Plan an explanation why (fg)' ≠ f'g' for a student who was absent today.

Answer 1

To prove that (fg)' ≠ f'g' you have to go back to the fundamental definition of a derivative:

To make typing easier let L mean lim h → 0

L (f(x+h)g(f+h) - f(x)g(x))/h

For any equation, you can always add a zero without changing the equation.

L (f(x+h)g(f+h) + 0 - f(x)g(x))/h

Note that if we add f(x)g(x+h) - f(x)g(x+h) is the same as adding zero.

L (f(x+h)g(f+h) + f(x)g(x+h) - f(x)g(x+h) - f(x)g(x))/h

Rearranging terms

L (g(x+h)(f(x+h) - f(x)) + f(x)(g(x+h) - g(x)))/h

Now we can distribute the limit to the two sums by limit definitions:

L g(x+h) • (f(x+h) - f(x))/h + L f(x) • (g(x+h) - g(x))/h

or

L g(x+h) • L (f(x+h) -f(x))/h + L (f(x) • L (g(x+h) - g(x))/h

Note the L (f(x+h) -f(x))/h = f ' (x) and L (g(x+h) - g(x))/h = g'(x)

Also L g(x+h) = g(x)

So rewriting with these definitions

(fg)' = g f ' + f g ' ≠ f 'g '

------

For some special, odd cases, (fg)' = f ' g' and some articles have fun showing which odd functions that works for. But generally (fg)' = g f ' + f g '.

Answer 2

(fg)' = fg' + gf'

the derivative of a product is the 1st term times derivative of the 2nd, plus the 2nd times derivative of the 1st

let f=x

and g = x^2,

then fg=x^3 and (fg)' =(x^3)' = 3x^2

not f'g'=(1)(2x^2) = 2x^2

or an even simpler case