Question

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) ?

Answer 1

\[f(x)=\sin ^{-1} ( (2x) / (1+x^2) ) \]

Answer 2

\[g(x)=\cos ^{-1} ( (1-x^2) / (1+x^2) ) \]

Answer 3

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

Answer 4

but seems to lengthy to do. :( ?

Answer 5

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

Answer 6

akanksha: what you mentioned before is right, but
(df(x)/dx)/ (dg(x)/dx) would be a better choice!

Answer 7

Ok right, you did the same thing. There is no other way to do it! :)

Answer 8

@Abdul , how did y ou do that ??

Answer 9

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.

Answer 10

Cool ! Thanks.