Differentiation Techniques

Maria Gaetana Agnesi

So, no one wants to do complicated limits to find derivatives. There are easier ways of course. There are a number of quick ways (rules, formulas) for finding derivatives of the Elementary Functions and their compositions. Here are some ways to introduce these rules; these are the subject of this week’s review of past posts.

Why Radians?

The Derivative I        Guessing the derivatives from the definition

The Derivative II      Using difference Quotient to graph and guess

The Derivative Rules I    The Power Rule

The Derivative Rules II       Another approach to the Product Rule from my friend Paul Foerster

The Derivative Rules III     The Quotient Rule developed using the Power Rule, an approach first suggested  by Maria Gaetana Agnesi (1718 – 1799) who was helping her brother learn the calculus.

Next week: The Chain Rule.






Power Rule Implies Chain Rule

Having developed the Product Rule d\left( uv \right)=u{v}'+{u}'v and the Power Rule \frac{d}{dx}{{x}^{n}}=n{{x}^{n-1}} for derivatives in your class, you can explore similar rules for the product of more than two functions and suddenly the Chain Rule will appear.

For three functions use the associative property of multiplication with the rule above:

d\left( uvw \right)=d\left( \left( uv \right)w \right)=u\cdot v\cdot dw+w\cdot d(uv)=u\cdot v\cdot dw+w\left( udv+vdu \right)

So expanding with a slight change in notation:

d\left( uvw \right)=uv{w}'+u{v}'w+u'vw

For four factors there is a similar result:

d\left( uvwz \right)=uvw{z}'+uv{w}'z+u{v}'wz+{u}'vwz

Exercise: Let {{f}_{i}} for i=1,2,3,...,n be functions. Write a general formula for the derivative of the product {{f}_{1}}{{f}_{2}}{{f}_{3}}\cdots {{f}_{n}} as above and in sigma notation


d\left( {{f}_{1}}{{f}_{2}}{{f}_{3}}\cdots {{f}_{n}} \right)={{f}_{1}}{{f}_{2}}{{f}_{3}}\cdots {{{f}'}_{n}}+{{f}_{1}}{{f}_{2}}{{{f}'}_{3}}\cdots {{f}_{n}}+{{f}_{1}}{{{f}'}_{2}}{{f}_{3}}\cdots {{f}_{n}}+\cdots +{{{f}'}_{1}}{{f}_{2}}{{f}_{3}}\cdots {{f}_{n}}

\displaystyle d\left( {{f}_{1}}{{f}_{2}}{{f}_{3}}\cdots {{f}_{n}} \right)=\sum\limits_{i=1}^{n}{\frac{{{f}_{1}}{{f}_{2}}{{f}_{3}}\cdots {{f}_{n}}}{{{f}_{i}}}{{{{f}'}}_{i}}}

This idea may now be used  to see the Chain Rule appear. Students may guess that d{{\left( f \right)}^{4}}=4{{\left( f \right)}^{3}}, but wait there is more to it.

Write {{\left( f \right)}^{4}}=f\cdot f\cdot f\cdot f\text{ }. Then from above

d{{\left( f \right)}^{4}}=d\left( f\cdot f\cdot f\cdot f\text{ } \right)=f\cdot f\cdot f\cdot {f}'+f\cdot f\cdot {f}'\cdot f+f\cdot {f}'\cdot f\cdot f+{f}'\cdot f\cdot f\cdot f

d{{\left( f \right)}^{4}}=4{{\left( f \right)}^{3}}{f}'\text{ }

Looks just like the power rule, but there’s that “extra” {f}'. Now you are ready to explain about the Chain Rule in the next class.

Foreshadowing the Chain Rule

I assigned another very easy but good problem this week. It was simple enough, but it gave a hint of things to come.

Use the Product Rule to find the derivative of {{\left( f\left( x \right) \right)}^{2}}.

Since we have not yet discussed the Chain Rule, the Product Rule was the only way to go.

\frac{d}{dx}{{\left( f \right)}^{2}}=\frac{d}{dx}\left( f\cdot f \right)=f\cdot {f}'+{f}'\cdot f=2f\cdot f'

 And likewise for higher powers:

\frac{d}{dx}{{f}^{3}}=\frac{d}{dx}\left( f\cdot f\cdot f \right)=f\cdot f\cdot {f}'+f\cdot {f}'\cdot f+{f}'\cdot f\cdot f=3{{f}^{2}}{f}'

If you just look at the answer, it is not clear where the {f}' comes from. But the result foreshadows the Chain Rule.

Then we used the new formula to differentiate a few expressions such as {{\left( 4x+7 \right)}^{2}} and {{\sin }^{2}}\left( x \right) and a few others.

Regarding the Chain Rule: I have always been a proponent of the Rule of Four, but I have never seen a good graphical explanation of the Chain Rule. (If someone has one, PLEASE send it to me – I’ll share it.)

Here is a rough verbal explanation that might help a little.

Consider the graph of y=\sin \left( x \right). On the interval [0,2\pi ] it goes through all its value in order once – from 0 to 1 to 0 to -1 and back to zero. Now consider the graph of y=\sin \left( 3x \right). On the interval \left[ 0,\tfrac{2\pi }{3} \right] it goes through all the same values in one-third of the time. Therefore, it must go through them three times as fast. So the rate of change of y=\sin \left( 3x \right) between 0 and \tfrac{2\pi }{3} must be three times the rate of change of y=\sin \left( x \right). So the rate of change of  must be 3\cos \left( 3x \right). Of course this rate of change is the slope and the derivative.

The Derivative Rules II

The Product Rule

Students naturally figure that the derivative of the product of two functions is the product of their derivatives. So first you must disabuse them of this idea. That is easy enough to do.

Consider two functions and their derivatives  f\left( x \right)={{x}^{7}}\text{ with }{f}'\left( x \right)=7{{x}^{6}} and g\left( x \right)={{x}^{5}}\text{ with }{g}'\left( x \right)=5{{x}^{4}}. So now f\left( x \right)g\left( x \right)={{x}^{12}} and \frac{d}{dx}\left( f\left( x \right)g\left( x \right) \right)=12{{x}^{11}}. Is this {f}'\left( x \right){g}'\left( x \right)=35{{x}^{10}}? No it is not!

But all is not lost. How can we get the correct answer from the original functions and their derivatives? Start with the 12; this comes from adding the 7 and 5 so the correct answer must be something along the lines of

From the expression we already have what can we put in the blank spaces to get the similar terms with {{x}^{11}}? How about the original functions?

And there is the product rule right there

You can use this same idea with other products.

You may also use the definition of derivative which you can find in most books, but bringing in zero in the form of -f\left( x+h)g\left( x \right) \right)+f\left( x+h)g\left( x \right) \right) is hardly something you would expect anyone to figure out by themselves. As I mentioned, I’m more into explaining than proving.

(This example is one of many I learned from Paul Foerster. Thanks again, Paul)

Here is another approach suggested by Dick Sisley. Thank you, Dick.

If students already know the Chain Rule:

Then–let h(x)= f(x)* f(x) = (f(x))^2 (this is a key equivalence.)

Next use the Chain Rule to get h'(x)= 2*f(x)*f ‘(x)= 2*(f ‘(x)*f(x)).

Now note that 2*(f ‘(x)*f(x))= f ‘(x)* f(x) + f ‘(x)* f(x)

The key step is then to let h(x) = f(x)*g(x) and ask students to use the result for f ‘(x)*f(x) to conjecture the result for f(x)*g(x).  There have always been some who come up with f ‘(x)*g(x) + f(x)*g'(x).  Others come up with other, non-equivalent conjectures.  But there is a way to evaluate the likelihood of every conjecture.

Use h(x)= f(x)*f(x)= x * x. We know the result should be 2*x.

Use h(x)= f(x)*f(x) = x^2 * x. We know the result should be 3*x^2.


We can experiment with products such as sin(x)*x^2.  If we use the correct conjecture pattern, we can test the reasonableness of the result using the numerical derivative feature of a graphing calculator on values the students select.

Updated 11-6-2013

Next The Quotient Rule.