How do you differentiate # y= 3y^4-4u+5 ;u=x^3-2x-5 # using the chain rule?

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Answer 1 · 2017-10-09T15:22:32

#dy/dx=(-12x^2+8)/(1-12y^3#

When we put the value of #u# in #y# we get

#y=3y^4-4(x^3-2x-5)+5#

#y=3y^4-4x^3+8x+25#

When we differentiate both sides with respect to #x# we apply chain rule.

#d/dxy=d/dx(3y^4-4x^3+8x+25)#

#dy/dx=12y^3d/dxy-12x^2d/dxx+8#

#dy/dx=12y^3dy/dx-12x^2+8#

#dy/dx-12y^3dy/dx=-12x^2+8#

#dy/dx(1-12y^3)=-12x^2+8#

Therefore

#dy/dx=(-12x^2+8)/(1-12y^3#