Answer
$$y'=\Big(\frac{2x}{x^2+1}+\tan2x\Big)\frac{2(x^2+1)}{\sqrt{\cos2x}}$$
Work Step by Step
$$y=\frac{2(x^2+1)}{\sqrt{\cos 2x}}$$
Take the natural logarithm of both sides: $$\ln y=\ln2(x^2+1)-\ln\sqrt{\cos2x}$$ $$\ln y =\ln2+\ln(x^2+1)-\ln(\cos 2x)^{1/2}$$ $$\ln y =\ln2+\ln(x^2+1)-\frac{1}{2}\ln\cos2x$$
Now take the derivative of both sides, and remember that $(\ln x)'=1/x$:
$$\frac{1}{y}\times y'=0+\Big(\frac{1}{x^2+1}\times(x^2+1)'\Big)-\frac{1}{2}\Big(\frac{1}{\cos2x}\times(\cos2x)'\Big)$$ $$\frac{y'}{y}=\frac{2x}{x^2+1}-\frac{1}{2}\Big(\frac{-\sin2x\times(2x)'}{\cos2x}\Big)$$ $$\frac{y'}{y}=\frac{2x}{x^2+1}-\frac{1}{2}\Big(\frac{-2\sin2x}{\cos2x}\Big)$$ $$\frac{y'}{y}=\frac{2x}{x^2+1}+\frac{\sin2x}{\cos2x}$$ $$\frac{y'}{y}=\frac{2x}{x^2+1}+\tan2x$$
Find $y'$: $$y'=\Big(\frac{2x}{x^2+1}+\tan2x\Big)y$$ $$y'=\Big(\frac{2x}{x^2+1}+\tan2x\Big)\frac{2(x^2+1)}{\sqrt{\cos2x}}$$
