Question

Differentiate:\[-\frac{d}{dx}f(x) = \frac{d}{dx}f(-x)\]I am trying to prove that the derivative of an odd function is even, but when I try to use the Chain Rule, it comes out to the derivative = 1

Answer 1

Did I miss a negative symbol somewhere?

Answer 2

maybe I misunderstand the problem, but if you derive the lefthand side you obtain the negative derivative of the general function right?

\[ -f'(x)\] ?

Answer 3

The question is just: Given that f(x) is odd, prove that f'(x) = f'(-x)

Answer 4

well for the right hand side I would use the chain rule, then you will get

\[ -f'(x)=-f'(-x) \] the negative cancels out.

Answer 5

(in fact I used the chain rule for both of them, but on the LHS it's just multiplying by 1) on the right hand side it's multiplying by -1

Answer 6

Are you sure it's not f(x)=-f(-x) to differentiate?

Answer 7

wouldn't make a difference, if you differentiate what you just have written you get the same

Answer 8

\[\frac{d}{dx}f(x) = -\frac{d}{dx}f(-x)\]right?

Answer 9

I know the definition of an odd function as the following:

\[ -f(x)=f(-x)\]

Answer 10

Yes, I just moved the negative sign over

Answer 11

good, so we're doing the same, yes, now differentiate, using the chain rule.

Answer 12

\[\frac{d}{dx}f(x) = -\frac{d}{dx}f(-x) \cdot \frac{d}{dx}f(x)\]am I applying the Chain Rule correctly?

Answer 13

The chain rule says, differentiate the outside with respect to the inside and then differentiate the inside and multiply, so what you write above doesn't look bad, but I believe this is more what happens:

\[ \Large \frac{d}{dx} f(x)=-\frac{d}{dx}f(-x)\cdot \frac{d}{dx}(-x)\]

Answer 14

Whoa, okay, I guess I don't fully understand the Chain Rule yet. I have no idea why you applied\[\frac{d}{dx}(-x)\]for the Chain Rule

Answer 15

\[ \frac{d}{dx}(5x+7)^8 =8(5x+8)^7 \cdot 5 \]
First I derived the outside of the function with respect to the inside, hence the inside stays the same, notice on both LHS and RHS we have 5x+7, but the function (the exponential) got derived. 8f(x)^7 and then I multiplied times the derivative of the inside which is 5.

Answer 16

oh sorry, I misspelled, it should be 5x+7 on both of course, my typing is bad today.

Answer 17

Ohh, I see now. I was using the Chain Rule similarly as I would with implicit differentiation

Answer 18

CORRECTION ! :

\[ \large \frac{d}{dx}(5x+7)^8 =8(5x+7)^7 \cdot 5\]

Answer 19

So we have\[\frac{d}{dx}f(x) = -\frac{d}{dx}f(-x) \cdot \frac{d}{dx}(-x)\]Not sure how to combine these terms or what I'm allowed to do and not do

Answer 20

The derivative of f(x) is just f'(x), similarly for f(-x) the derivative is f'(-x) and the derivative of x would be one as you know, but you differentiate -x here, so the derivative becomes minus one.

Answer 21

therefore the minus cancels out, as predicted by the comment above we had to prove.

Answer 22

Usually I see derivatives in the form:\[\frac{dy}{dx}\]so this kind of form really threw me off...is this the same as\[\frac{d}{dx}y\]?

Answer 23

yes this is Leibniz's Notation for derivatives, it's a bit confusing at first, the term

\[ \large \frac{d}{dx} \] is Leibniz differential Operator, there are many equal ways of writing something here now:

\[ \large f'(x)= \frac{dy}{dx}=\frac{d}{dx}y = \frac{d(f(x))}{dx} \]

Answer 24

you will see plenty of those.

Answer 25

So isn't\[\frac{dy}{dx} = 1\]?

Answer 26

No why should it be? The derivative of a single variable is one, if you differentiate it to respect of that variable

\[ y= x \] Apply Leibniz Operator

\[ \frac{dy}{dx}=\frac{dx}{dx}=1 \]

Answer 27

y is a function, it can be anything. From linear to complex.

Answer 28

If f(x) = y
How would I write f(-x) in terms of y?

Answer 29

Same, you don't have a function specified there, remember that whatever you denote into the brackets you will write in the actual function

\[ f(-x)=y\] this is an even function now

Answer 30

But you can write -f(x) as -y, right?

Answer 31

yes you can do that

\[ -y = f(-x) \\
y=-f(-x)\]

Is this what you are trying to do? Differentiate

\[ \frac{dy}{dx}=f'(x)=f'(-x)\]

Answer 32

Something like that, I'm just trying to understand the general concepts

Answer 33

same result as above, so it works