroughly without getting too technical. Here, I concentrate on the basic concepts and processes so that students will be able to complete examination and assignment and homework questions with alacrity (if not, aplomb!).

Implicit Differentiation ~ Three Examples

Hopefully, you will have seen my previous video which explains what implicit differentiation is, and how to 'do it.'

In this video I explain how to find the derivative of y = sin(3x + 4y) using implicit differentiation.

I recently searched the Internet looking for a website or webpage that provided a good set of questions on which students might practise their skills. One page that I discovered contained a lovely collection of 16 questions as well as links to their worked solutions. Let me therefore recommend Duane Kouba's webpage at the University of California, Davis.

I have chosen three of his examples to use as the basis for these three (demonstration) videos ... of which this is the first.

In this video I explain how to find the derivative of y = x²y³ + x³y² using implicit differentiation.

In this video I explain how to find the derivative of x² - xy + y² = 3 using implicit differentiation.

How to Differentiate a Function to the Power of a Function

Sudhanshu (a YouTube subscriber) asked me to produce a video explaining how to differentiate a function to the power of another function, in other words, y = f(x)^g(x). This is the result!

Rather than provide all the theory, I demonstrate the process using three functions:

• y = cosθ^tanθ
• y = ln(x²)^sin(2x)
• y = x˟

In each case we convert the function f(x) into an exponential function by using the fact that exponentials and logarithms are inverse functions. This means that f(x) = e^ln[f(x)].  Once this conversion has been made, we now have to find the derivative of an exponential function y = e^{ln[f(x)].g(x)} ... and "its bark is worse than its bite!" In other words, it is easier than it looks!

This requires that we use the chain rule, since we have a function (exponential base e) of a function {ln[f(x)].g(x)}. Note that we will also require the product rule because ln[f(x)].g(x) is a product of two functions.

Thank you for asking this question, Sudhanshu. I hope the video answers your question and helps others as well.

If you want a good variety of examples to practise your skills in finding derivatives, including some of the form f(x)^g(x), download my FREE PDF FILE which should give you sufficient revision practice for quite a while. You might like to work through them either on your own or with friends. You may download the file from here.

Amazing Way to Graph the Gradient Function (Derivative)

It is likely that you have never seen this in all your schooling. Certainly, in our schools it seems to be an unknown skill.

Usually, when we are asked to draw a gradient function (graph of the derivative of a function), we are not asked for great accuracy.

We are expected to find the stationary points (locations of horizontal tangents). We are also expected to identify whether the gradient is positive or negative between each of those points of zero gradient. If the original graph has no identified points and no scale on the axes, any more detail than that is not required.

What you are about to see is a graphical/geometric method for drawing quite accurate gradient functions (graphs of derivatives), using a very simple geometric technique (parallel lines) and triangulation concept (gradient = rise/run).

I read about it many years ago in a publication of the NSW Department of Education called The Mathematics Teacher. The contributer was G I Miller from Corowa (near the NSW/Victorian border). I tried to find out more about this person but could not. I think he deserves some cudos for having shared this method.