Solution 1.1:2d
From Förberedande kurs i matematik 2
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- | {{ | + | If we write <math>\sqrt{x}</math> in power form <math>x^{1/2}</math>, we see that the square root is a function having the appearance of <math>x^n</math> and its derivative is therefore equal to |
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- | {{ | + | {{Displayed math||<math>f^{\,\prime}(x) = \frac{d}{dx}\,\sqrt{x} = \frac{d}{dx}\,x^{1/2} = \tfrac{1}{2}x^{1/2-1} = \tfrac{1}{2}x^{-1/2}\,\textrm{.}</math>}} |
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+ | The answer can also be written as | ||
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+ | {{Displayed math||<math>f^{\,\prime}(x) = \frac{1}{2\sqrt{x}}</math>}} | ||
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+ | since <math>x^{-1/2} = \bigl(x^{1/2}\bigr)^{-1} = \bigl(\sqrt{x}\,\bigr)^{-1} = \frac{1}{\sqrt{x}}\,</math>. |
Current revision
If we write \displaystyle \sqrt{x} in power form \displaystyle x^{1/2}, we see that the square root is a function having the appearance of \displaystyle x^n and its derivative is therefore equal to
\displaystyle f^{\,\prime}(x) = \frac{d}{dx}\,\sqrt{x} = \frac{d}{dx}\,x^{1/2} = \tfrac{1}{2}x^{1/2-1} = \tfrac{1}{2}x^{-1/2}\,\textrm{.} |
The answer can also be written as
\displaystyle f^{\,\prime}(x) = \frac{1}{2\sqrt{x}} |
since \displaystyle x^{-1/2} = \bigl(x^{1/2}\bigr)^{-1} = \bigl(\sqrt{x}\,\bigr)^{-1} = \frac{1}{\sqrt{x}}\,.