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

Oct 9, 2017

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

#### Explanation:

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

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

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

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

$\frac{d}{\mathrm{dx}} y = \frac{d}{\mathrm{dx}} \left(3 {y}^{4} - 4 {x}^{3} + 8 x + 25\right)$

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

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

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

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

Therefore

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