How do I transform $f(x)=\log(1+e^x)$ such that graph rotates $90^{\circ}$ on the $x$-$y$ axis I am looking for a function $f(x)$ that is of a specific shape on the $x$-$y$ axis.
I have a function $f(x)=\log(1+e^x)$ that has right shape. I want it rotated $90^\circ$ on $x$-$y$ axis. 
How can I get an $f(x)$ that is essentially a $90^\circ$ rotation of $f(x)=\log(1+e^x)$?
 A: The rotation $R:\mathbb{R}^2 \to \mathbb{R}^2$ with angle $\pi/2$ is defined by
$$
R(x,y)=(-y,x) \quad \forall (x,y) \in \mathbb{R}^2.
$$
Your problem is to find a function $g: \text{dom}(g)\subset \mathbb{R} \to \mathbb{R}$ whose graph $G(g)$ is the image through $R$ of the graph $G(f)$ of $f$, i.e.
$$
G(g)=\{(s,g(s)): s \in \text{dom}(g)\ \} =\{(-f(x),x):\ x \in \mathbb{R}\ \}.
$$
Since
$$
f'(x)=\frac{e^x}{1+e^x}>0 \quad \forall x \in \mathbb{R},
$$
the function $f: \mathbb{R} \to (0,\infty)$ is strictly increasing and therefore invertible, and for every $s \in (-\infty,0)$ we have
$$
s=-f(x) \iff x=f^{-1}(-s)=\ln(e^{-s}-1).
$$ 
It follows that $g$ is the invertible function defined by
$$
g: (-\infty,0) \to \mathbb{R},\ g(x)=\ln(e^{-x}-1).
$$
A: You have $y=f(x)$
You want to rotate by $90$ degrees, which means you want the $x$ to become the $y$ and the $y$ to become $-x$.
Take $f^{-1}$ the function so that $\forall x\in \Bbb R,f^{-1}(f(x))=f(f^{-1}(x))=x$
If you trace $f^{-1}$ you have swapped $x$ and $y$, that is you've a reflection with respect to the line $y=x$. So now you just need to replace $x$ by $-x$. And since $x$ is now on the vertical axis, you need to change the sign of the "input" of the function. So you need to trace $x\mapsto f^{-1}(-x)$
A: This approach is for an arbitrary angle change, and then applied to a $90^\circ$ rotation.
We use the formula for the rotation matrix:
$$\pmatrix{x'\\y'} = \pmatrix{\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta}\pmatrix{x\\y}$$
The vector on the LHS is the new, rotated point; the vector on the RHS is the input point.
For your equation, we have $x=x$, $y=\ln(1+e^x)$.  So, we put that in the RHS vector:
$$\pmatrix{x'\\y'} = \pmatrix{\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta}\pmatrix{x\\\ln(1+e^x)}$$
Performing matrix multiplication:
$$\pmatrix{x'\\y'} = \pmatrix{x\cos\theta -\ln(1+e^x)\sin\theta \\ x\sin\theta + \ln(1+e^x)\cos\theta}$$
Now, we insert the desired value for $\theta$, and simplify:
$$\begin{align}
\pmatrix{x'\\y'} &= \pmatrix{x\cos(90^\circ) -\log(1+e^x)\sin(90^\circ) \\ x\sin(90^\circ) + \ln(1+e^x)\cos(90^\circ)}\\
&=\pmatrix{x\cdot0  -\log(1+e^x)\cdot1 \\ x\cdot1 + \ln(1+e^x)\cdot0}\\
&=\pmatrix{-\ln(1+e^x) \\ x}
\end{align}$$
This gives us a set of simultaneous equations to solve:
$$\begin{cases}
x' = -\ln(1+e^x)\\
y' = x
\end{cases}$$
We want to solve for $y'$ in terms of $x'$:
$$x' = -\ln(1+e^{y'})$$
$$-x' = \ln(1+e^{y'})$$
$$e^{-x'} =1+e^{y'}$$
$$e^{y'} = e^{-x'} -1$$
$$y' = \ln(e^{-x'} -1)$$
This process gets the same answer as just switching $y$ and $x$ in the original equation, but it lets us perform other rotations (for example, a $45^\circ$ rotation) instead of just a $90^\circ$ rotation.
