How do you differentiate y= 3y^4-4u+5 ;u=x^3-2x-5 y=3y44u+5;u=x32x5 using the chain rule?

1 Answer
Oct 9, 2017

dy/dx=(-12x^2+8)/(1-12y^3dydx=12x2+8112y3

Explanation:

When we put the value of uu in yy we get

y=3y^4-4(x^3-2x-5)+5y=3y44(x32x5)+5

y=3y^4-4x^3+8x+25y=3y44x3+8x+25

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

d/dxy=d/dx(3y^4-4x^3+8x+25)ddxy=ddx(3y44x3+8x+25)

dy/dx=12y^3d/dxy-12x^2d/dxx+8dydx=12y3ddxy12x2ddxx+8

dy/dx=12y^3dy/dx-12x^2+8dydx=12y3dydx12x2+8

dy/dx-12y^3dy/dx=-12x^2+8dydx12y3dydx=12x2+8

dy/dx(1-12y^3)=-12x^2+8dydx(112y3)=12x2+8

Therefore

dy/dx=(-12x^2+8)/(1-12y^3dydx=12x2+8112y3