Question

how to find the derivative of x + y2 = ln(x/y) ? please show all workings and simplify!


the answer is supposed to be y^2-x/xy

Answer 1

Answer:

Step-by-step explanation:

If your function is

$x+y^2=ln(\frac{x}{y} )$, that is definitely not the answer you should get after taking the derivative implicitely.  Rewrite your function to simplify a bit:

$x+y^2=ln(x)-ln(y)$

Take the derivative of x terms like "normal", but taking the derivative of y with respect to x has to be offset by dy/dx.  Doing that gives you:

$1+2y\frac{dy}{dx}=\frac{1}{x}-\frac{1}{y}\frac{dy}{dx}$

Collect the terms with dy/dx on one side and everything else on the other side:

$2y\frac{dy}{dx}+\frac{1}{y}\frac{dy}{dx}=\frac{1}{x}-1$

Now factor out the common dy/dx term, leaving this:

$\frac{dy}{dx}(2y+\frac{1}{y})=\frac{1}{x}-1$Now divide on the left to get dy/dx alone:

$\frac{dy}{dx}=\frac{\frac{1}{x}-1 }{2y+\frac{1}{y} }$

Simplify each set of fractions to get:

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

Bring the lower fraction up next to the top one and flip it upside down to multiply:

$\frac{dy}{dx}=\frac{1-x}{x}$×$\frac{y}{2y^2+1}$

Simplifying that gives you the final result:

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

or you could multiply in the x on the bottom, as well.  Same difference as far as the solution goes.  You'd use this formula to find the slope of a function at a point by subbing in both the x and the y coordinates so it doesn't matter if you do the distribution at the very end or not.  You'll still get the same value for the slope.