# 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.

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.

# Far Out!

A monster problem for Halloween.

A while ago I suggested you look at $\displaystyle \underset{x\to \infty }{\mathop{\lim }}\,\frac{\ln \left( {{x}^{5}} \right)}{{{x}^{0.02}}}$ , which using the dominance idea is zero. Of course your students may try graphing or a table. Here’s the graph done by a TI-Nspire CAS. Note the scales.

This is not the way to go. Since the function is increasing near the origin, but the limit at infinity is zero there must be a maximum point where the function starts decreasing. And as the expression can never be negative once x > 1, there must be a point of inflection where the graph becomes concave up and can thereafter approach the x-axis from above as a horizontal asymptote. The maximum can be found by hand which makes for some great algebra manipulation practice:

$\displaystyle \frac{d}{dx}\left( \frac{\ln \left( {{x}^{5}} \right)}{{{x}^{0.02}}} \right)=\frac{{{x}^{0.02}}\tfrac{5{{x}^{4}}}{{{x}^{5}}}-\ln \left( {{x}^{5}} \right)\left( 0.02{{x}^{-0.98}} \right)}{{{x}^{0.04}}}$

$\displaystyle \frac{d}{dx}\left( \frac{\ln \left( {{x}^{5}} \right)}{{{x}^{0.02}}} \right)=\frac{{{x}^{-0.98}}\left( 5-\left( 0.10 \right)\ln \left( x \right) \right)}{{{x}^{0.04}}}=\frac{50-\ln \left( x \right)}{10{{x}^{1.02}}}$

Setting this equal to zero and solving gives $x={{e}^{50}}\approx 5.185\times {{10}^{21}}$

The second derivative is $\displaystyle \frac{{{d}^{2}}}{d{{x}^{2}}}\left( \frac{\ln \left( {{x}^{5}} \right)}{{{x}^{0.02}}} \right)=\frac{-510+10.2\ln \left( x \right)}{100{{x}^{2.02}}}$

and is zero when x$\displaystyle {{e}^{\frac{520}{10.2}}}\approx 1.382\times {{10}^{22}}$

Okay, I skipped a few steps here, but you can challenge your students with that. Since we’re really interested in the solution here more than the solving ,this is really a good place to use a CAS calculator.

The first line in the figure above defines the function to save typing it each time. The second line finds the x-coordinate of the maximum point (how do we know this is a maximum?) and the third finds the x-coordinate of the point of inflection.  Much simpler this way!

Take a minute to consider the numbers. They are BIG! In fact, if the units on our graph paper are centimeters, then the maximum point is a little over 5,480 light-years away from the origin! The point of inflection is about 2.665 times farther at more than 14,607 light-years away!

Meanwhile the maximum value is only 91.9699 cm. That’s right centimeters, less than a meter. And the y-coordinate of the point of inflection is about 91.9524 cm. A drop of 0.0175 cm. in a horizontal distance of a little over 9,127 light-years.

Some problems are a lot less scary if done with technology.

# Derivative Rules III

The Quotient Rule

This approach to the quotient rule is credited to Maria Gaetana Agnesi (1718 – 1799) who wrote the first known mathematics textbook Analytical Institutions (1748) to help her brothers learn algebra.

The quotient rule can also be proven from the definition of derivative. But here is a simpler approach – as a corollary of the product rule.

Begin by letting $\displaystyle h\left( x \right)=\frac{f\left( x \right)}{g\left( x \right)}$.

Then

$f\left( x \right)=h\left( x \right)g\left( x \right)$ and ${f}'\left( x \right)=g\left( x \right){h}'\left( x \right)+h\left( x \right){g}'\left( x \right)$.

Then solving for ${h}'\left( x \right)$:

$\displaystyle {h}'\left( x \right)=\frac{{f}'\left( x \right)-h\left( x \right){g}'\left( x \right)}{g\left( x \right)}$

$\displaystyle =\frac{{f}'\left( x \right)-\frac{f\left( x \right)}{g\left( x \right)}{g}'\left( x \right)}{g\left( x \right)}$

$\displaystyle =\frac{g\left( x \right){f}'\left( x \right)-f\left( x \right){g}'\left( x \right)}{{{\left( g\left( x \right) \right)}^{2}}}$

Mnemonics

I’m not really one for mnemonics. I cannot spell SOHCOHTOA without saying to myself, “sine, opposite over hypotenuse; cosine, adjacent …” It seems better to me anyway to have student just memorize the formulas in words using the correct terms:

The derivative of a product is the first factor times the derivative of the second plus the second factor times the derivative of the first.

The derivative of a quotient is the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator all divided by the square of the denominator.

But whatever works for you. Lo Di Hi