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Sagot :
Let's start by analyzing the given function [tex]\( y \)[/tex]:
[tex]\[ y = \sqrt{x} + \frac{1}{\sqrt{x}} \][/tex]
To prove the statement [tex]\( 2x \frac{dy}{dx} + y = 2\sqrt{x} \)[/tex], we need to follow several steps.
### Step 1: Compute [tex]\(\frac{dy}{dx}\)[/tex]
We differentiate [tex]\( y \)[/tex] with respect to [tex]\( x \)[/tex].
[tex]\[ y = \sqrt{x} + \frac{1}{\sqrt{x}} \][/tex]
The first term, [tex]\(\sqrt{x}\)[/tex], can be rewritten as [tex]\( x^{1/2} \)[/tex], and the derivative of [tex]\( x^{1/2} \)[/tex] is:
[tex]\[ \frac{d}{dx}(\sqrt{x}) = \frac{d}{dx}(x^{1/2}) = \frac{1}{2} x^{-1/2} = \frac{1}{2\sqrt{x}} \][/tex]
The second term, [tex]\(\frac{1}{\sqrt{x}}\)[/tex], can be rewritten as [tex]\( x^{-1/2} \)[/tex], and the derivative of [tex]\( x^{-1/2} \)[/tex] is:
[tex]\[ \frac{d}{dx} \left( \frac{1}{\sqrt{x}} \right) = \frac{d}{dx} \left( x^{-1/2} \right) = -\frac{1}{2} x^{-3/2} = -\frac{1}{2x^{3/2}} \][/tex]
Thus, the derivative [tex]\(\frac{dy}{dx}\)[/tex] is:
[tex]\[ \frac{dy}{dx} = \frac{1}{2\sqrt{x}} - \frac{1}{2x^{3/2}} \][/tex]
### Step 2: Calculate [tex]\( 2x \frac{dy}{dx} \)[/tex]
Next, we multiply [tex]\(\frac{dy}{dx}\)[/tex] by [tex]\(2x\)[/tex]:
[tex]\[ 2x \frac{dy}{dx} = 2x \left( \frac{1}{2\sqrt{x}} - \frac{1}{2x^{3/2}} \right) \][/tex]
Distribute [tex]\(2x\)[/tex]:
[tex]\[ 2x \frac{dy}{dx} = 2x \cdot \frac{1}{2\sqrt{x}} - 2x \cdot \frac{1}{2x^{3/2}} \][/tex]
Simplify each term:
[tex]\[ = x \cdot \frac{1}{\sqrt{x}} - x \cdot \frac{1}{x^{3/2}} \][/tex]
[tex]\[ = \sqrt{x} - \frac{x}{x^{3/2}} \][/tex]
[tex]\[ = \sqrt{x} - \frac{1}{x^{1/2}} \][/tex]
[tex]\[ = \sqrt{x} - \frac{1}{\sqrt{x}} \][/tex]
### Step 3: Add [tex]\( y \)[/tex] to [tex]\( 2x \frac{dy}{dx} \)[/tex]
Now, add [tex]\( y \)[/tex] to the expression we found for [tex]\( 2x \frac{dy}{dx} \)[/tex]:
[tex]\[ 2x \frac{dy}{dx} + y = \left( \sqrt{x} - \frac{1}{\sqrt{x}} \right) + \left( \sqrt{x} + \frac{1}{\sqrt{x}} \right) \][/tex]
Combine the terms:
[tex]\[ = \sqrt{x} + \sqrt{x} - \frac{1}{\sqrt{x}} + \frac{1}{\sqrt{x}} \][/tex]
Notice that the negative and positive [tex]\(\frac{1}{\sqrt{x}}\)[/tex] terms cancel each other out:
[tex]\[ = 2\sqrt{x} \][/tex]
### Final Result
We have shown that:
[tex]\[ 2x \frac{dy}{dx} + y = 2\sqrt{x} \][/tex]
Therefore, the original statement is proven to be true.
[tex]\[ y = \sqrt{x} + \frac{1}{\sqrt{x}} \][/tex]
To prove the statement [tex]\( 2x \frac{dy}{dx} + y = 2\sqrt{x} \)[/tex], we need to follow several steps.
### Step 1: Compute [tex]\(\frac{dy}{dx}\)[/tex]
We differentiate [tex]\( y \)[/tex] with respect to [tex]\( x \)[/tex].
[tex]\[ y = \sqrt{x} + \frac{1}{\sqrt{x}} \][/tex]
The first term, [tex]\(\sqrt{x}\)[/tex], can be rewritten as [tex]\( x^{1/2} \)[/tex], and the derivative of [tex]\( x^{1/2} \)[/tex] is:
[tex]\[ \frac{d}{dx}(\sqrt{x}) = \frac{d}{dx}(x^{1/2}) = \frac{1}{2} x^{-1/2} = \frac{1}{2\sqrt{x}} \][/tex]
The second term, [tex]\(\frac{1}{\sqrt{x}}\)[/tex], can be rewritten as [tex]\( x^{-1/2} \)[/tex], and the derivative of [tex]\( x^{-1/2} \)[/tex] is:
[tex]\[ \frac{d}{dx} \left( \frac{1}{\sqrt{x}} \right) = \frac{d}{dx} \left( x^{-1/2} \right) = -\frac{1}{2} x^{-3/2} = -\frac{1}{2x^{3/2}} \][/tex]
Thus, the derivative [tex]\(\frac{dy}{dx}\)[/tex] is:
[tex]\[ \frac{dy}{dx} = \frac{1}{2\sqrt{x}} - \frac{1}{2x^{3/2}} \][/tex]
### Step 2: Calculate [tex]\( 2x \frac{dy}{dx} \)[/tex]
Next, we multiply [tex]\(\frac{dy}{dx}\)[/tex] by [tex]\(2x\)[/tex]:
[tex]\[ 2x \frac{dy}{dx} = 2x \left( \frac{1}{2\sqrt{x}} - \frac{1}{2x^{3/2}} \right) \][/tex]
Distribute [tex]\(2x\)[/tex]:
[tex]\[ 2x \frac{dy}{dx} = 2x \cdot \frac{1}{2\sqrt{x}} - 2x \cdot \frac{1}{2x^{3/2}} \][/tex]
Simplify each term:
[tex]\[ = x \cdot \frac{1}{\sqrt{x}} - x \cdot \frac{1}{x^{3/2}} \][/tex]
[tex]\[ = \sqrt{x} - \frac{x}{x^{3/2}} \][/tex]
[tex]\[ = \sqrt{x} - \frac{1}{x^{1/2}} \][/tex]
[tex]\[ = \sqrt{x} - \frac{1}{\sqrt{x}} \][/tex]
### Step 3: Add [tex]\( y \)[/tex] to [tex]\( 2x \frac{dy}{dx} \)[/tex]
Now, add [tex]\( y \)[/tex] to the expression we found for [tex]\( 2x \frac{dy}{dx} \)[/tex]:
[tex]\[ 2x \frac{dy}{dx} + y = \left( \sqrt{x} - \frac{1}{\sqrt{x}} \right) + \left( \sqrt{x} + \frac{1}{\sqrt{x}} \right) \][/tex]
Combine the terms:
[tex]\[ = \sqrt{x} + \sqrt{x} - \frac{1}{\sqrt{x}} + \frac{1}{\sqrt{x}} \][/tex]
Notice that the negative and positive [tex]\(\frac{1}{\sqrt{x}}\)[/tex] terms cancel each other out:
[tex]\[ = 2\sqrt{x} \][/tex]
### Final Result
We have shown that:
[tex]\[ 2x \frac{dy}{dx} + y = 2\sqrt{x} \][/tex]
Therefore, the original statement is proven to be true.
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