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

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Question

How do you differentiate # g(x) = sqrtarccos(x^2-1) 
 
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Ezekiel Alexander · Accepted Answer

Use implicit differentiation and the chain rule to find that

#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))#

Adam Adkins · Answer

undefined To differentiate $ g(x) = \sqrt{\arccos(x^2-1)} $, we can use the chain rule:

Let $ u = \arccos(x^2-1) $. Then $ y = \sqrt{u} $.

Using the chain rule:
$$ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} $$

First, find $ \frac{du}{dx} $:
$$ \frac{du}{dx} = \frac{d}{dx}[\arccos(x^2-1)] $$

Applying the chain rule again:
$$ \frac{du}{dx} = \frac{-1}{\sqrt{1 - (x^2-1)^2}} \cdot \frac{d}{dx}(x^2-1) $$

$$ \frac{du}{dx} = \frac{-1}{\sqrt{1 - (x^2-1)^2}} \cdot 2x $$

Now, find $ \frac{dy}{du} $:
$$ \frac{dy}{du} = \frac{d}{du}[\sqrt{u}] $$

$$ \frac{dy}{du} = \frac{1}{2\sqrt{u}} $$

Combine both:
$$ \frac{dy}{dx} = \frac{1}{2\sqrt{u}} \cdot \frac{-1}{\sqrt{1 - (x^2-1)^2}} \cdot 2x $$

$$ \frac{dy}{dx} = \frac{-x}{\sqrt{u(1 - (x^2-1)^2)}} $$

Finally, substitute $ u = \arccos(x^2-1) $:
$$ \frac{dy}{dx} = \frac{-x}{\sqrt{\arccos(x^2-1)(1 - (x^2-1)^2)}} $$

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