find-the-derivative-of-the-function-y2tanx1sinx12yleftfrac2tan-x1-sin-xrightfrac

Question

Find the derivative of the function y=(2tan⁡x1−sin⁡x)12y=\left(\frac{2	an x}{1-\sin x}\$$right)^{\frac12}.$$

Accepted Answer

y′=sin⁡x+sec⁡2x−sec⁡2xsin⁡x2(tan⁡x(1−sin⁡$$x)3)12y^{\prime}$$=\frac{\sin x+\$$sec^2x$$-\$$sec^2x$$\sin x}{\sqrt2\left(	an x\left(1-\sin x\$$right)^3$$\$$right)^{\frac12}$$}

Answer

y′=sin⁡x−sec⁡2x+sec⁡2xsin⁡x2(tan⁡x(1−sin⁡$$x)3)12y^{\prime}$$=\frac{\sin x-\$$sec^2x$$+\$$sec^2x$$\sin x}{\sqrt2\left(	an x\left(1-\sin x\$$right)^3$$\$$right)^{\frac12}$$}

Answer

y′=sin⁡x+sec⁡2x−sec⁡2xsin⁡x(tan⁡x(1−sin⁡$$x)3)2y^{\prime}$$=\frac{\sin x+\$$sec^2x$$-\$$sec^2x$$\sin x}{\left(	an x\left(1-\sin x\$$right)^3$$\$$right)^2$$}

Answer

y′=sin⁡x−sec⁡2x+sec⁡2xsin⁡x(tan⁡x(1−sin⁡$$x)3)2y^{\prime}$$=\frac{\sin x-\$$sec^2x$$+\$$sec^2x$$\sin x}{\left(	an x\left(1-\sin x\$$right)^3$$\$$right)^2$$}

Find the derivative of the function y=(2tanx1−sinx)12y=\(\left\)(\(\frac{2\tan x}{1-\sin x}\]\right\))^{\(\frac\)12}.

Find the derivative of the function $y = {(\frac{2 \tan x}{1 - \sin x})}^{\frac{1}{2}}$ .