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Mathematics 16 Online
OpenStudy (anonymous):

Find d/dx (2^cosx)

OpenStudy (zarkon):

\[\frac{d}{dx}a^{f(x)}=a^{f(x)}f'(x)\ln(a)\]

OpenStudy (anonymous):

how did you find that?

OpenStudy (anonymous):

-ln 2 sin x (2^cos x)

OpenStudy (anonymous):

oooohhh i get it thanks guys!

myininaya (myininaya):

\[y=2^{\cos(x)}\] Take ln( ) of both sides! :) so we can bring down that function(x). \[\ln(y)=\cos(x) \ln(2)\] Now it should be pretty easy to differentiate both sides! :) \[\frac{y'}{y}=-\sin(x) \ln(2)\] To find y' multiply y on both sides. \[y'=-y \sin(x) \ln(2)\] But \[y=2^{\cos(x)}\] so we can write \[y'=-2^{\cos(x)} \sin(x) \ln(2)\]

OpenStudy (anonymous):

Thank you so much. I understand that so much more clearly

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