Here's the question you clicked on:
akanksha
Differentiate a func with wrt another ? f(x) = arcsin((2x)/(1+x^2)) and g(x) = arccos((1-x^2)/(1+x^2)) , then df(x)/dg(x) = ? How do I do this ? df(x)/dx * dx/dg(x) ?
\[f(x)=\sin ^{-1} ( (2x) / (1+x^2) ) \]
\[g(x)=\cos ^{-1} ( (1-x^2) / (1+x^2) ) \]
I 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%281-x^2%29%2F%281%2Bx^2%29%29%29
but seems to lengthy to do. :( ?
df(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
akanksha: what you mentioned before is right, but (df(x)/dx)/ (dg(x)/dx) would be a better choice!
Ok right, you did the same thing. There is no other way to do it! :)
@Abdul , how did y ou do that ??
OK, 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.