If cos(x2)+tan−1(y)=1\cos(x^2) + \tan^{-1}(y) = 1cos(x2)+tan−1(y)=1, what is dydx\frac{dy}{dx}dxdy?
dydx=2x(1+y2)sin(x2)\displaystyle\frac{dy}{dx} = 2x(1+y^2)\sin(x^2)dxdy=2x(1+y2)sin(x2)
dydx=−2x(1+y2)sin(x2)\displaystyle\frac{dy}{dx} = -2x(1+y^2)\sin(x^2)dxdy=−2x(1+y2)sin(x2)
dydx=2xsin(x2)1+y2\displaystyle\frac{dy}{dx} = \frac{2x\sin(x^2)}{1+y^2}dxdy=1+y22xsin(x2)
dydx=2x(1+y2)cos(x2)\displaystyle\frac{dy}{dx} = 2x(1+y^2)\cos(x^2)dxdy=2x(1+y2)cos(x2)
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AI Tutor 1:23 AM
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