How do you differentiate $g(x) = sqrtarccos(x^2-1)$ ?

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Answer 1 · 2016-03-02T19:54:10

$d/dxsqrt(arccos(x^2-1))=x/(|x|sqrt(arccos(x^2-1))sqrt(2-x^2))$

First, let's see what the derivative of the $arccos$ function is by using implicit differentiation:

Let $y = arccos(x)$

$=>cos(y) = x$

$=>d/dxcos(y) = d/dxx$

$=>-sin(y)dy/dx = 1$

$=>dy/dx = -1/sin(y)$

$=-1/sin(arccos(x))$

$=-1/sqrt(1-x^2)$
(For the last step, try drawing a right triangle where $cos(theta) = x$ and then see what $sin(theta)$ is equal to)#

With that, the remainder of the problem can be done using the chain rule:

$d/dxsqrt(arccos(x^2-1))$

$= 1/2(arccos(x^2-1))^(-1/2)(d/dxarccos(x^2-1))$

$=1/(2sqrt(arccos(x^2-1)))*(-1)/sqrt(1-(x^2-1)^2)(d/dx(x^2-1))$

$=(2x)/(2sqrt(arccos(x^2-1))sqrt((1+x^2-1)(1-x^2+1))$

$=x/(|x|sqrt(arccos(x^2-1))sqrt(2-x^2))$

How do you differentiate g(x) = sqrtarccos(x^2-1) g(x) = sqrtarccos(x^2-1) ?