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Find dy/dx of ,y=1/sin(2x/1+x^2)

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Find dy/dx of ,y=1/sin(2x/1+x^2)

Posted by Sourav Attri 3 years, 8 months ago

CBSE > Class 12 > Mathematics

1 answers

Shreya Rani 3 years, 8 months ago

dy/dx =d [sin(2x/1+x^2)^-1] /dx =-1/sin^2(2x/1+x^2). d(2x/ 1+x^2)/dx =-1/sin^2(2x/1+x^2). [(1+x^2).d(2x)/dx - 2x.d(1+x^2)/dx)] / (1+x^2)^2 =1/sin^2(2x/1+x^2).[(1+x^2).2 - 2x.2x] / (1+x^2)^2 =1/sin^2(2x/1+x^2).(2+2x^2 - 4x^2) /(1+x^2)^2

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