The derivative of $y = {sin}^{- 1} (\frac{x}{2})$ $y=sin^-1(x/2)$ is?

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Answer 1 · 2018-06-13T07:31:10

$\frac{d y}{d x} = \frac{1}{2 \sqrt{1 - \frac{x^{2}}{4}}}$ $dy/dx=1/(2sqrt(1-x^2/4))$

If $y = {sin}^{- 1} (f (x))$ $y=sin^-1(f(x))$ then $dy/dx=(f'(x))/sqrt(1-f(x)^2)$

$f(x)=x/2$
$f'(x)=1/2$
$f(x)^2=x^2/4$

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

Answer 2 · 2018-06-13T07:37:20

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

$y=sin^(-1)(x/2)$

$=>x/2=siny$

differentiate $wrt x$

$1/2=cosy(dy)/(dx)$

$:. (dy)/(dx)=1/(2cosy)--(1)$

but

$sin^2u+cos^2u=1$

$:. cos^2y==1-sin^2y$

$cosy=sqrt(1-(x/2)^2$

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

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

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

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

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

The derivative of y=sin^-1(x/2)y=sin−1(x2)y=sin^-1(x/2) is?