How do you differentiate $f(x) = e^-x sinx-e^xcosx$ ?

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Answer 1 · 2015-11-11T14:48:36

$f'(x)=(e^{x}-e^{-x})(sin(x)-cos(x))$

$f'(x)=frac{d}{dx}(e^{−x}sin(x)−e^xcos(x))$

$=frac{d}{dx}(e^{−x}sin(x))−frac{d}{dx}(e^xcos(x))$

$=[e^{−x}frac{d}{dx}(sin(x))+sin(x)frac{d}{dx}(e^{−x})]$

$−[e^xfrac{d}{dx}(cos(x))+cos(x)frac{d}{dx}(e^x)]$

$=[e^{−x}(cos(x))+sin(x)(-e^{−x})]$

$−[e^x(-sin(x))+cos(x)(e^x)]$

$=e^{−x}(cos(x)-sin(x))-e^x(cos(x)-sin(x))$

$=(e^{−x}-e^{x})(cos(x)-sin(x))$

How do you differentiate f(x) = e^-x sinx-e^xcosxf(x) = e^-x sinx-e^xcosx?