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
12{{x}^{11}}=7{{x}^{6}}\_\_\_\_\_+5{{x}^{4}}\_\_\_\_\_

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?
12{{x}^{11}}=7{{x}^{6}}\underline{{{x}^{5}}}+5{{x}^{4}}\underline{{{x}^{7}}}

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)

Next The Quotient Rule.

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