# Thread: implicit differentiation of log of polynomial

1. ## implicit differentiation of log of polynomial

hello.
find derivative of:
y = ln(5x² + y²)
Using method described at:
Solutions to Implicit Differentiation Problems
I get:
y' = D{ln(5x² + y²)}
y' = 1/(5x² + y²) D{(5x² + y²)}
y' = 1/(5x² + y²) (10x + 2y y')
y' = 10x/(5x² + y²) + 2y y'/(5x² + y²)
y' - 2y y'/(5x² + y²) = 10x/(5x² + y²)
y'(5x² + y²)/(5x² + y²) - 2y y'/(5x² + y²) = 10x/(5x² + y²)
(y'(5x² + y²) - 2y y')/(5x² + y²) = 10x/(5x² + y²)
y'(5x² + y² - 2y)/(5x² + y²) = 10x/(5x² + y²)
y' = 10x/(5x² + y²) (5x² + y²)/(5x² + y² - 2y)
y' = 10x/(5x² + y² - 2y)

I like to use online tools to check my work. There are several that purport to solve this type of equation. They all seem to agree that the answer is:
y' = 10x/(5x² + y²)
Of course, they don't show steps, so I can't tell if I'm just wrong or if the online tools are wrong (or if I'm using them wrong).

Where's the error?

2. Originally Posted by Jim Pfoss
hello.
find derivative of:
y = ln(5x² + y²)
Using method described at:
Solutions to Implicit Differentiation Problems
I get:
y' = D{ln(5x² + y²)}
y' = 1/(5x² + y²) D{(5x² + y²)}
y' = 1/(5x² + y²) (10x + 2y y')
y' = 10x/(5x² + y²) + 2y y'/(5x² + y²)
y' - 2y y'/(5x² + y²) = 10x/(5x² + y²)
y'(5x² + y²)/(5x² + y²) - 2y y'/(5x² + y²) = 10x/(5x² + y²)
(y'(5x² + y²) - 2y y')/(5x² + y²) = 10x/(5x² + y²)
y'(5x² + y² - 2y)/(5x² + y²) = 10x/(5x² + y²)
y' = 10x/(5x² + y²) (5x² + y²)/(5x² + y² - 2y)
y' = 10x/(5x² + y² - 2y)

I like to use online tools to check my work. There are several that purport to solve this type of equation. They all seem to agree that the answer is:
y' = 10x/(5x² + y²)
Of course, they don't show steps, so I can't tell if I'm just wrong or if the online tools are wrong (or if I'm using them wrong).

Where's the error?

Those online tools suck big time: you are right...and next time trust more your brains and education than lame online "mathematical" tools.

Tonio

3. ## thanks

Thanks Tonio,

I may be crazy, but at least I'm not stupid!