How do you implicitly differentiate #5y^2-x=1/sin(x)#?

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Answer 1 · 2015-12-31T05:18:22

#dy/dx=(sin^2x-cosx)/(10ysin^2x)#

#d/dx(5y^2)=10y*y'#

#d/dx(x)=1#

#d/dx((sinx)^-1)=-sin^-2x*d/dx(sinx)=-sin^-2xcosx=-cosx/sin^2x#

Add all the derivatives (sum rule).

#10y*y'-1=-cosx/sin^2x#

Solve for #y'#.

#10y*y'=1-cosx/sin^2x#

#10y*y'=(sin^2x-cosx)/sin^2x#

#y'=(sin^2x-cosx)/(10ysin^2x)=dy/dx#