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I am having another problem with basic math in my homework. Help is much appreciated.

The answer is: $\frac 1{\sqrt{4-x^2}}$

The problem is: Find $\frac {\mathrm{d}}{\mathrm dx} \sin^{-1} (x/2)$

What I have:

$\frac 1{\sqrt{1-(x/2)^2}} \cdot \frac12$

$\frac 1{\sqrt{4-x^2}} \cdot \frac 12$.

How do I move from what I have to the answer. Essentially, I don't know how to get rid of the 1/2. I typed in the answer in WolframAlpha and it gave me the answer. I clicked "Show Steps" and it showed my work until the last step putting a 2 in the denominator but not showing how to get rid of the two for the answer.


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I've edited your question for better readability. You should checked whether I did not change anything important. (I undestood that some of your x's meant $x$ and some of them were $\times$. I hope that was correct.) – Martin Sleziak Nov 24 '11 at 20:24
BTW since you write that you used WolframAlpha, adding the link with your input would be nice, something like this: – Martin Sleziak Nov 24 '11 at 20:27
Also, it's not clear to me why the "inverse" tag exists. – Dylan Moreland Nov 24 '11 at 20:27
@Dylan, presumably because it's a question about the inverse sine function. But if the tag offends you, just edit it out, I'm sure no one will mind. – Gerry Myerson Nov 24 '11 at 22:48
@GerryMyerson Oh, I picked up on that. But I can't imagine someone thinking, "I'd like to look at all the questions about inverses". I'll poke around on meta. – Dylan Moreland Nov 24 '11 at 23:23
up vote 7 down vote accepted

Just a bit of factoring: $$\eqalign{ {1\over\sqrt{1-(x/2)^2}}\cdot{1\over 2} &={1\over\sqrt{1-{x^2\over4}}}\cdot{1\over 2} \cr &={1\over\sqrt{{1\over 4}(4-{x^2})}}\cdot{1\over 2} \cr &={1\over\sqrt{{1\over 4}}\sqrt{4-{x^2}}}\cdot{1\over 2} \cr &={1\cdot {1\over 2}\over{1\over 2}{\sqrt{4-{x^2}}}} \cr &={1\over {\sqrt{4-{x^2}}}} .} $$

The $1/2$'s cancelled in the last step.

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You could also "push" the $2$ into the radicand, making it 4. Then distribute that 4 across $1-(x/2)^2$: $$\sqrt{4(1-(x/2)^2)}=\sqrt{4-4\cdot(x/2)^2}=\sqrt{4-(2\cdot(x/2))^2}=\sqrt{4-x^‌​2}$$ – David Mitra Nov 24 '11 at 20:37

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