Simple logarithmic differentiation

Question

Given the equation $\frac{dy}{dx} = \frac{-x}{y}$

How can you solve for $\frac{d^2y}{d^2x}$ using logarithmic differentiation?

Here is my work:

$ln(\frac{dy}{dx}) = ln(\frac{-x}{y})$

$ln(\frac{dy}{dx}) = ln(-x) - ln(y)$

(Take derivative of both sides)

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

(Then solve for $\frac{d^2y}{d^2x}$)

$\frac{d^2y}{d^2x} = \frac{\frac{dy}{dx}}{x} - \frac{\frac{dy}{dx}}{y}$

(Then plug in $\frac{-x}{y}$ for $\frac{dy}{dx}$)

$\frac{d^2y}{d^2x} = \frac{-x}{yx} + \frac{x}{y^2}$

And simplify to get:

$\frac{d^2y}{d^2x} = \frac{-1}{y} + \frac{x}{y^2}$

$ = \frac{(x-y)}{y^2}$

However, when I solve the derivative using the quotient rule, I get:

$\frac{x\frac{-x}{y} - y}{y^2}$

$= \frac{\frac{-x^2}{y}-y}{y^2}$

What am I doing wrong?

Answer 1

You made a mistake. In particular, differentiating both sides of $$\log(\frac{dy}{dx}) = \log(-x)-\log(y)$$ gives $$\frac{\frac{d^2y}{dx^2}}{\frac{dy}{dx}} = \frac{1}{x} - \frac{1}{y} \color{red}{\frac{dy}{dx}}.$$ Here $\frac{d}{dx} \log(y) = \frac{1}{y} \color{red}{\frac{dy}{dx}}$ comes from the chain rule.

Answer 2

That equation can be rewritten with another notation like y'y = -x

simply claculate the dervative of that expression

$$y"y+y'²= -1 $$

where y'= dy/dx and y" = d²y/dx²

Rewriting this last equation with your notation

$$d²y/dx²=-(x²+y²)/y³$$