How do you find the derivative of the function: $y=arccos(1/x)$ ?

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Answer 1 · 2016-11-08T21:44:03

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

We can write $y=arccos(1/x) <=>cosy=1/x$
$:. cosy=x^-1$

We can then differentiate implicitly:

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

Using the identity $sin^2A + cos^2A -= 1$ we have

$sin^2y + (1/x)^2 = 1$
$:. sin^2y = 1 - 1/x^2$
$siny = sqrt(1 - 1/x^2)$

And substituting this we have:

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

How do you find the derivative of the function: y=arccos(1/x)y=arccos(1/x)?