Find the log³³√2x⁵?
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Answer:
Firstly, let #y=\sqrtsin(2x)#
and let #u=sin(2x)#
this means #y=u^{\frac{1}{2}}#
Therefore #\frac{dy}{dx}=\frac{dy}{du}*\frac{du}{dx}#
#\frac{dy}{dx}=\frac{1}{2}u^{-\frac{1}{2}} * 2cos(2x)#
Which goes to
#\frac{dy}{dx}=\frac{cos2x}{sqrt{u}#
bring back #u=sin2x#
to get
#\frac{dy}{dx}=\frac{cos2x}{sqrt{sin2x}}#
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