Solution 1.2:3f
From Förberedande kurs i matematik 2
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| - | {{ | + | We have no differentiation rule for a function raised to another function, but instead we use the formula |
| - | < | + | |
| - | {{ | + | {{Displayed math||<math>a^b = e^{\ln a^b} = e^{b\ln a}\,,</math>}} |
| + | |||
| + | which, in our case, gives | ||
| + | |||
| + | {{Displayed math||<math>x^{\tan x} = e^{\tan x\cdot\ln x}\,\textrm{.}</math>|(*)}} | ||
| + | |||
| + | Now, we obtain the derivative by first using the chain rule | ||
| + | |||
| + | {{Displayed math||<math>\frac{d}{dx}\,e^{\bbox[#FFEEAA;,1.5pt]{\tan x\cdot\ln x}} = {}\rlap{e^{\bbox[#FFEEAA;,1.5pt]{\tan x\cdot\ln x}}\cdot \bigl( \bbox[#FFEEAA;,1.5pt]{\tan x\cdot\ln x}\bigr)'}\phantom{e^{\tan x\cdot \ln x}\bigl((\tan x)'\cdot\ln x + \tan x\cdot (\ln x)'\bigr)}</math>}} | ||
| + | |||
| + | and then the product rule | ||
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| + | {{Displayed math||<math>\begin{align} | ||
| + | \phantom{\frac{d}{dx}\,e^{\bbox[#FFEEAA;,1.5pt]{\tan x\cdot\ln x}}}{} | ||
| + | &= e^{\tan x\cdot \ln x}\bigl((\tan x)'\cdot\ln x + \tan x\cdot (\ln x)'\bigr)\\[5pt] | ||
| + | &= e^{\tan x\cdot\ln x}\Bigl(\frac{1}{\cos^2\!x}\cdot\ln x + \tan x\cdot\frac{1}{x} \Bigr)\\[5pt] | ||
| + | &= e^{\tan x\cdot\ln x}\Bigl(\frac{\ln x}{\cos^2\!x} + \frac{\tan x}{x}\Bigr)\\[5pt] | ||
| + | &= x^{\tan x}\Bigl(\frac{\ln x}{\cos^2\!x} + \frac{\tan x}{x}\Bigr)\,, | ||
| + | \end{align}</math>}} | ||
| + | |||
| + | where we have used (*) in reverse. | ||
Current revision
We have no differentiation rule for a function raised to another function, but instead we use the formula
| \displaystyle a^b = e^{\ln a^b} = e^{b\ln a}\,, |
which, in our case, gives
| \displaystyle x^{\tan x} = e^{\tan x\cdot\ln x}\,\textrm{.} | (*) |
Now, we obtain the derivative by first using the chain rule
| \displaystyle \frac{d}{dx}\,e^{\bbox[#FFEEAA;,1.5pt]{\tan x\cdot\ln x}} = {}\rlap{e^{\bbox[#FFEEAA;,1.5pt]{\tan x\cdot\ln x}}\cdot \bigl( \bbox[#FFEEAA;,1.5pt]{\tan x\cdot\ln x}\bigr)'}\phantom{e^{\tan x\cdot \ln x}\bigl((\tan x)'\cdot\ln x + \tan x\cdot (\ln x)'\bigr)} |
and then the product rule
| \displaystyle \begin{align}
\phantom{\frac{d}{dx}\,e^{\bbox[#FFEEAA;,1.5pt]{\tan x\cdot\ln x}}}{} &= e^{\tan x\cdot \ln x}\bigl((\tan x)'\cdot\ln x + \tan x\cdot (\ln x)'\bigr)\\[5pt] &= e^{\tan x\cdot\ln x}\Bigl(\frac{1}{\cos^2\!x}\cdot\ln x + \tan x\cdot\frac{1}{x} \Bigr)\\[5pt] &= e^{\tan x\cdot\ln x}\Bigl(\frac{\ln x}{\cos^2\!x} + \frac{\tan x}{x}\Bigr)\\[5pt] &= x^{\tan x}\Bigl(\frac{\ln x}{\cos^2\!x} + \frac{\tan x}{x}\Bigr)\,, \end{align} |
where we have used (*) in reverse.
