Differentiating :calculus theory

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SUMMARY

The forum discussion focuses on verifying the identity arctan(x) + arctan(1/x) = π/2 using calculus theory. Participants emphasize the importance of differentiating the left-hand side of the identity, leading to the conclusion that the derivative f'(x) = 1/(1+x²) - 1/(x²+1) equals zero for all x ≠ 0. This indicates that the function is constant across its domain, confirming the identity holds true for x > 0. The discussion clarifies the necessity of selecting a specific value of x to determine the constant.

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chapsticks
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Homework Statement




Verify the identity:

arctanx + arctan(1/x)=∏/2

using calculus theory.
(Hint: Differentiate the left hand side of the identity)

Homework Equations



?

The Attempt at a Solution


is this correct?

tan(arctanx + arctan(1/x))
= [tan(arctan(x)) + tan(arctan(1/x))][1 - tan(arctan(x))*tan(arctan(1/x))]

= [x + 1/x]/[1 - x*1/x] = (x + 1/x)/0 = oo

tan pi/2 = oo
 
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mmm... they wanted you to do this:
\frac{1}{1+x^2}+ etc...
 
where did that come from I'm confused??
 
chapsticks said:

Homework Statement




Verify the identity:

arctanx + arctan(1/x)=∏/2

using calculus theory.
(Hint: Differentiate the left hand side of the identity)

Homework Equations



?

The Attempt at a Solution


is this correct?

tan(arctanx + arctan(1/x))
= [tan(arctan(x)) + tan(arctan(1/x))][1 - tan(arctan(x))*tan(arctan(1/x))]

= [x + 1/x]/[1 - x*1/x] = (x + 1/x)/0 = oo

tan pi/2 = oo

Start by using the hint, which is to differentiate the left side.
 
f'(x)=1/(1+x2) + -1/(x2+1) =0

for all x≠0

function is constant on domain

is this right?
 
chapsticks said:
f'(x)=1/(1+x2) + -1/(x2+1) =0

for all x≠0

function is constant on domain

is this right?

Yes, it is. I guessing they meant to specify x>0. Now just pick a nice value of x to put into figure out what the constant is.
 

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