To find the derivative of the function \(\( f(x) = \sqrt{x^2 + a^2} - \sqrt{x^2 - a} \)\) with respect to \(\( x \)\), we can follow these steps:
1. Express the Function and Differentiate:
The function is:
\(\[
f(x) = \sqrt{x^2 + a^2} - \sqrt{x^2 - a}
\]\)
2. Differentiate Each Term Separately:
We will use the chain rule to differentiate each term separately.
3. Differentiate \(\( \sqrt{x^2 + a^2} \)\):
Let \(\( u = x^2 + a^2 \)\). Then the term becomes \(\( \sqrt{u} \)\).
\(\[
\frac{d}{dx}\sqrt{u} = \frac{1}{2\sqrt{u}} \cdot \frac{du}{dx}
\]\)
Since \(\( u = x^2 + a^2 \)\), we have:
\(\[
\frac{du}{dx} = 2x
\]\)
So,
\(\[
\frac{d}{dx}\sqrt{x^2 + a^2} = \frac{1}{2\sqrt{x^2 + a^2}} \cdot 2x = \frac{x}{\sqrt{x^2 + a^2}}
\]\)
4. Differentiate \(\( \sqrt{x^2 - a} \)\):
Let \(\( v = x^2 - a \)\). Then the term becomes \(\( \sqrt{v} \)\).
\(\[
\frac{d}{dx}\sqrt{v} = \frac{1}{2\sqrt{v}} \cdot \frac{dv}{dx}
\]\)
Since \(\( v = x^2 - a \)\), we have:
\(\[
\frac{dv}{dx} = 2x
\]\)
So,
\(\[
\frac{d}{dx}\sqrt{x^2 - a} = \frac{1}{2\sqrt{x^2 - a}} \cdot 2x = \frac{x}{\sqrt{x^2 - a}}
\]\)
5. Combine the Derivatives:
Now we can combine the derivatives of both terms:
\(\[
f'(x) = \frac{x}{\sqrt{x^2 + a^2}} - \frac{x}{\sqrt{x^2 - a}}
\]\)
Therefore, the derivative of the function \(\( f(x) = \sqrt{x^2 + a^2} - \sqrt{x^2 - a} \)\) with respect to \(\( x \)\) is:
\(\[
f'(x) = \frac{x}{\sqrt{x^2 + a^2}} - \frac{x}{\sqrt{x^2 - a}}
\]\)
