1

Question: Suppose that $g$ is any even differentiable function defined for all real numbers (not necessarily a polynomial). 

<p>Prove that $\frac{dg}{dx}$ is an odd function.</p>

I have seen this post but I want help on the specific part I am stuck on

What I have attempted (my two attempts) :

First attempt

Consider the definition of a derivative: $$\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}$$

If g(x) is a function which is even then $g(-x)=g(x)$ and $g(x-h) = g(-x+h)$

$$g'(x) \lim_{h \rightarrow 0}\frac{g(-x+h)-g(-x)}{h}$$

$$g'(x) \lim_{h \rightarrow 0}\frac{g(x-h)-g(x)}{h}$$

Now I am stuck trying to deduce the deriviative

Second attempt

Consider a function $g(x)$ which is even then $g(x) = g(-x)$ then

$$ g'(x) = \frac{d}{dx}[g(-x)] $$

Then by chain rule

$$ g'(x) = -g'(-x) $$

but if a function is odd $f(-x)=-f(x)$

$$ g'(x) = -(-g'(x)) $$

$$ g'(x) = g'(x) $$

This is not correct but where did I go wrong?

Community
  • 1
bigfocalchord
  • 5,985
  • Hint: since $h$ is approaching $0$ from either side, you can effectively replace $h$ with $-h$ in the limit (or, if you prefer, let $h_2 = -h$). – Théophile Mar 31 '16 at 08:42
  • dude second attempt is 100% correct. just check the domain to be symmetric and from the problem we get that the domain is whole real line so all good. i don't get what's wrong with that. it's totally correct – K.K.McDonald Mar 31 '16 at 08:42
  • 1
    On your second attempt, after doing the chain rule, sub in -x for x. Your problem is you assume g' is odd when you're supposed to be proving it, which you're only one step away from. – Kaynex Mar 31 '16 at 08:42
  • @K.K.McDonald I think i am wrong because I have not proven the derivative is a odd function? that is why – bigfocalchord Mar 31 '16 at 08:44
  • @Kaynex wait why can i sub in -x for x? – bigfocalchord Mar 31 '16 at 08:50

1 Answers1

1

First attempt: $$f'(x) = \lim_{h\rightarrow 0} \frac{f(x + h) - f(x)}{h}$$ Let's assume $f$ is even. Subbing in -x for x: $$f'(-x) = \lim_{h\rightarrow 0} \frac{f(-x + h) - f(-x)}{h}$$ $$f'(-x) = \lim_{h\rightarrow 0} \frac{f(x - h) - f(x)}{h}$$ Make the substitution w = -h: $$f'(-x) = \lim_{w\rightarrow 0} \frac{f(x + w) - f(x)}{-w}$$ $$f'(-x) = -f'(x)$$ Second attempt: $$f'(x) = \frac{d}{dx}[f(-x)]$$ $$f'(x) = -f'(-x)$$ Letting x be -x: $$f'(-x) = -f'(x)$$

Kaynex
  • 2,498
  • 1
  • 15
  • 16