If ln(siny)+x3=2\ln(\sin y) + x^3 = 2ln(siny)+x3=2, then dydx=\frac{dy}{dx} =dxdy=
dydx=−3x2siny\displaystyle\frac{dy}{dx} = -\frac{3x^2}{\sin y}dxdy=−siny3x2
dydx=−3x2tany\displaystyle\frac{dy}{dx} = -3x^2 \tan ydxdy=−3x2tany
dydx=−3x2coty\displaystyle\frac{dy}{dx} = -3x^2 \cot ydxdy=−3x2coty
dydx=3x2tany\displaystyle\frac{dy}{dx} = 3x^2 \tan ydxdy=3x2tany
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AI Tutor 11:50 PM
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