How find $dy/dx$ $y= \frac{x^2+\sin2x}{2x+\cos^2 x}$ $$dy/dx\space\space y= \frac{x^2+\sin2x}{2x+\cos^2x}$$
 A: We need the quotient rule and the chain rule to find $\frac {dy}{dx} = y'$ given $$y= \frac{x^2+\sin2x}{2x+\cos^2x}$$
Given a quotient of functions: $f(x) = \frac{g(x)}{h(x)}$
$$f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2}$$
In your case, we have $$f(x) = \frac{x^2+\sin2x}{2x+\cos^2x}$$
So put $g(x) = x^2 + \sin 2x,\;$ and $\;h(x) = 2x + \cos^2x$.
Now, $g'(x) = 2x + 2\cos 2x,\;$ and $\;h'(x) = 2 - 2\cos x \sin x = 2 - \sin(2x)$.
So $$f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2}$$ $$f'(x) = \frac{dy}{dx} = \frac{(2x + 2\cos 2x)(2x + \cos^2 x) + (x^2 + \sin 2x)(2 - \sin 2x)}{\left(2x + \cos^2 x\right)^2}$$
The rest is of the work is merely algebraic simplification.
A: Remember the quotient rule:  if $f(x)=\frac {g(x)}{h(x)}$ then $f'(x)=\frac {(h(x)g'(x)-g(x)h'(x)}{h^2(x)}$.  Can you take the derivative of the numerator and denominator, then plug into this formula?
A: Using the formula 
if $y=\frac {f(x)}{g(x)} $ then $ \frac{\mathrm d}{\mathrm d x}y=\left( \frac {f'(x)}{f(x)} - \frac {g'(x)}{g(x)} \right)\frac {f(x)}{g(x)}$
$$  \frac{\mathrm d y}{\mathrm d x}=\left(  \frac{(2x+ 2 \cos 2x)}{\mathrm (x²+\sin2x)} - \frac{(2-2\cos x \sin x)}{\mathrm (2x+\cos^2x)} \right) \frac{x²+\sin2x}{2x+\cos^2x}$$
