What is the derivative of this function ${sin}^{- 1} (\frac{x}{4})$ $sin^-1(x/4)$ ?

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Answer 1 · 2016-11-28T19:36:19

$\frac{d}{d x} {sin}^{- 1} (\frac{x}{4}) = \frac{1}{4 \sqrt{1 - \frac{x^{2}}{16}}}$ $d/dx sin^-1(x/4)= 1/(4sqrt(1-x^2/16))$

Let $y = {sin}^{- 1} (\frac{x}{4}) \Leftrightarrow sin y = \frac{x}{4}$ $y=sin^-1(x/4) <=> siny=x/4$

Differentiate Implicitly:

$cos y \frac{d y}{d x} = \frac{1}{4}$ $cosydy/dx = 1/4$ .... [1]

Using the fundamental Trig Identity ${sin}^{2} A + {cos}^{2} A \equiv 1$ $sin^2A+cos^2A-=1$

${sin}^{2} y + {cos}^{2} y = 1$ $sin^2y+cos^2y=1$
$:. (x/4)^2+cos^2y=1$
$:. x^2/16+cos^2y=1$
$:. cos^2y=1-x^2/16$
$:. cosy=sqrt(1-x^2/16)$

Substituting into [1]:

$sqrt(1-x^2/16)dy/dx = 1/4$
$:. dy/dx = 1/(4sqrt(1-x^2/16))$

What is the derivative of this function sin^-1(x/4)sin−1(x4)sin^-1(x/4)?