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 3 years ago
Differentiate a func with wrt another ? f(x) = arcsin((2x)/(1+x^2)) and g(x) = arccos((1x^2)/(1+x^2)) , then df(x)/dg(x) = ?
How do I do this ? df(x)/dx * dx/dg(x) ?
 3 years ago
Differentiate a func with wrt another ? f(x) = arcsin((2x)/(1+x^2)) and g(x) = arccos((1x^2)/(1+x^2)) , then df(x)/dg(x) = ? How do I do this ? df(x)/dx * dx/dg(x) ?

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akanksha
 3 years ago
Best ResponseYou've already chosen the best response.0\[f(x)=\sin ^{1} ( (2x) / (1+x^2) ) \]

akanksha
 3 years ago
Best ResponseYou've already chosen the best response.0\[g(x)=\cos ^{1} ( (1x^2) / (1+x^2) ) \]

akanksha
 3 years ago
Best ResponseYou've already chosen the best response.0I worked it out as http://www.wolframalpha.com/input/?i=derivative+of+arcsin%28%282x%29%2F%281%2Bx^2%29%29+*+%281%2F+derivative+of+arccos%28%281x^2%29%2F%281%2Bx^2%29%29%29

akanksha
 3 years ago
Best ResponseYou've already chosen the best response.0but seems to lengthy to do. :( ?

AbdulRehman
 3 years ago
Best ResponseYou've already chosen the best response.1df(x)/dx=2/(1+x^2) and dg(x)/dx=2/(1+x^2) dx/dg(x)=(1+x^2)/2 so df(x)/dg(x)=1

amogh
 3 years ago
Best ResponseYou've already chosen the best response.0akanksha: what you mentioned before is right, but (df(x)/dx)/ (dg(x)/dx) would be a better choice!

amogh
 3 years ago
Best ResponseYou've already chosen the best response.0Ok right, you did the same thing. There is no other way to do it! :)

akanksha
 3 years ago
Best ResponseYou've already chosen the best response.0@Abdul , how did y ou do that ??

amogh
 3 years ago
Best ResponseYou've already chosen the best response.0OK, you have another way to do it. \[x = \tan \theta\] Thus \[f(x) = \sin^{1}\sin(2 \theta)\]\[= 2 \theta\] \[g(x) = \cos^{1} \cos(2 \theta)\]\[= 2 \theta\] so the answer is 1.
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