Vtyjkyuk

Let
$$
f(x) = arcsinleft(frac2x1+x^2right).
$$
What is the value of $f'(1)$?

The function splits into
$$
f(x) = begincases
2arctan(x), & textif $-1 leq x leq 1$,\
pi - 2arctan(x), & textif $x > 1$.
endcases
$$

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?

Note that $arcsin$ is differentiable in $(-1,1)$, but it is not differentiable at $pm1$. Since $frac2times11+1^2=1$, it would be strange indeed if $f(x)=arcsinleft(frac2x1+x^2right)$ was differentiable at $1$.

Notice that
beginalign*
f'(x) &= frac1sqrt1-left(frac2x1+x^2right)^2fracddxleft(frac2x1+x^2right)
\&= frac1sqrtleft(frac1+x^21+x^2right)^2-left(frac2x1+x^2right)^2cdotfrac(1+x^2)(2) - (2x)(2x)(1+x^2)^2
\&= sqrtfrac(1+x^2)^2(1-x^2)^2cdot frac2(1-x^2)(1+x^2)^2
endalign*
(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
$$
f'(x) = frac21+x^2cdot frac1-x^2
$$
As $xto 1$, 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,
$$
lim_xto 1^- f'(x) = 1qquadtextandqquadlim_xto 1^+ f'(x) = -1
$$
It follows that $f$ cannot be differentiable at $1$.

Recall that

$$(arcsin u)'=frac1sqrt1-u^2$$

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

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