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

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Answer 1 · 2017-10-09T15:22:32

$\frac{d y}{d x} = \frac{- 12 x^{2} + 8}{1 - 12 y^{3}}$ $dy/dx=(-12x^2+8)/(1-12y^3$

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

$y = 3 y^{4} - 4 (x^{3} - 2 x - 5) + 5$ $y=3y^4-4(x^3-2x-5)+5$

$y = 3 y^{4} - 4 x^{3} + 8 x + 25$ $y=3y^4-4x^3+8x+25$

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

$\frac{d}{d x} y = \frac{d}{d x} (3 y^{4} - 4 x^{3} + 8 x + 25)$ $d/dxy=d/dx(3y^4-4x^3+8x+25)$

$\frac{d y}{d x} = 12 y^{3} \frac{d}{d x} y - 12 x^{2} \frac{d}{d x} x + 8$ $dy/dx=12y^3d/dxy-12x^2d/dxx+8$

$\frac{d y}{d x} = 12 y^{3} \frac{d y}{d x} - 12 x^{2} + 8$ $dy/dx=12y^3dy/dx-12x^2+8$

$\frac{d y}{d x} - 12 y^{3} \frac{d y}{d x} = - 12 x^{2} + 8$ $dy/dx-12y^3dy/dx=-12x^2+8$

$\frac{d y}{d x} (1 - 12 y^{3}) = - 12 x^{2} + 8$ $dy/dx(1-12y^3)=-12x^2+8$

Therefore

$\frac{d y}{d x} = \frac{- 12 x^{2} + 8}{1 - 12 y^{3}}$ $dy/dx=(-12x^2+8)/(1-12y^3$

How do you differentiate y=3y4−4u+5;u=x3−2x−5 y= 3y^4-4u+5 ;u=x^3-2x-5 using the chain rule?