How do you find the derivative of inverse trig functions #y= arctan(x^2-1)^(1/2) + arc csc(x)# when x1?

Question

How do you find the derivative of inverse trig functions #y= arctan(x^2-1)^(1/2) + arc csc(x)#  when x>1?

Emma Aldridge · Accepted Answer

#y^' = 0#

Assuming the equation was meant to be read as

#y = arctan(sqrt(x^2-1)) + arc csc(x)#

We can say the derivative will be the sum of the two other derivatives

#u = arctan(sqrt(x^2-1))# #v = arc csc(x)#

#v# will be easier to differentiate, know, knowing that

#csc(arc csc(x)) = x#

We differentiate both sides and use the chain rule, so

#1 = csc^'(arc csc(x))*arc csc^'(x)# #1/csc^'(arc csc(x)) = arc csc^'(x)#

We know that the derivative of the cossecant is #-csc(x)cot(x)#, so, in other words

#1/(-csc(arc csc(x))cot(arc csc(x))) = arc csc^'(x)#

Since they're both rational functions we can bring them up to the numerator

#-tan(arc csc(x))/x = arc csc^'(x)#

Using the pythagorean identity #1 + cot^2(theta) = csc^2(theta)# we have

#1 + 1/(tan^2(arc csc(x))) = x^2#

#1/(tan^2(arc csc(x))) = x^2 - 1#

#tan(arc csc(x)) = 1/sqrt(x^2 - 1)# so

#arc csc^'(x) = -1/(xsqrt(x^2-1)) #

(There are authors that say it like this, there are people that have that leading x as #|x|#, considering we're working with #x > 1#, the absolute value bars aren't needed regardless of your philosophy if they should be there or not).

Now, we do #u = arctan(sqrt(x^2-1))#, for the sake of brevity, I'll skip the proof of the arctan's derivative, as the process is much like the one used for the arccsc, except it'd involve the #tan^2(theta) + 1 = sec^2(theta)# identity.

So we have

#u = arctan(sqrt(x^2-1))# #u^' = arctan^'(sqrt(x^2-1))*sqrt((x^2-1))^'*2x#

It's important to remember we need to apply the chain rule twice here.

#u^' = 1/(1+(sqrt(x^2-1))^2)*1/(2sqrt(x^2-1))*2x# #u^' = 1/(1+x^2-1)*1/(2sqrt(x^2-1))*2x# #u^' = 1/x^2*1/(2sqrt(x^2-1))*2x# #u^' = 1/(xsqrt(x^2-1))#

Summing them up we have

#y^' = 1/(xsqrt(x^2-1)) - 1/(xsqrt(x^2-1)) = 0#

Which makes sense if you look at the graph for #x > 0#. For smaller values the absolute value probably would have mattered and would have made the derivative non-zero.

Aurora Alvarado · Answer

undefined To find the derivative of the given function, we apply the chain rule and the derivative formulas for inverse trigonometric functions:

1. For $ y = \arctan((x^2 - 1)^{1/2}) $:
   - Let $ u = (x^2 - 1)^{1/2} $.
   - Then, $ y = \arctan(u) $.
   - Applying the chain rule, the derivative is $ \frac{dy}{dx} = \frac{1}{1 + u^2} \cdot \frac{du}{dx} $.
   - Now, find $ \frac{du}{dx} $ using the power rule: $ \frac{du}{dx} = \frac{1}{2}(x^2 - 1)^{-1/2} \cdot 2x = \frac{x}{(x^2 - 1)^{1/2}} $.
   - Substitute back into the derivative: $ \frac{dy}{dx} = \frac{1}{1 + u^2} \cdot \frac{x}{(x^2 - 1)^{1/2}} = \frac{x}{(x^2 - 1) + x^2} = \frac{x}{2x^2 - 1} $.

2. For $ y = \text{arccsc}(x) $:
   - Recall that $ \text{arccsc}(x) = \arcsin\left(\frac{1}{x}\right) $.
   - Applying the derivative of $ \arcsin $, we have $ \frac{d}{dx}(\arcsin(u)) = \frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx} $.
   - So, $ \frac{dy}{dx} = \frac{1}{\sqrt{1 - (1/x)^2}} \cdot \left(-\frac{1}{x^2}\right) = -\frac{1}{|x|\sqrt{x^2 - 1}} $.

Therefore, the derivative of the given function is:

$$ \frac{dy}{dx} = \frac{x}{2x^2 - 1} - \frac{1}{|x|\sqrt{x^2 - 1}} $$

for $ x > 1 $.

How do you find the derivative of inverse trig functions #y= arctan(x^2-1)^(1/2) + arc csc(x)# when x&gt;1?

How do you find the derivative of inverse trig functions #y= arctan(x^2-1)^(1/2) + arc csc(x)# when x>1?