How do you implicitly differentiate #3y + y^4/x^2 = 2#?

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1 Answer
Shwetank Mauria
Aug 5, 2016

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

Explanation:

When we implicitly differentiate a function #f(x,y)=0#, whenever we differentiate w.r.t. we use chain rule and when we differentiate w.r.t. #y# and then multiply it by #(dy)/(dx)#.

As such as we have #3y+y^4/x^2=2#

#3xx1xx(dy)/(dx)+(4y^3xx(dy)/(dx)-y^4xx2x)/x^4=0# or

#3(dy)/(dx)+(4y^3(dy)/(dx)-2xy^4)/x^4=0# or

#3(dy)/(dx)+4y^3/x^4(dy)/(dx)-2y^4/x^3=0# or

#(dy)/(dx)[3+(4y^3)/x^4]=(2y^4)/x^3# or

#(dy)/(dx)=((2y^4)/x^3)/(3+(4y^3)/x^4)# or

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

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