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