Given $f(x) = \arcsin\left(\frac{2x}{1+x^2}\right)$, find $f'(1)$.

Question

Let $$ f(x) = \arcsin\left(\frac{2x}{1+x^2}\right). $$ What is the value of $f'(1)$?

The function splits into $$ f(x) = \begin{cases} \phantom{\pi-\,}2\arctan(x), & \text{if $-1 \leq x \leq 1$},\\ \pi - 2\arctan(x), & \text{if $x > 1$}. \end{cases} $$

I differentiated the function corresponding to $1$ but my answer is coming to be $1$, but the book tells me the answer does not exist. How can we say the derivative doesn't exist?

Answer 1

Note that $\arcsin$ is differentiable in $(-1,1)$, but it is not differentiable at $\pm1$. Since $\frac{2\times1}{1+1^2}=1$, it would be strange indeed if $f(x)=\arcsin\left(\frac{2x}{1+x^2}\right)$ was differentiable at $1$.

Answer 2

Recall that

$$(\arcsin u)'=\frac1{\sqrt{1-u^2}}$$

and since the derivative tends to $\infty$ as $u\to 1^-$ the function is not differentiable at that point.

That case is similar to $f(x)=\sqrt x$ which has vertical tangent at $x=0$.

Answer 3

Notice that \begin{align*} f'(x) &= \frac{1}{\sqrt{1-\left(\frac{2x}{1+x^2}\right)^2}}\frac{d}{dx}\left(\frac{2x}{1+x^2}\right) \\&= \frac{1}{\sqrt{\left(\frac{1+x^2}{1+x^2}\right)^2-\left(\frac{2x}{1+x^2}\right)^2}}\cdot\frac{(1+x^2)(2) - (2x)(2x)}{(1+x^2)^2} \\&= \sqrt{\frac{(1+x^2)^2}{(1-x^2)^2}}\cdot \frac{2(1-x^2)}{(1+x^2)^2} \end{align*} (Skipping a bit of the algebra here). You would like to cancel the square and square root, and in the case of $(1+x^2)$, you can. But $1-x^2$ could be positive or negative, so $\sqrt{(1-x^2)^2} = |1-x^2|$, not $1-x^2$. Therefore $$ f'(x) = \frac{2}{1+x^2}\cdot \frac{1-x^2}{|1-x^2|} $$ As $x\to 1$, the first factor tends to $2$, unambiguously. But the second factor is $\pm 1$ depending on whether the numerator is positive or negative. That is, $$ \lim_{x\to 1^-} f'(x) = 1\qquad\text{and}\qquad\lim_{x\to 1^+} f'(x) = -1 $$ It follows that $f$ cannot be differentiable at $1$.