How do you differentiate #y^5-xcosy+y^3/x=11#?

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Answer 1 · 2018-02-06T10:08:51

#dy/dx = (x^2cosy+y^3)/(x^3siny+5x^2y^4+3xy^2)#

#y^5-xcosy+y^3/x=11#

#5y^4y'-(-xsiny*y'+1*cosy)+(x*3y^2y'-y^3*1)/(x^2) =0#

#5y^4y'+xsiny*y' -cosy + (3y^2y')/x-y^3/x^2=0#

#y'(5y^4+xsiny+(3y^2)/x) = y^3/x^2+cosy#

#y' = (y^3/x^2+cosy)/(5y^4+xsiny+(3y^2)/x)#

#= (y^3+x^2cosy)/(x^2(5y^4+xsiny+(3y^2)/x))#

#= (x^2cosy+y^3)/(x^3siny+5x^2y^4+3xy^3)#