Solution 1.2:3c
From Förberedande kurs i matematik 2
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| - | {{ | + | We can write the expression as |
| - | < | + | |
| - | {{ | + | {{Displayed math||<math>\frac{1}{x\sqrt{1-x^{2}}} = \bigl(x\sqrt{1-x^2}\,\bigr)^{-1}\,\textrm{,}</math>}} |
| - | {{ | + | |
| - | < | + | and then we see that we have "something raised to -1", which can be differentiated one step by using the chain rule, |
| - | {{ | + | |
| + | {{Displayed math||<math>\begin{align} | ||
| + | \frac{d}{dx}\,\bigl( \bbox[#FFEEAA;,1.5pt]{x\sqrt{1-x^2}}\,\bigr)^{-1} | ||
| + | &= {}\rlap{-1\cdot \bigl( \bbox[#FFEEAA;,1.5pt]{x\sqrt{1-x^2}}\,\bigr)^{-2}\bigl(\bbox[#FFEEAA;,1.5pt]{x\sqrt{1-x^2}}\,\bigr)'}\phantom{-\frac{1}{x^2(1-x^2)}\Bigl(\sqrt{1-x^2} + x\cdot\frac{1}{2\sqrt{1-x^2}}(1-x^2)'\Bigr)}\\[5pt] | ||
| + | &= -\frac{1}{\bigl(x\sqrt{1-x^2}\bigr)^2}\cdot \bigl(x\sqrt{1-x^2}\bigr)'\\[5pt] | ||
| + | &= -\frac{1}{x^2(1-x^2)}\cdot\bigl(x\sqrt{1-x^2}\bigr)'\,\textrm{.} | ||
| + | \end{align}</math>}} | ||
| + | |||
| + | The next step is to differentiate the product <math>x\cdot\sqrt{1-x^2}</math> using the product rule, | ||
| + | |||
| + | {{Displayed math||<math>\begin{align} | ||
| + | \phantom{\frac{d}{dx}\,\bigl( \bbox[#FFEEAA;,1.5pt]{x\sqrt{1-x^2}}\,\bigr)^{-1}}{} | ||
| + | &= {}\rlap{-\frac{1}{x^2(1-x^2)}\cdot \Bigl( (x)'\sqrt{1-x^2} + x(\sqrt{1-x^2})'\Bigr)}\phantom{-\frac{1}{x^2(1-x^2)}\Bigl(\sqrt{1-x^2} + x\cdot\frac{1}{2\sqrt{1-x^2}}(1-x^2)'\Bigr)}\\[5pt] | ||
| + | &= -\frac{1}{x^2(1-x^2)}\cdot \Bigl(1\cdot\sqrt{1-x^2} + x\cdot (\sqrt{1-x^2})'\Bigr)\,\textrm{.} | ||
| + | \end{align}</math>}} | ||
| + | |||
| + | The expression <math>\sqrt{1-x^2}</math> is of the type "square root of something", so we use the chain rule to differentiate, | ||
| + | |||
| + | {{Displayed math||<math>\begin{align} | ||
| + | \phantom{\frac{d}{dx}\,\bigl( \bbox[#FFEEAA;,1.5pt]{x\sqrt{1-x^2}}\,\bigr)^{-1}}{} | ||
| + | &= -\frac{1}{x^2(1-x^2)}\Bigl(\sqrt{1-x^2} + x\cdot\frac{1}{2\sqrt{1-x^2}}(1-x^2)'\Bigr)\\[5pt] | ||
| + | &= -\frac{1}{x^2(1-x^2)}\Bigl(\sqrt{1-x^2} + x\cdot\frac{1}{2\sqrt{1-x^2}}\cdot ( -2x)\Bigr)\\[5pt] | ||
| + | &= -\frac{1}{x^2(1-x^2)}\Bigl(\sqrt{1-x^2} - \frac{x^2}{\sqrt{1-x^2}}\Bigr)\,\textrm{.} | ||
| + | \end{align}</math>}} | ||
| + | |||
| + | We write the expression on the right over a common denominator, | ||
| + | |||
| + | {{Displayed math||<math>\begin{align} | ||
| + | \phantom{\frac{d}{dx}\,\bigl( \bbox[#FFEEAA;,1.5pt]{x\sqrt{1-x^2}}\,\bigr)^{-1}}{} | ||
| + | &= {}\rlap{-\frac{1}{x^2(1-x^2)}\cdot \frac{\bigl(\sqrt{1-x^2}\bigr)^2-x^2}{\sqrt{1-x^2}}}\phantom{-\frac{1}{x^2(1-x^2)}\Bigl(\sqrt{1-x^2} + x\cdot\frac{1}{2\sqrt{1-x^2}}(1-x^2)'\Bigr)}\\[5pt] | ||
| + | &= -\frac{1}{x^2(1-x^2)}\cdot \frac{1-x^2-x^2}{\sqrt{1-x^2}}\\[5pt] | ||
| + | &= -\frac{1-2x^2}{x^2(1-x^2)^{3/2}}\,\textrm{.} | ||
| + | \end{align}</math>}} | ||
| + | |||
| + | |||
| + | Note: When we make simplifications of the form <math>\bigl(\sqrt{1-x^2} \bigr)^2 = 1-x^2</math>, we assume that both sides are well defined (i.e. in this case that <math>x</math> lies between -1 and 1). | ||
Current revision
We can write the expression as
| \displaystyle \frac{1}{x\sqrt{1-x^{2}}} = \bigl(x\sqrt{1-x^2}\,\bigr)^{-1}\,\textrm{,} |
and then we see that we have "something raised to -1", which can be differentiated one step by using the chain rule,
| \displaystyle \begin{align}
\frac{d}{dx}\,\bigl( \bbox[#FFEEAA;,1.5pt]{x\sqrt{1-x^2}}\,\bigr)^{-1} &= {}\rlap{-1\cdot \bigl( \bbox[#FFEEAA;,1.5pt]{x\sqrt{1-x^2}}\,\bigr)^{-2}\bigl(\bbox[#FFEEAA;,1.5pt]{x\sqrt{1-x^2}}\,\bigr)'}\phantom{-\frac{1}{x^2(1-x^2)}\Bigl(\sqrt{1-x^2} + x\cdot\frac{1}{2\sqrt{1-x^2}}(1-x^2)'\Bigr)}\\[5pt] &= -\frac{1}{\bigl(x\sqrt{1-x^2}\bigr)^2}\cdot \bigl(x\sqrt{1-x^2}\bigr)'\\[5pt] &= -\frac{1}{x^2(1-x^2)}\cdot\bigl(x\sqrt{1-x^2}\bigr)'\,\textrm{.} \end{align} |
The next step is to differentiate the product \displaystyle x\cdot\sqrt{1-x^2} using the product rule,
| \displaystyle \begin{align}
\phantom{\frac{d}{dx}\,\bigl( \bbox[#FFEEAA;,1.5pt]{x\sqrt{1-x^2}}\,\bigr)^{-1}}{} &= {}\rlap{-\frac{1}{x^2(1-x^2)}\cdot \Bigl( (x)'\sqrt{1-x^2} + x(\sqrt{1-x^2})'\Bigr)}\phantom{-\frac{1}{x^2(1-x^2)}\Bigl(\sqrt{1-x^2} + x\cdot\frac{1}{2\sqrt{1-x^2}}(1-x^2)'\Bigr)}\\[5pt] &= -\frac{1}{x^2(1-x^2)}\cdot \Bigl(1\cdot\sqrt{1-x^2} + x\cdot (\sqrt{1-x^2})'\Bigr)\,\textrm{.} \end{align} |
The expression \displaystyle \sqrt{1-x^2} is of the type "square root of something", so we use the chain rule to differentiate,
| \displaystyle \begin{align}
\phantom{\frac{d}{dx}\,\bigl( \bbox[#FFEEAA;,1.5pt]{x\sqrt{1-x^2}}\,\bigr)^{-1}}{} &= -\frac{1}{x^2(1-x^2)}\Bigl(\sqrt{1-x^2} + x\cdot\frac{1}{2\sqrt{1-x^2}}(1-x^2)'\Bigr)\\[5pt] &= -\frac{1}{x^2(1-x^2)}\Bigl(\sqrt{1-x^2} + x\cdot\frac{1}{2\sqrt{1-x^2}}\cdot ( -2x)\Bigr)\\[5pt] &= -\frac{1}{x^2(1-x^2)}\Bigl(\sqrt{1-x^2} - \frac{x^2}{\sqrt{1-x^2}}\Bigr)\,\textrm{.} \end{align} |
We write the expression on the right over a common denominator,
| \displaystyle \begin{align}
\phantom{\frac{d}{dx}\,\bigl( \bbox[#FFEEAA;,1.5pt]{x\sqrt{1-x^2}}\,\bigr)^{-1}}{} &= {}\rlap{-\frac{1}{x^2(1-x^2)}\cdot \frac{\bigl(\sqrt{1-x^2}\bigr)^2-x^2}{\sqrt{1-x^2}}}\phantom{-\frac{1}{x^2(1-x^2)}\Bigl(\sqrt{1-x^2} + x\cdot\frac{1}{2\sqrt{1-x^2}}(1-x^2)'\Bigr)}\\[5pt] &= -\frac{1}{x^2(1-x^2)}\cdot \frac{1-x^2-x^2}{\sqrt{1-x^2}}\\[5pt] &= -\frac{1-2x^2}{x^2(1-x^2)^{3/2}}\,\textrm{.} \end{align} |
Note: When we make simplifications of the form \displaystyle \bigl(\sqrt{1-x^2} \bigr)^2 = 1-x^2, we assume that both sides are well defined (i.e. in this case that \displaystyle x lies between -1 and 1).
