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Lösung 1.1:2f

Aus Online Mathematik Brückenkurs 2

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We can rewrite the function using a trigonometric addition formula,

f(x)=cosx+3=cosxcos3−sinxsin3. 

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

f(x)=ddxcosxcos3−sinxsin3=cos3ddxcosx−sin3ddxsinx=cos3(−sinx)−sin3cosx.

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

f(x)=−sinxcos3+cosxsin3=−sinx+3.


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“