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Implicit Differentiation With Trig Functions

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If cos⁡(x2)+tan⁡−1(y)=1\cos(x^2) + \tan^{-1}(y) = 1, what is dydx\frac{dy}{dx}?

A

dydx=2x(1+y2)sin⁡(x2)\displaystyle\frac{dy}{dx} = 2x(1+y^2)\sin(x^2)

B

dydx=−2x(1+y2)sin⁡(x2)\displaystyle\frac{dy}{dx} = -2x(1+y^2)\sin(x^2)

C

dydx=2xsin⁡(x2)1+y2\displaystyle\frac{dy}{dx} = \frac{2x\sin(x^2)}{1+y^2}

D

dydx=2x(1+y2)cos⁡(x2)\displaystyle\frac{dy}{dx} = 2x(1+y^2)\cos(x^2)

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