Find x'_y, the derivative of x with respect to y, given y
Find $x'_{y}$, the derivative of $x$ with respect to $y$, given $y = e^{x}+x$.
Explanation of Method
The derivative of an inverse that has no closed form.
Step 1: Differentiate the given relation
\[ y = e^{x}+x \;\Longrightarrow\; y'_{x} = e^{x}+1. \]
Step 2: Apply the inverse function rule
Where $y'_{x} \neq 0$, the inverse function is differentiable with
\[ x'_{y} = \frac{1}{y'_{x}} = \frac{1}{e^{x}+1}. \]
Step 3: Why the inverse exists
Since $e^{x} \gt 0$, we have $y'_{x} = e^{x}+1 \gt 1 \gt 0$ for every real $x$: the function is strictly increasing, hence injective, and it maps $\mathbb{R}$ onto $\mathbb{R}$. So a differentiable inverse $x = x(y)$ exists everywhere, even though it cannot be written with elementary functions.
Step 4: Verification
\[x = 0:\quad y = 1,\quad x'_{y} = \frac{1}{1+1} = \tfrac12\]
\[\text{Check: } y(0.01) = e^{0.01}+0.01 \approx 1.020050 \;\Longrightarrow\; \frac{0.01-0}{1.020050-1} \approx 0.4988 \approx \tfrac12 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,x'_{y} = \dfrac{1}{e^{x}+1}\,}\]
Remark
Remark. The answer is expressed in terms of $x$ rather than $y$, which is the honest thing to do here: solving $y = e^{x}+x$ for $x$ requires the Lambert $W$ function. The identity $x'_{y}\,y'_{x} = 1$ is exactly the chain rule applied to $x(y(x)) = x$. Note that $0 \lt x'_{y} \lt 1$ always, so the inverse is a contraction — the graph of $y$ is steeper than $45^{\circ}$ everywhere.