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Sagot :
To find the derivative of the function [tex]\( f(x) = \frac{1}{\sqrt{x}} \)[/tex] with respect to [tex]\( x \)[/tex] from first principles, we need to use the definition of the derivative. The definition states:
[tex]\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \][/tex]
Let’s apply this definition step-by-step.
1. Step 1: Define [tex]\( f(x) \)[/tex] and [tex]\( f(x + h) \)[/tex]:
[tex]\[ f(x) = \frac{1}{\sqrt{x}} \][/tex]
[tex]\[ f(x + h) = \frac{1}{\sqrt{x + h}} \][/tex]
2. Step 2: Form the difference quotient:
[tex]\[ \frac{f(x + h) - f(x)}{h} = \frac{\frac{1}{\sqrt{x + h}} - \frac{1}{\sqrt{x}}}{h} \][/tex]
To simplify this expression, we need to find a common denominator for the fractions in the numerator:
[tex]\[ \frac{1}{\sqrt{x + h}} \text{ and } \frac{1}{\sqrt{x}} \][/tex]
The common denominator is [tex]\( \sqrt{x + h} \cdot \sqrt{x} \)[/tex]. Thus, the difference quotient becomes:
[tex]\[ \frac{f(x + h) - f(x)}{h} = \frac{\frac{\sqrt{x} - \sqrt{x + h}}{\sqrt{x(x + h)}}}{h} \][/tex]
3. Step 3: Simplify the numerator:
[tex]\[ \frac{\sqrt{x} - \sqrt{x + h}}{\sqrt{x(x + h)}} = \frac{(\sqrt{x} - \sqrt{x + h})(\sqrt{x} + \sqrt{x + h})}{\sqrt{x(x + h)}(\sqrt{x} + \sqrt{x + h})} \][/tex]
The numerator simplifies using the difference of squares formula:
[tex]\[ = \frac{x - (x + h)}{\sqrt{x(x + h)}(\sqrt{x} + \sqrt{x + h})} = \frac{x - x - h}{\sqrt{x(x + h)}(\sqrt{x} + \sqrt{x + h})} = \frac{-h}{\sqrt{x(x + h)}(\sqrt{x} + \sqrt{x + h})} \][/tex]
4. Step 4: Include [tex]\( h \)[/tex] divisor:
[tex]\[ \frac{f(x + h) - f(x)}{h} = \frac{-h}{h \sqrt{x(x + h)}(\sqrt{x} + \sqrt{x + h})} = \frac{-1}{\sqrt{x(x + h)}(\sqrt{x} + \sqrt{x + h})} \][/tex]
5. Step 5: Take the limit as [tex]\( h \)[/tex] approaches 0:
As [tex]\( h \)[/tex] approaches 0, [tex]\( x + h \)[/tex] approaches [tex]\( x \)[/tex], and [tex]\( \sqrt{x + h} \)[/tex] approaches [tex]\( \sqrt{x} \)[/tex]:
[tex]\[ \lim_{h \to 0} \frac{-1}{\sqrt{x(x + h)}(\sqrt{x} + \sqrt{x + h})} = \frac{-1}{\sqrt{x \cdot x}(\sqrt{x} + \sqrt{x})} = \frac{-1}{x (\sqrt{x} + \sqrt{x})} = \frac{-1}{x \cdot 2\sqrt{x}} = \frac{-1}{2x^{3/2}} \][/tex]
Therefore, the derivative of the function [tex]\( f(x) = \frac{1}{\sqrt{x}} \)[/tex] with respect to [tex]\( x \)[/tex] is:
[tex]\[ f'(x) = -\frac{1}{2x^{3/2}} \][/tex]
[tex]\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \][/tex]
Let’s apply this definition step-by-step.
1. Step 1: Define [tex]\( f(x) \)[/tex] and [tex]\( f(x + h) \)[/tex]:
[tex]\[ f(x) = \frac{1}{\sqrt{x}} \][/tex]
[tex]\[ f(x + h) = \frac{1}{\sqrt{x + h}} \][/tex]
2. Step 2: Form the difference quotient:
[tex]\[ \frac{f(x + h) - f(x)}{h} = \frac{\frac{1}{\sqrt{x + h}} - \frac{1}{\sqrt{x}}}{h} \][/tex]
To simplify this expression, we need to find a common denominator for the fractions in the numerator:
[tex]\[ \frac{1}{\sqrt{x + h}} \text{ and } \frac{1}{\sqrt{x}} \][/tex]
The common denominator is [tex]\( \sqrt{x + h} \cdot \sqrt{x} \)[/tex]. Thus, the difference quotient becomes:
[tex]\[ \frac{f(x + h) - f(x)}{h} = \frac{\frac{\sqrt{x} - \sqrt{x + h}}{\sqrt{x(x + h)}}}{h} \][/tex]
3. Step 3: Simplify the numerator:
[tex]\[ \frac{\sqrt{x} - \sqrt{x + h}}{\sqrt{x(x + h)}} = \frac{(\sqrt{x} - \sqrt{x + h})(\sqrt{x} + \sqrt{x + h})}{\sqrt{x(x + h)}(\sqrt{x} + \sqrt{x + h})} \][/tex]
The numerator simplifies using the difference of squares formula:
[tex]\[ = \frac{x - (x + h)}{\sqrt{x(x + h)}(\sqrt{x} + \sqrt{x + h})} = \frac{x - x - h}{\sqrt{x(x + h)}(\sqrt{x} + \sqrt{x + h})} = \frac{-h}{\sqrt{x(x + h)}(\sqrt{x} + \sqrt{x + h})} \][/tex]
4. Step 4: Include [tex]\( h \)[/tex] divisor:
[tex]\[ \frac{f(x + h) - f(x)}{h} = \frac{-h}{h \sqrt{x(x + h)}(\sqrt{x} + \sqrt{x + h})} = \frac{-1}{\sqrt{x(x + h)}(\sqrt{x} + \sqrt{x + h})} \][/tex]
5. Step 5: Take the limit as [tex]\( h \)[/tex] approaches 0:
As [tex]\( h \)[/tex] approaches 0, [tex]\( x + h \)[/tex] approaches [tex]\( x \)[/tex], and [tex]\( \sqrt{x + h} \)[/tex] approaches [tex]\( \sqrt{x} \)[/tex]:
[tex]\[ \lim_{h \to 0} \frac{-1}{\sqrt{x(x + h)}(\sqrt{x} + \sqrt{x + h})} = \frac{-1}{\sqrt{x \cdot x}(\sqrt{x} + \sqrt{x})} = \frac{-1}{x (\sqrt{x} + \sqrt{x})} = \frac{-1}{x \cdot 2\sqrt{x}} = \frac{-1}{2x^{3/2}} \][/tex]
Therefore, the derivative of the function [tex]\( f(x) = \frac{1}{\sqrt{x}} \)[/tex] with respect to [tex]\( x \)[/tex] is:
[tex]\[ f'(x) = -\frac{1}{2x^{3/2}} \][/tex]
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