What am I doing wrong finding the derivative of $\frac{3-x}{2}\sqrt{1-2x-x^2}+2\arcsin{\frac{1+x}{\sqrt{2}}}$?

Question

$$y=\frac{3-x}{2}\sqrt{1-2x-x^2}+2\arcsin{\frac{1+x}{\sqrt{2}}}$$ For convenience, let $$A=\frac{3-x}{2}\sqrt{1-2x-x^2},$$ $$B=2\arcsin{\frac{1+x}{\sqrt{2}}}.$$

$$y'=A'+B'$$ $$A'=(-\frac{1}{2})(\sqrt{1-2x-x^2})+(-2x)(\frac{1}{2}\cdot \frac{1}{{\sqrt{1-2x-x^2}}})(\frac{3-x}{2})$$ $$A'=(-\frac{1}{2})(\sqrt{1-2x-x^2})+(-x)(\frac{1}{{\sqrt{1-2x-x^2}}})(\frac{3-x}{2})$$ $$A'=(-\frac{1-2x-x^2}{2\sqrt{1-2x-x^2}})+(\frac{x^2-3x}{{2\sqrt{1-2x-x^2}}})$$ $$A'=(-\frac{1-2x-x^2}{2\sqrt{1-2x-x^2}})+(\frac{x^2-3x}{{2\sqrt{1-2x-x^2}}})$$ $$A'=\frac{2x^2-x-1}{{2\sqrt{1-2x-x^2}}}$$

$$B'=2\bigg(\arcsin{\frac{1+x}{\sqrt{2}}}\bigg)'$$ $$B'=2 \bigg( \frac{1+x}{\sqrt{2}} \bigg)' \bigg(\arcsin{\frac{1+x}{\sqrt{2}}}\bigg)'$$ $$B'=\sqrt{2} \bigg( \frac{1}{\sqrt{1- \frac{1+2x+x^2}{2}} } \bigg)$$ $$B'=\sqrt{\frac{4}{1-2x-x^2}}$$ $$B'=\frac{4}{2\sqrt{1-2x-x^2}}$$

$$y'=A'+B'=\frac{2x^2-x+3}{2\sqrt{1-2x-x^2}}$$

The answer in the book is $$y'=\frac{x^2}{\sqrt{1-2x-x^2}}$$

Answer 1

You got A' wrong.

$$ A = \frac{3-x}{2}\cdot(\sqrt{1-2x-x^2}) = f\cdot g$$

where $ f =\frac{3-x}{2} $ and $ g = \sqrt{1-2x-x^2} $.

Looking at $$A'= f'g + g'f$$the first part

$$ f'g = (-\frac{1}{2})(\sqrt{1-2x-x^2})$$ is correct, but the second part is $$ g'f = (-2x -2)(\frac{1}{2}\cdot \frac{1}{{\sqrt{1-2x-x^2}}})(\frac{3-x}{2})$$ rather than $(-2x)(\frac{1}{2}\cdot \frac{1}{{\sqrt{1-2x-x^2}}})(\frac{3-x}{2})$ as you wrote.

Answer 2

In first $A'$ instead of $(-2x)$ you have $(-2-2x)$

Answer 3

In the first line of your derivation of $A'$: after the first summand, your second summand should be $$\color{blue}{(-2 -2x)}\left(\frac{1}{2}\cdot \frac{1}{{\sqrt{1-2x-x^2}}}\right)\left(\frac{3-x}{2}\right)$$

So in fact, $$A'=\left(-\frac{1}{2}\right)\left(\sqrt{1-2x-x^2}\right)+ \color{blue}{(-2 -2x)}\left(\frac{1}{2}\cdot \frac{1}{{\sqrt{1-2x-x^2}}}\right)\left(\frac{3-x}{2}\right)$$