What is the derivative with respect to x of . . . . xy'/y^2
I can get as far as derivative of xy' = y' + xy"
But I dont know what to do about the bit on the bottom . .
I'm guessing i need the quotient rule somehow . .
Cheers
What is the derivative with respect to x of . . . . xy'/y^2
I can get as far as derivative of xy' = y' + xy"
But I dont know what to do about the bit on the bottom . .
I'm guessing i need the quotient rule somehow . .
Cheers
You have to use the quotient rule: The derivative of $\displaystyle \frac{u}{v}$ is equal to $\displaystyle \frac{\frac{du}{dx} \, v - \frac{dv}{dx} \, u}{v^2}$.
In your problem:
$\displaystyle u = xy' \Rightarrow \frac{du}{dx} = y' + x \, y''$, as you correctly found (using the product rule).
$\displaystyle v = y^2$. To get $\displaystyle \frac{dv}{dx}$ you have to use the chain rule: $\displaystyle \frac{dv}{dx} = 2y\, y'$.