# Thread: Implicit natural log differentiation

1. ## Implicit natural log differentiation

How would i go about the steps of solving for dy/dx implicitly

$\displaystyle \ln x^4y^4 = x^3 + y^3$

2. Originally Posted by x5pyd3rx
How would i go about the steps of solving for dy/dx implicitly

$\displaystyle \ln x^4y^4 = x^3 + y^3$

$\displaystyle \frac{4x^3}{x^4}y^4 + \ln(x^4)(4y^3)\frac{dy}{dx}=3x^2+3y^2\frac{dy}{dx}$

3. Originally Posted by vince
$\displaystyle \frac{4x^3}{x^4}y^4 + \ln(x^4)(4y^3)\frac{dy}{dx}=3x^2+3y^2\frac{dy}{dx}$
He has given you the solution, but not much of an explanation. Let's look just at $\displaystyle \ln(x^4)(y^4)$

This is a product rule so we need to use $\displaystyle f'g+fg'$
$\displaystyle f=\ln(x^4)$
$\displaystyle f'=\frac{1}{x^4}*{4x^3}$ This is a chain rule using the fact that the derivative of $\displaystyle \ln(x) = \frac{1}{x}$

$\displaystyle g=y^4$
For $\displaystyle g'$ think about $\displaystyle y$ as a function of $\displaystyle x$. In other words, if we take a hypothetical $\displaystyle y=(x^2+1)^4$, the derivative would be $\displaystyle 4(x^2+1)^3*2x$. This is precisely $\displaystyle g'=4y^3*\frac{dy}{dx}$