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Mathematics 19 Online
OpenStudy (anonymous):

d/dy tan inverse (y/x)

OpenStudy (aravindg):

Apply chain rule.

OpenStudy (anonymous):

please solve that

OpenStudy (aravindg):

Don't expect direct worked out answers here. I can guide you to the answer though.

OpenStudy (anonymous):

plzzzzzzzzzzzzzzzzzz

OpenStudy (aravindg):

Treat y/x=k And first differentiate tan inverse part. Then chain rule.

OpenStudy (aravindg):

Try what I said.

OpenStudy (anonymous):

ok thn got that

OpenStudy (aravindg):

So you got the final answer?

OpenStudy (anonymous):

yes

OpenStudy (aravindg):

If you can write it here I can check it. Else if you are confident then its okay :)

OpenStudy (anonymous):

1/x(xsquare +y square)

OpenStudy (aravindg):

Not exactly derivative of tan-1 y/x \[\dfrac{1}{1+(y/x)^2} \times \dfrac{1}{x}\] That leaves us with x on numerator instead of x on denominator.

OpenStudy (anonymous):

sorry integration of that

OpenStudy (anonymous):

integration of cos square

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