Question

Problem Set 1. 1C-2: Let f(x) = (x−a)g(x). Use the definition of the derivative to calculate that f'(a) = g(a), assuming that g is continuous.

Answer 1

I don't know what I'm doing wrong. I tried the following:

$$f’(x) = \frac{f(x + \Delta x) – f(x)}{\Delta x} = \frac{(x + \Delta x\ –\ a)g(x+ \Delta x)-(x-a)g(x)}{\Delta x}$$

I could distribute the function g inside the parenthesis but I'm going to end up with terms that don't allow me to get rid of \Delta x of the denominator.

Thanks !!

Answer 2

How about writing it like this (let me use h instead of \( \Delta x \) , it's easier to type.
\[ f'(x) = \lim_{h\rightarrow 0 }\frac{ ( h + (x-a))g(x+h) - (x-a) g(x)}{h} \\
= \lim_{h\rightarrow 0 }\frac{h g(x+h) + (x-a) g(x+h) - (x-a) g(x) }{h} \\
= \lim_{h\rightarrow 0 }\frac{h g(x+h) + (x-a) (g(x+h) - g(x) )}{h} \\
\lim_{h\rightarrow 0 }\frac{h g(x+h) }{h} + (x-a) \lim_{h\rightarrow 0 }\frac{g(x+h) - g(x)}{h}
\]

Answer 3

Thanks for answering @phi

There are no more steps after the last one you've written, right? Can the leftmost expression be left in that way, even when\(\ h\) cannot be removed from the denominator?

Answer 4

Yes, there are more steps. The question asks that you show
\[ f'(a) = g(a) \]

***
Can the leftmost expression be left in that way, even when h cannot be removed from the denominator?
***
I do not understand. in the left-most term you have
\[ \frac{\cancel{ h} \ g(x+h)}{\cancel{ h}} = g(x+h) \]
as long as h is *not* zero (though we can let h approach as close to 0 as we like)

the right-most term is (by definition) g'(x)

Answer 5

*the right-most term is (by definition) g'(x) times (x-a)

Answer 6

|: I confused the right side with the left side.

Perfect I got it now.

Thank you!!

Answer 7

@phi