Sure! Let's solve the problem step-by-step to prove the given equation.
Given [tex]\( y = \sqrt{x} + \frac{1}{\sqrt{x}} \)[/tex], we need to prove that:
[tex]\[ 2x \frac{dy}{dx} + y = 2\sqrt{x}. \][/tex]
1. Find the derivative [tex]\(\frac{dy}{dx}\)[/tex]:
First, let's differentiate [tex]\( y \)[/tex] with respect to [tex]\( x \)[/tex].
[tex]\[
y = \sqrt{x} + \frac{1}{\sqrt{x}}
\][/tex]
To differentiate [tex]\( y \)[/tex]:
[tex]\[
\frac{dy}{dx} = \frac{d}{dx} (\sqrt{x}) + \frac{d}{dx} \left(\frac{1}{\sqrt{x}}\right)
\][/tex]
We know that:
[tex]\[
\frac{d}{dx} (\sqrt{x}) = \frac{1}{2\sqrt{x}}
\][/tex]
and
[tex]\[
\frac{d}{dx} \left(\frac{1}{\sqrt{x}}\right) = \frac{d}{dx} (x^{-\frac{1}{2}}) = -\frac{1}{2} x^{-\frac{3}{2}} = -\frac{1}{2x^{\frac{3}{2}}}
\][/tex]
Therefore:
[tex]\[
\frac{dy}{dx} = \frac{1}{2\sqrt{x}} - \frac{1}{2x^{\frac{3}{2}}}
\][/tex]
2. Calculate [tex]\( 2x \frac{dy}{dx} \)[/tex]:
[tex]\[
2x \frac{dy}{dx} = 2x \left(\frac{1}{2\sqrt{x}} - \frac{1}{2x^{\frac{3}{2}}}\right)
\][/tex]
Distribute [tex]\( 2x \)[/tex]:
[tex]\[
2x \frac{dy}{dx} = 2x \cdot \frac{1}{2\sqrt{x}} - 2x \cdot \frac{1}{2x^{\frac{3}{2}}}
\][/tex]
Simplify each term:
[tex]\[
2x \cdot \frac{1}{2\sqrt{x}} = \frac{2x}{2\sqrt{x}} = \frac{x}{\sqrt{x}} = \sqrt{x}
\][/tex]
and
[tex]\[
2x \cdot \frac{1}{2x^{\frac{3}{2}}} = \frac{2x}{2x^{\frac{3}{2}}} = \frac{x}{x^{\frac{3}{2}}} = \frac{1}{x^{\frac{1}{2}}} = \frac{1}{\sqrt{x}}
\][/tex]
Hence:
[tex]\[
2x \frac{dy}{dx} = \sqrt{x} - \frac{1}{\sqrt{x}}
\][/tex]
3. Combine [tex]\( 2x \frac{dy}{dx} \)[/tex] and [tex]\( y \)[/tex]:
Recall [tex]\( y = \sqrt{x} + \frac{1}{\sqrt{x}} \)[/tex]. Then:
[tex]\[
2x \frac{dy}{dx} + y = (\sqrt{x} - \frac{1}{\sqrt{x}}) + (\sqrt{x} + \frac{1}{\sqrt{x}})
\][/tex]
Notice that the terms [tex]\(- \frac{1}{\sqrt{x}} \)[/tex] and [tex]\(+ \frac{1}{\sqrt{x}}\)[/tex] cancel each other out, so we get:
[tex]\[
\sqrt{x} - \frac{1}{\sqrt{x}} + \sqrt{x} + \frac{1}{\sqrt{x}} = \sqrt{x} + \sqrt{x} = 2\sqrt{x}
\][/tex]
Thus, we have proven that:
[tex]\[ 2x \frac{dy}{dx} + y = 2\sqrt{x} \][/tex]
This completes the proof.
