Implicit Differentiation - Logarithm

Question

$x\log(x) + y\log(y) = 1$

$\dfrac{dy}{dx}= ?$

I calculated $\frac{dy}{dx}= -\frac{1+\log(x)}{1+\log(y)}$

however, the correct answer seems to be $-\log(x)/\log(y)$

I'm confused, can someone help?

Answer 1

I got $$\log(x)+1+y'\log(y)+y'=0$$

Answer 2

Assuming that $\log(x)=\log_e(x)=\ln(x)$, we start off with our equation:

$$x\log(x)+y\log(y)=1$$

Next we perform implicit differentiation to both sides:

$$ \frac{dy}{dx}(x\log(x)+y\log(y))=\frac{dy}{dx}(1)\\ 1+\log(x)+\frac{dy}{dx}1+\frac{dy}{dx}\log(y)=0 $$

Now we isolate $\frac{dy}{dx}$ to one side:

$$ 1+\log(x)+\frac{dy}{dx}1+\frac{dy}{dx}\log(y)=0\\ \frac{dy}{dx}1+\frac{dy}{dx}\log(y)=-\log(x)-1\\ \frac{dy}{dx}(1+\log(y))=-\log(x)-1\\ \frac{dy}{dx}=\frac{-\log(x)-1}{1+\log(y)}\\ $$

Finally we clean things up and simplify:

$$\begin{align} \frac{dy}{dx}&=\frac{-\log(x)-1}{1+\log(y)}\\ &\bbox[5px,border:2px solid black] {=-\frac{1+\log(x)}{1+\log(y)} } \end{align}$$

We see that your solution is indeed correct.