## akanksha Group Title 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) ? 3 years ago 3 years ago

1. akanksha Group Title

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

2. akanksha Group Title

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

3. akanksha Group Title
4. akanksha Group Title

but seems to lengthy to do. :( ?

5. AbdulRehman Group Title

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

6. amogh Group Title

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

7. amogh Group Title

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

8. akanksha Group Title

@Abdul , how did y ou do that ??

9. amogh Group Title

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.

10. akanksha Group Title

Cool ! Thanks.