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Question
hi, im trying to find the derivative of (1/2)tan^-1(x/2) - (1/3)tan^-1(x/3)
also does tan to the power of -1 mean inverse tan?
I dont no how to approach this problem and i would appreciate any help
The -1 means the iverse. If x = tan(Θ), then Θ = tan^-1(x).
If we integrate x = tan(Θ), we get dx = sec˛(Θ)dΘ.
Solving for dΘ requires looking at the triangle.
We'll give the triangle a far side of x.
Since that tan is the far side / the near side,
the near side would be 1.
The hypoteneuse is √(A˛+B˛) where A and B are the sides.
Since the two sides are 1 and x, the hypoteneuse is √(1+x˛).
In terms of x, that makes sec˛(Θ) into 1+x˛.
Taking our equation dx = sec˛(Θ) dΘ,
that means dx = (1+x˛)dΘ. Divide both sides by 1+x˛ to put the terms
on the right side of the equation and the result it
[1/(1+x˛)]dx = dΘ.
Note that the 1/2 and the 1/3 out front would just come along
as constants and would be times the derivative.
However, to differentiate the two terms,
one with x/2 and the other with x/3,
we need to multiply by the derivative of those two terms when done.
The first needs to be times the derivative of x/2, which is 1/2.
The second needs to be times the derivative of x/3, which is 1/3.
Overall, what we get, then, is
(1/2)sec˛(x/2)d(x/2)/dx + (1/3)sec˛(x/3)d(x/3)dx
= (1/2)sec˛(x/2)(1/2) + (1/3)sec˛(x/3)(1/3)
= (1/4)sec˛(x/2) + (1/9)sec˛(x/3).
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Answers by Expert:
Scotto
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Any kind of calculus question you want. I also have answered some questions in Physics (mass, momentum, falling bodies), Chemistry (charge, reactions, symbols, molecules), and Biology.
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