Why Does Differentiating Arctan(x^2-1) + Arccsc(x) Yield Zero?

AI Thread Summary
The discussion centers on the derivative of the function y = Arctan(x^2-1) + Arccsc(x) for x > 1, which the textbook claims equals zero. Participants clarify that the derivative should indeed yield zero due to the relationship between the inverse tangent and cosecant functions, specifically that Arctan(√(x^2-1)) + Arccsc(x) equals π/2. One user expresses confusion over their initial calculations, while another confirms the correct reasoning. The exchange highlights the importance of recognizing specific trigonometric identities in calculus problems. Understanding these relationships is crucial for solving similar derivative problems effectively.
Simfish
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Hi, I kow about the chain rule and such; but there I cannot solve a problem in my calc textbook.

y = Arctan(x^2-1) + Arccscx, x > 1.

Answer on back says 0; but my answer didn't go to 0. BTW, this is just one problem from a HW assignment so this shouldn't count as cheating.

Thanks!

-Simfish
 
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The derivative should be zero because

\tan^{-1} \sqrt{x^2-1} + \csc^{-1}x = \frac {\pi}{2}
 
Sleazy Tide!
Why didn't I see that..
 
arildno said:
Sleazy Tide!
Why didn't I see that..

LoL! Oh, you would have seen it - I just happened to see it first.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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