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In my assignment I need to analyze the function

$f(x)=\arcsin \frac{1-x^2}{1+x^2}$

And so I need to do the first derivative and my result is:

$-\dfrac{4x}{\left(x^2+1\right)^2\sqrt{1-\frac{\left(1-x^2\right)^2}{\left(x^2+1\right)^2}}}$

But in the solution of this assignment it says

$f'(x)=-\frac{2x}{|x|(1+x^2)}$

I don't understand how they get this. I checked my answer on online calculator and it is the same.

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    $\begingroup$ $$(x^2+1)\sqrt{1-\frac{(1-x^2)^2}{(x^2+1)^2}} = \sqrt{(x^2+1)^2 - (1-x^2)^2}$$ $\endgroup$ Commented Jan 22, 2017 at 21:31
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    $\begingroup$ Wait, doesn't that then equal 2*|x| $\endgroup$
    – Nebeski
    Commented Jan 22, 2017 at 21:32

3 Answers 3

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You should have developed your result a bit more to obtain the assignment solution.

\begin{align} -\frac{4x}{(x^2+1)^2\sqrt{1-\frac{(1-x^2)^2}{(x^2+1)^2}}}&=-\frac{4x}{(x^2+1)^2\sqrt{\frac{x^4+1+2x^2-1-x^4+2x^2}{(x^2+1)^2}}} \\ &=-\frac{4x}{(x^2+1)^2\sqrt{\frac{4x^2}{(x^2+1)^2}}}\\ &=-\frac{4x}{(x^2+1)^2\frac{2|x|}{(x^2+1)}}\\ &=-\frac{2x}{|x|(x^2+1) } \end{align} Voila!

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Divide et impera.

Under the square root you have $$ 1-\frac{(1-x^2)^2}{(1+x^2)^2}= \frac{(1+x^2)^2-(1-x^2)^2}{(1+x^2)^2}=\frac{4x^2}{(1+x^2)^2} $$ so the big square root is actually $$ \frac{2|x|}{1+x^2} $$ Thus your formula becomes $$ -\frac{4x}{(1+x^2)^2}\frac{1+x^2}{2|x|}=-\frac{2x}{|x|(1+x^2)} $$

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Use that $(1+x^2)^2\sqrt{1-\frac{(1-x^2)^2}{(1+x^2)^2}}=(1+x^2)\sqrt{(1+x^2)^2-\frac{(1+x^2)^2(1-x^2)^2}{(1+x^2)^2}}=(1+x^2)\sqrt{((1+x^2)+(1-x^2))((1+x^2)-(1-x^2))}=(1+x^2)\sqrt{2\cdot2x^2}$

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