Erstellt von Niamh Ryan
vor mehr als 7 Jahre
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This is the one of the most frequently used rules of differentiation. \[ \text{If }y=x^n \text{ then} \frac{dy}{dx} = nx^{n-1}\] This can be used to differentiate surds or fractions involving powers of \( x\) . For example, \( \sqrt{x} \) can also be written as \( x^{\frac{1}{2}}\) , allowing it to be easily differentiated using the above rule. \( \frac{3}{x}\) can also be written as \( 3x^{-1}\) , so \[\frac{d}{dx}(\frac{3}{x})=-3x^{-2}\]
\(\frac{d}{dx} \) is a linear operator. This means it can be separated out over addition or subtraction and constants can be moved outside the operation. Therefore, \[\frac{d}{dx}(x^m + x^n) = \frac{d}{dx}(x^m) +\frac{d}{dx}(x^n)\] and \[ \frac{d}{dx}(kx^m) = k\frac{d}{dx}(x^m)\] where \(k\) is any constant.
Example 1: Differentiate the following function: \[y=3\sqrt{x} + \frac{4}{x^2}\] Answer: As demonstrated on the previous page, we can treat the two terms in this function as if they were separate functions and their constants do not change with differentiation. First, rewrite the function using exponents instead of surds and fractions: \[ y=3x^{\frac{1}{2}} + 4x^{-2}\] and then differentiatie:\[\frac{dy}{dx} = 3(\frac{1}{2})x^{\frac{1}{2} - 1}+4(-2)x^{-2-1}= \frac{3}{2}x^{-\frac{1}{2}}-8x^{-3}\] This can be simplified to give: \[\frac{dy}{dx}=\frac{3}{2 \sqrt{x}} - \frac{8}{x^{-3}}\]
Example 2: Differentiate the following function: \[y=\frac{x^{2}-3x-4}{x+1}\] Answer: Where there's a fraction with \(x\) terms on top and bottom, factorise the top before continuing. This gives:\[y=\frac{(x-4)(x+1)}{x+1} \] Clearly \((x+1)\) can be factored from top and bottom to give\[y=x-4\] which is the same as \[y=x^1-4x^0\] and then differentiate:\[\frac{dy}{dx} = 1\]
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