Question

X^y=y^x find derivitive

Answer 1

Answer:

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

Step-by-step explanation:

Take natural log of both sides first.

$x^y=y^x$

$\ln(x^y)=\ln(y^x)$

Taking the natural log of both sides allows you to bring down the powers.

$y\ln(x)=x\ln(y)$

I'm going to differentiate both sides using the power rule.

$(y)'(\ln(x))+(\ln(x))'y=(x)'(\ln(y))+(\ln(y))'x$

Now recall (ln(x))'=(x)'/x=1/x while (ln(y))'=(y)'/y=y'/y.

$y'(\ln(x))+\frac{1}{x}y=1(\ln(y))+\frac{y'}{y}x$

Simplifying a bit:

$y' \ln(x)+\frac{y}{x}=\ln(y)+\frac{y'}{y}x$

Now going to gather my terms with y' on one side while gathering other terms without y' on the opposing side.

Subtracting y'ln(x) and ln(y) on both sides gives:

$\frac{y}{x}-\ln(y)=-y'\ln(x)+\frac{y'}{y}x$

Now I'm going to factor out the y' on the right hand side:

$\frac{y}{x}-\ln(y)=(-\ln(x)+\frac{x}{y})y'$

Now we get to get y' by itself by dividing both sides by (-ln(x)+x/y):

$\frac{\frac{y}{x}-\ln(y)}{-\ln(x)+\frac{x}{y}}=y'$

Now this looks nasty to write mini-fractions inside a bigger fraction.

So we are going to multiply top and bottom by xy giving us:

$\frac{y^2-yx\ln(y)}{-xy\ln(x)+x^2}=y'$

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