Assertion (A) If $y=f(x)=(|x|-|x-1|)^2$, then $\left(\frac{d y}{d x}\right)_{x=1}=1$

Reason (R) $\mathop {\lim }\limits_{x \to a} \frac{f(x)-f(a)}{x-a}$ exist, then it is called derivative of $f(x)$ at $x=a$.

If $x^2+y^2=t-\frac{1}{t}$ and $x^4+y^4=t^2+\frac{1}{t^2}$, then $\frac{d y}{d x}=$

If $y=(a x+b) \cos x$, then

$$ y_2+y_1 \sin 2 x+y\left(1+\sin ^2 x\right)= $$

If $5 f(x)+3 f\left(\frac{1}{x}\right)=x+2$ and $y=x f(x)$, then $\frac{d y}{d x}$ at $x=1$ is equal to