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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.

Thanks.

  • 0
    @Dylan: I think I complained about that one in that long long thread on tags...2011-11-24

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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.

  • 5
    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}$2011-11-24