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Question Number 103149 by bemath last updated on 13/Jul/20

u s i n g f i r s t p r i n c i p a l

y = \ln (\sin \sqrt{x}) \to y^{'} = ?

Answered by bobhans last updated on 13/Jul/20

y^{'} = \lim_{△ x \to 0} \frac{\ln (\sin \sqrt{x + △ x}) - \ln (\sin \sqrt{x})}{△ x}

y^{'} = \lim_{△ x \to 0} \frac{\ln (\frac{\sin \sqrt{x + △ x}}{\sin \sqrt{x}})}{△ x}

Answered by mathmax by abdo last updated on 14/Jul/20

y (x) = \ln (\sin (\sqrt{x})) \Rightarrow y^{^{'}} (x) = \frac{{(\sin (\sqrt{x}))}^{^{'}}}{\sin (\sqrt{x})} = \frac{\cos (\sqrt{x})}{2 \sqrt{x} \sin (\sqrt{x})} = \frac{1}{2 \sqrt{x}} cotan (\sqrt{x})