Lösung 1.1:2f

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We can rewrite the function using a trigonometric addition formula:
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We can rewrite the function using a trigonometric addition formula,
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{{Displayed math||<math>f(x) = \cos\Bigl(x+\frac{\pi}{3}\Bigr) = \cos x\cdot\cos \frac{\pi}{3} - \sin x\cdot\sin\frac{\pi}{3}\,\textrm{.}</math>}}
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<math>f\left( x \right)=\cos \left( x+\frac{\pi }{3} \right)=\cos x\centerdot \cos \frac{\pi }{3}-\sin x\centerdot \sin \frac{\pi }{3}</math>
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If we now differentiate this expression, <math>\cos (\pi/3)</math> and <math>\sin (\pi/3)</math> are constants and we obtain
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If we now differentiate this expression,
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<math>\cos \frac{\pi }{3}</math>
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and
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<math>\sin \frac{\pi }{3}</math>
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are constants and we obtain
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<math>\begin{align}
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& {f}'\left( x \right)=\frac{d}{dx}\left( \cos x\centerdot \cos \frac{\pi }{3}-\sin x\centerdot \sin \frac{\pi }{3} \right) \\
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& =\cos \frac{\pi }{3}\centerdot \frac{d}{dx}\cos x-\sin \frac{\pi }{3}\centerdot \frac{d}{dx}\sin x \\
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& =\cos \frac{\pi }{3}\centerdot \left( -\sin x \right)-\sin \frac{\pi }{3}\centerdot \cos x \\
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\end{align}</math>
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{{Displayed math||<math>\begin{align}
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f^{\,\prime}(x)
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&= \frac{d}{dx}\,\Bigl(\cos x\cdot\cos\frac{\pi}{3} - \sin x\cdot\sin\frac{\pi}{3} \Bigr)\\[5pt]
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&= \cos\frac{\pi}{3}\cdot\frac{d}{dx}\,\cos x - \sin\frac{\pi}{3}\cdot\frac{d}{dx}\,\sin x\\[5pt]
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&= \cos\frac{\pi}{3}\cdot (-\sin x) - \sin\frac{\pi}{3}\cdot\cos x\,\textrm{.}
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\end{align}</math>}}
If we then use the addition formula in reverse, this gives
If we then use the addition formula in reverse, this gives
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{{Displayed math||<math>\begin{align}
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f^{\,\prime}(x)
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&= -\Bigl(\sin x\cdot\cos\frac{\pi}{3} + \cos x\cdot\sin\frac{\pi}{3}\Bigr)\\[5pt]
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&= -\sin\Bigl(x+\frac{\pi}{3}\Bigr)\,\textrm{.}
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\end{align}</math>}}
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<math>\begin{align}
 
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& {f}'\left( x \right)=-\left( \sin x\centerdot \cos \frac{\pi }{3}+\cos x\centerdot \sin \frac{\pi }{3} \right) \\
 
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& =-\sin \left( x+\frac{\pi }{3} \right) \\
 
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\end{align}</math>
 
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NOTE: In the next section, we will go through differentiation rules which make it possible to differentiate the expression directly without rewriting in this way.
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Note: In the next section, we will go through differentiation rules which make it possible to differentiate the expression directly without rewriting in this way.

Version vom 12:47, 14. Okt. 2008

We can rewrite the function using a trigonometric addition formula,

\displaystyle f(x) = \cos\Bigl(x+\frac{\pi}{3}\Bigr) = \cos x\cdot\cos \frac{\pi}{3} - \sin x\cdot\sin\frac{\pi}{3}\,\textrm{.}

If we now differentiate this expression, \displaystyle \cos (\pi/3) and \displaystyle \sin (\pi/3) are constants and we obtain

\displaystyle \begin{align}

f^{\,\prime}(x) &= \frac{d}{dx}\,\Bigl(\cos x\cdot\cos\frac{\pi}{3} - \sin x\cdot\sin\frac{\pi}{3} \Bigr)\\[5pt] &= \cos\frac{\pi}{3}\cdot\frac{d}{dx}\,\cos x - \sin\frac{\pi}{3}\cdot\frac{d}{dx}\,\sin x\\[5pt] &= \cos\frac{\pi}{3}\cdot (-\sin x) - \sin\frac{\pi}{3}\cdot\cos x\,\textrm{.} \end{align}

If we then use the addition formula in reverse, this gives

\displaystyle \begin{align}

f^{\,\prime}(x) &= -\Bigl(\sin x\cdot\cos\frac{\pi}{3} + \cos x\cdot\sin\frac{\pi}{3}\Bigr)\\[5pt] &= -\sin\Bigl(x+\frac{\pi}{3}\Bigr)\,\textrm{.} \end{align}


Note: In the next section, we will go through differentiation rules which make it possible to differentiate the expression directly without rewriting in this way.

Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_1.1:2f“