Is the derivative of an even function always an odd function?

[JasonRox]

If a function is even, prove that the derivative is odd.

Look at a graph of x^2 we can clearly see why.

This is how I would approach this...

If we solve d/dx x^n and n is an even integer, we get the derivative nx^(n-1). Since n is even, n-1 is odd.

Because n-1 is odd, the derivative nx^(n-1) becomes odd because f'(-x)= - f'(x). Therefore the derivative of an even function becomes an odd function.

Note: The TA couldn't solve this...

I excluded the proof of d/dx x^n = nx^(n-1) because it is not necessary because I know how to do that.

 

[NateTG]

Since f(x) is even we have:
f(x)=f(-x)
now, let's take the derivative of both sides. The LHS is easy:
\frac{d}{dx}f(x)=f'(x)
and we can do the RHS using the chain rule:
\frac{d}{dx}f(-x)=- f'(-x)
so, the original equation turns into:
f'(x)=-f'(-x)
multiply both sides by -1
-f'(x)=f'(-x)
which indicates that the function is odd.

 

[HallsofIvy]

Not all functions are powers of x!

If f(x)= sin(x), an odd function, its derivative is cos(x), an even function.

NateTG's response is the way to go.

 

[JasonRox]

Thanks NateTG.

 

[izilude]

Not all functions are powers of x!

If f(x)= sin(x), an odd function, its derivative is cos(x), an even function.

NateTG's response is the way to go.

Yes but this would be sufficient if the function can be expanded in a power series. sin x = x - x^3/3! + x^5/5! +...

d sinx/dx = 1 - x^2/2! + x^4/4 +... = cos x.

Thus, his proof is correct for all analytic functions.