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When differentiating $\sin^{-1}(x/2)$, I got $\frac{1}{2}(4-x^2)^{-1/2}$ but the answer I'm given does not include being multiplied by half.

Can anyone explain if the answer I'm given is right and why they did not multiply the equation by the derivative of the function in the inverse trig?

J. Doe
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3 Answers3

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You have to use the chain rule, and then simply it algebraically. $$\eqalign{ & y = si{n^{ - 1}}(x/2) \cr & y' = {1 \over {\sqrt {1 - {{(x/2)}^2}} }}*{1 \over 2} = {1 \over {2\sqrt {1 - {{{x^2}} \over 4}} }} = {1 \over {\sqrt {4 - {x^2}} }} \cr}$$

  • thanks for answering my question. But i am confused as to how you moved from the second last step to the last step. I would be grateful if you could clear my confusion. – J. Doe Apr 26 '15 at 15:46
  • Just algebra. $${1 \over {2\sqrt {1 - {{{x^2}} \over 4}} }} = {1 \over {\sqrt 4 \sqrt {1 - {{{x^2}} \over 4}} }} = {1 \over {\sqrt {4\left( {1 - {{{x^2}} \over 4}} \right)} }} = {1 \over {\sqrt {4 - {x^2}} }}$$ –  Apr 26 '15 at 15:48
  • genius, thanks man – J. Doe Apr 26 '15 at 15:49
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$$\left(\sin^{-1}(x/2)\right)' = \frac 12\cdot \frac 1{\sqrt{1-\left(\frac x2\right)^2}} = \frac 1{2\sqrt{1-\left(\frac {x^2}{4}\right)}} $$

$$ \overset{\large*}{=}\frac 1{\sqrt{4\left(1-\left(\frac {x^2}{4}\right)\right)}}= \frac 1{\sqrt{4 - x^2}}$$

*Note that $2 = \sqrt 4$.

Jordan Glen
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0

Let $y=\sin^{-1}(x/2)$.

$\sin(y)=x/2$

Differentiate both sides with respect to $x$:

$\cos(y)*y'=1/2$

$y'=1/(2\cos(y))=\frac{1}{2} * \frac{1}{\cos(\sin^{-1}(x/2))}=\frac{1}{2} * \frac{1}{\sqrt{2^2-x^2}/2}=\frac{1}{\sqrt{4-x^2}}$

James Pak
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