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Question
1. If A and B are constants, show that the derivative with respect to x of each of the following functions
2sin^-1 [(x-B)/(A-B)]^1/2 and 2tan^-1 [(x-B)/(A-x)]^1/2 is [(A-x)(x-B)]^-1/2
I'll assume in both cases that the exponet is done before the trig funtion. If not, let me know.
The derivative d(sin^-1(u(x)))/dx, is equal to
(1/squareroot(1-u²))du/dx.
Here, u(x) = [(x-B)/(A-B)]^(1/2) so
du/dx = (1/2)[(x-A)/(A-B)]^(-1/2)[1/(A-B)].
The derivative d{tan^-1[u(x)]}/dx is equal to [1/(1+u²)]du/dx
where u(x)=[(x-B)/(A-x)]^(1/2).
du/dx=(1/2)[(x-B)/(A-x)]^(-1/2)[(A-x)+(x-B)]/(A-x)²].
This can be reduced even farther by combining terms.
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