# Question #66265

May 20, 2015

$4 {x}^{3} + \ln {y}^{4} - 4 {y}^{2} = 10$

$\frac{d}{\mathrm{dx}} \left(4 {x}^{3} + \ln {y}^{4} - 4 {y}^{2}\right) = \frac{d}{\mathrm{dx}} \left(10\right)$

$12 {x}^{2} + \frac{1}{y} ^ 4 \cdot 4 {y}^{3} \frac{\mathrm{dy}}{\mathrm{dx}} - 8 y \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$12 {x}^{2} + \frac{4}{y} \frac{\mathrm{dy}}{\mathrm{dx}} - 8 y \frac{\mathrm{dy}}{\mathrm{dx}} = 0$ Now, clear the fraction (mult by $y$)

$12 {x}^{2} y + 4 \frac{\mathrm{dy}}{\mathrm{dx}} - 8 {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{12 {x}^{2} y}{8 {y}^{2} - 4} = \frac{3 {x}^{2} y}{2 {y}^{2} - 1}$

May 20, 2015

I assume lny^4=$\ln \left({y}^{4}\right)$, otherwise, you should have written ${\ln}^{4} \left(y\right)$, but please, re-write the question LaTeX-ly

First of all, we need to write the gradient of $F \left(x\right) = 4 {x}^{3} + 4 \ln \left(\left\mid y \right\mid\right) - 4 {y}^{2} - 10$, which is well defined $\forall \left(x , y\right) : y \ne 0$

It is $\left(12 {x}^{2} , \frac{4}{y} - 8 y\right)$, and it's always nonzero (trivial), so, for Dini's theorem, we can write $\frac{\mathrm{dy}}{\mathrm{dx}}$ whenever ${F}_{y} \ne 0$, so whenever $y \ne \pm \frac{\sqrt{2}}{2}$ and it is defined as $\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{{F}_{x}}{{F}_{y}} = - \left(\frac{12 {x}^{2} y}{4 - 8 {y}^{2}}\right)$.

Notice that ${F}_{x} \left(0\right) = 0$ and $F \left(x , -\right)$ is even $\forall x$, so you have ${y}_{1} = y \left(x\right)$ and ${y}_{2} = - y \left(x\right)$ for the implicit function theorem.

graph{4x^3 - ln(y^4) -4y^2=10 [-1.709, 1.709, -0.853, 0.856]}