How do you find the derivative of $cos [arcsin (x)]$ $Cos[arcsin(x)]$ ?

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How do you find the derivative of $cos [arcsin (x)]$ $Cos[arcsin(x)]$ ?

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Answer 1 · 2017-06-08T20:31:53

$\frac{d}{d x} cos (arcsin (x)) = - \frac{x}{\sqrt{1 - x^{2}}}$ $d/(dx) cos(arcsin(x)) = -x/sqrt(1-x^2)$

Explanation:

Note that:

${cos}^{2} θ + {sin}^{2} θ = 1$ $cos^2 theta + sin^2 theta = 1$

So for any $θ$ $theta$ , we have:

$cos (θ) = \pm \sqrt{1 - sin (θ)}$ $cos(theta) = +-sqrt(1-sin(theta))$

Note that for any real $x \in [- 1, 1]$ $x in [-1, 1]$ , by definition:

$arcsin (x) \in [- \frac{π}{2}, \frac{π}{2}]$ $arcsin(x) in [-pi/2, pi/2]$

Further note that if $θ \in [- \frac{π}{2}, \frac{π}{2}]$ $theta in [-pi/2, pi/2]$ then:

$cos θ \geq 0$ $cos theta >= 0$

Hence:

$cos (arcsin (x)) = \sqrt{1 - x^{2}}$ $cos(arcsin(x)) = sqrt(1-x^2)$

for any $x \in [- 1, 1]$ $x in [-1, 1]$

Outside this interval $cos (arcsin (x))$ $cos(arcsin(x))$ takes non-real complex values.

So if we are dealing with this purely as a real valued function, then this is the only definition we need.

Then:

$\frac{d}{d x} cos (arcsin (x)) = \frac{d}{d x} {(1 - x^{2})}^{\frac{1}{2}}$ $d/(dx) cos(arcsin(x)) = d/(dx) (1-x^2)^(1/2)$

$\frac{d}{d x} cos (arcsin (x)) = \frac{1}{2} {(1 - x^{2})}^{- \frac{1}{2}} \cdot (- 2 x)$ $color(white)(d/(dx) cos(arcsin(x))) = 1/2(1-x^2)^(-1/2)*(-2x)$

$\frac{d}{d x} cos (arcsin (x)) = - \frac{x}{\sqrt{1 - x^{2}}}$ $color(white)(d/(dx) cos(arcsin(x))) = -x/sqrt(1-x^2)$

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How do you find the derivative of $cos [arcsin (x)]$ $Cos[arcsin(x)]$ ?

1 Answer

Explanation:

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How do you find the derivative of $cos [arcsin (x)]$ $Cos[arcsin(x)]$ ?