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akanksha

  • 4 years ago

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) ?

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  1. akanksha
    • 4 years ago
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    \[f(x)=\sin ^{-1} ( (2x) / (1+x^2) ) \]

  2. akanksha
    • 4 years ago
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    \[g(x)=\cos ^{-1} ( (1-x^2) / (1+x^2) ) \]

  3. akanksha
    • 4 years ago
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    but seems to lengthy to do. :( ?

  4. AbdulRehman
    • 4 years ago
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    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

  5. amogh
    • 4 years ago
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    akanksha: what you mentioned before is right, but (df(x)/dx)/ (dg(x)/dx) would be a better choice!

  6. amogh
    • 4 years ago
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    Ok right, you did the same thing. There is no other way to do it! :)

  7. akanksha
    • 4 years ago
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    @Abdul , how did y ou do that ??

  8. amogh
    • 4 years ago
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    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.

  9. akanksha
    • 4 years ago
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    Cool ! Thanks.

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