Partial derivatives of $\ln(x+y)$

I am having trouble with because in my guide the answer is different.

I want to calculate the total derivative of the function: $f(x,y)=\ln(x+y)$

By definition: The Total derivative/Chain rule for functions of functions.

If $\omega=f(x,y)$ a continuous function. Then the total derivative is:

$$\frac{\partial \omega }{\partial t} = \frac{\partial \omega }{\partial x}\frac{d x }{d t} + \frac{\partial \omega }{\partial y}\frac{d y }{d t} $$

For our case $t=x+y$ and $$\omega= f(x,y)=\ln(x+y)$$ I got:

$$\frac{1}{x+y}$$

But in the book, the correct answer is:

$$\frac{x+y}{x+y}$$

What am I doing wrong?

1 Answer

The book definitely got that wrong. Since $\omega=\ln t$, $\partial_t\omega=\frac{1}{t}$. The chain rule agrees:$$\partial_x\omega=\partial_y\omega=\partial_t\omega=\frac{1}{t}\implies\partial_t\omega=\frac{1}{t}\frac{d}{dt}(x+y)=\frac{1}{t}.$$