Chapter 4, Problem 33RE

To determine

To calculate : All the values of $x$ for which the function is differentiable.

Answer to Problem 33RE

Function is differentiable for $x<1$

Explanation of Solution

Given information : Given function is $y=\sqrt{\frac{1-x}{1+x^2}}$

Formula used : Differentiation with respect to $x$

Calculation : Firstly, we will find the domain of $y$

For this we will take $\frac{1-x}{1+x^2}\ge 0$

  $1-x\ge 0$

  $1\ge x$

  $x\le 1$

The domain of $y$ is $x\le 1$

Now, we find the derivative of

  $y$ to check if there are any restrictions on its domain.

  $y=\sqrt{\frac{1-x}{1+x^2}}$

  $y={\left(\frac{1-x}{1+x^2}\right)}^{\frac{1}{2}}$

  $y\text{'}=\frac{1}{2}{\left(\frac{1-x}{1+x^2}\right)}^{\frac{1}{2}}.\frac{-1(1+x^2)-(1-x)(2x)}{{(1+x^2)}^2}$

  $y\text{'}=\frac{1}{2}{\left(\frac{1+x^2}{1-x}\right)}^{\frac{1}{2}}.\frac{-1-x^2-2x+2x^2}{{\left(1+x^2\right)}^2}$

  $y\text{'}=\frac{1}{2}.\frac{{(1+x^2)}^{\frac{1}{2}}}{\sqrt{1-x}}.\frac{x^2-2x-1}{{(1+x^2)}^2}$

  $y\text{'}=\frac{x^2-2x-1}{2\sqrt{1-x}}.\frac{1}{{(1+x^2)}^{\frac{3}{2}}}$

Now, ${(1+x^2)}^{\frac{3}{2}}$ has a domain of all real numbers and is never equal to 0 and $\sqrt{1-x}$ has a restricted domain and cannot be equal to 0. Therefore, set $1-x$ to be greater than 0 to find the domain of $y\text{'}$

  $\begin{array}{l}1-x>0\\ 1>x\\ x<1\end{array}$

So, $y$ is differentiable only for values of $x$ that are in domain of $y$ and $y\text{'}$ .Therefore, function is differentiable for $x<1$