The derivative of an odd function is always an even function.

Answer

To prove that the derivative of an odd function is always an even function, we will follow these steps:

### Step 1: Define Odd and Even Functions
An odd function \( f(x) \) satisfies the condition:
\[ f(-x) = -f(x) \]
for all \( x \).

An even function \( g(x) \) satisfies the condition:
\[ g(-x) = g(x) \]
for all \( x \).

### Step 2: Define the Derivative
The derivative of a function \( f(x) \) is defined as:
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]
We can also express the derivative at \( -x \):
\[ f'(-x) = \lim_{h \to 0} \frac{f(-x+h) - f(-x)}{h} \]

### Step 3: Substitute for Odd Function
Since \( f(x) \) is an odd function, we can substitute \( -x \) into the function:
\[ f(-x) = -f(x) \]
Now, substituting \( -x \) into the derivative:
\[ f'(-x) = \lim_{h \to 0} \frac{f(-x+h) - f(-x)}{h} \]
Using the property of odd functions, we have:
\[ f(-x+h) = -f(x-h) \]
and
\[ f(-x) = -f(x) \]
Thus, we can rewrite the derivative:
\[ f'(-x) = \lim_{h \to 0} \frac{-f(x-h) + f(x)}{h} \]

### Step 4: Simplify the Expression
Rearranging gives:
\[ f'(-x) = -\lim_{h \to 0} \frac{f(x-h) - f(x)}{-h} \]
This can be rewritten as:
\[ f'(-x) = -f'(x) \]

### Step 5: Show that the Derivative is Even
To show that \( f'(-x) = f'(x) \), we need to manipulate our expression:
\[ f'(-x) = -f'(x) \]
This means that if we take the negative of \( f'(-x) \) we get \( f'(x) \), which implies that the derivative \( f'(x) \) is even:
\[ f'(-x) = f'(x) \]

### Conclusion
Thus, we have shown that the derivative of an odd function \( f(x) \) is always an even function \( f'(x) \).

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