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

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

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Answer 1 · 2016-08-28T12:02:09

$\frac{d}{d x} tan (arcsin (x)) = \frac{1}{{(1 - x^{2})}^{\frac{3}{2}}}$ $d/(dx) tan(arcsin(x))= 1/(1-x^2)^(3/2)$

Explanation:

Let $t = arcsin (x)$ $t = arcsin(x)$

Then:

$x = sin (t)$ $x = sin(t)$

So:

$tan (arcsin (x)) = tan (t) = \frac{sin (t)}{cos (t)} = \frac{x}{\sqrt{1 - x^{2}}}$ $tan(arcsin(x)) = tan(t) = sin(t)/cos(t) = x/sqrt(1-x^2)$

So:

$\frac{d}{d x} tan (arcsin (x))$ $d/(dx) tan(arcsin(x))$

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

$= {(1 - x^{2})}^{- \frac{1}{2}} + x \cdot (- \frac{1}{2}) {(1 - x^{2})}^{- \frac{3}{2}} \cdot (- 2 x)$ $= (1-x^2)^(-1/2) + x*(-1/2)(1-x^2)^(-3/2)*(-2x)$

$= (1 - x^{2}) {(1 - x^{2})}^{- \frac{3}{2}} + x^{2} {(1 - x^{2})}^{- \frac{3}{2}}$ $= (1-x^2)(1-x^2)^(-3/2) +x^2(1-x^2)^(-3/2)$

$= \frac{1}{{(1 - x^{2})}^{\frac{3}{2}}}$ $= 1/(1-x^2)^(3/2)$

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

1 Answer

Explanation:

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Science

Math

Humanities

... and beyond

How do you find the derivative of tan(arcsin(x))tan(arcsin(x))?

1 Answer

Explanation:

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Impact of this question

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