Calculus 1 1 True or False (1) f(x)= ) 1,xis rational 0,xis irrational is integrable on any bounded closed interval [a,b] (2)The derivative of y=y(x) determined by the formula arctan(y)/(x)=lnsqrt (x^2+y^2) is y'=(x+y)/(x-y)

Let's analyze each statement one by one:<br /><br />(1) The function $f(x)=\{ \begin{matrix} 1,xis\quad rational\\ 0,xis\quad irrational\end{matrix} $ is not integrable on any bounded closed interval $[a,b]$. This is because the function is not continuous on the interval, as it jumps between two different values at each irrational and rational point. For a function to be integrable, it must be continuous on the interval.<br /><br />(2) To find the derivative of $y=y(x)$ determined by the formula $arctan\frac {y}{x}=ln\sqrt {x^{2}+y^{2}}$, we need to differentiate both sides of the equation with respect to $x$. <br /><br />Differentiating the left side with respect to $x$ gives us $\frac{1}{1+\left(\frac{y}{x}\right)^2} \cdot \frac{y}{x^2} + \frac{y'}{x}$. <br /><br />Differentiating the right side with respect to $x$ gives us $\frac{1}{2\sqrt{x^2+y^2}} \cdot \left(2x + \frac{2xy}{\sqrt{x^2+y^2}}\right)$. <br /><br />Simplifying the right side, we get $\frac{x+y}{x-y}$. <br /><br />Therefore, the derivative of $y=y(x)$ determined by the formula $arctan\frac {y}{x}=ln\sqrt {x^{2}+y^{2}}$ is $y'=\frac {x+y}{x-y}$.