I'm updating this question with inline LaTeX versions of the images included to make it more self-contained.
What mistake am I making when applying the following theorem to the following question. (I also include an example for illustration purposes)?
Theorem Therorem 7.1: If $f(x,y)$ is continuously differentiable in domain D and if C is a differentiable curve in D defined by $x(t), y(t), a \le t \le b$, then $f_c(t)$ is differentiable and we have $f'_c(t_0)=f_x(x_0,y_0)\;x'(t_0)+f_y(x_0,y_0)\;y'(t_0)$ where $x_0=x(t_0), y_0=y(t_0)$
Example Example: Let $f(x,y)=x^2+2xy$ and $x(t)=cos(t)$ and $y(t)=sin(t)$
Then $f_x=2x+2y$ and $f_y=2x$
Further, $x'(t)=-sin(t)$ and $y'(t)=cos(t)$
Thus, if $t_0=\frac{1}{4}\pi$ we have $x_0=\frac{\sqrt{2}}{2}$ and $y_0=\frac{\sqrt{2}}{2}$
and so $f_x(x_0,y_0)=2\sqrt{2}$ and $f_y(x_0,y_0)=\sqrt{2}$
Hence $f'_c(\frac{1}{4}\pi)=2\sqrt{2}*\frac{-\sqrt{2}}{2}+\sqrt{2}*\frac{\sqrt{2}}{2}=-1$
In this case we could also calculate $f'_c(t)$ directly, obtaining:
$f'_c(t)=-2cos(t)sin(t)+2cos^2(t)-2sin^2(t)$
setting $t =\frac{1}{4}\pi$ we find that $f'_c(\frac{1}{4}\pi)=-1$
I need to do the same but for the following function:
Question $f(x,y)=log(1+x^2+y^2) - 2\arctan{y}$ where $x=log(1+t^2)$ and $y=e^t$
The question is to find the point of the curve corresponding to $t=0$ and compute $f'_c(0)$ by the chain rule. Then check the answer by differentiating $f_c(t)$ and setting $t=0$
The answer given is $0$. So, I do the check first and I do indeed get this answer when I differentiate $f_c(t)$ and set $t=0$
So now I hope to get this answer when applying the chain rule:
So I have:
$f_x = \frac{2x}{log(1+x^2+y^2)}$ and $f_y=\frac{2y}{log(1+x^2+y^2)}-\frac{2}{y^2+1}$ and $x'(t)=\frac{2t}{1+t^2}$ and $y'(t)=e^t$
When $t=t_0=0$ we have $x_0=0, y_0=1, x'(t_0)=0$ and $ y'(t_0)=1$
and also $f_x(x_0,y_0)=0$ and $f_y(x_0,y_0)=\frac{2}{log(2)}-1$
So applying the chain rule formula gives:
$f'_c(t_0)=0*0+(\frac{2}{log(2)}-1)*1$
which is not $0$.
Can anyone see where I have made my mistake?