I'm not really BAD at math, even though I'm not good. But What am I doing wrong here?
$U_x$ denotes partial derivative of $U$ with respect to $x$
Let $U(x,y)=F(f(x)+g(y))$ and $V(x,y)=lnU_x-lnU_y$
Find $V_x$
Answer sheet says that the answer is $\frac{f''(x)}{f'(x)}$ But How come? I couldn't end up with that answer, what is wrong with my approach?
$$V_x=\frac{U_{xx}}{U_x}-\frac{U_{yx}}{U_y}$$
$$U_{xx}=F_{xx}(f(x)+g(y))f'(x)+F_x(f(x)+g(y))f''(x)\\ U_{yx}=F_{yx}(f(x)+g(y))f'(x)g'(y)\\ V_x=\frac{F_{xx}(f(x)+g(y))}{F_x(f(x)+g(y))}+\frac{f''(x)}{f'(x)}-\frac{F_{yx}(f(x)+g(y))f'(x)}{F_y(f(x)+g(y))}$$
Isn't this the answer
You made a mistake
$U_x=f'(x)F'(f(x)+g(y))$
$U_{xx}=f''(x)F'(f(x)+g(y))+f'(x)^2F''(f(x)+g(y))$
$U_y=g'(y)F'(f(x)+g(y))$
$U_{yx}=g'(y)f'(x)F''(f(x)+g(y))$
Then conclude with your result $V_x=\frac{U_{xx}}{U_x}-\frac{U_{yx}}{U_y}$