How do you find the derivative of $arcsin (x/2)$ ?

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Answer 1 · 2016-12-01T23:40:30

$1/sqrt(4-x^2)$

Let $y=arcsin(x/2)$ . Rearranging this using the properties of inverse functions yields:

$sin(y)=x/2$

Differentiate both sides with respect to $x$ . The left hand side will need the chain rule.

The right hand side's derivative is $1/2$ since $x/2=1/2x$ .

$cos(y)*dy/dx=1/2$

We can write $cos(y)$ in terms of $sin(y)$ , which we know equals $x/2$ . From $sin^2(y)+cos^2(y)=1$ , we see that $cos(y)=sqrt(1-sin^2(y))$ .

$sqrt(1-sin^2(y))*dy/dx=1/2$

Since $sin(y)=x/2$ , we see that $sin^2(y)=x^2/4$ .

$sqrt(1-x^2/4)*dy/dx=1/2$

$sqrt((4-x^2)/4)*dy/dx=1/2$

Taking the $sqrt4$ out of the denominator:

$1/2sqrt(4-x^2)*dy/dx=1/2$

Multiplying both sides by $2$ :

$sqrt(4-x^2)*dy/dx=1$

$dy/dx=1/sqrt(4-x^2)$

Thus the derivative of $arcsin(x/2)$ is $1/sqrt(4-x^2)$ .

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