# Applying the chain rule: differentiating $f(x+s)$

This might sound trivial, but I'm confused about using the chain rule on univariate real-valued functions. On Wikipedia, I see that

$${\frac {dz}{dx}}={\frac {dz}{dy}}\cdot {\frac {dy}{dx}}$$

My question: suppose I have a real-valued function $$f(x+s)$$ and I would like to find $$\frac{\partial f}{\partial s}$$. Is it

$$\frac{\partial f(x+s)}{\partial s} = \frac{\partial f(x+s)}{\partial (x+s)} \frac{\partial (x+s)}{\partial (s)} = \frac{\partial f(x+s)}{\partial (x+s)}$$

or

$$\frac{\partial f(x+s)}{\partial s} = \frac{\partial f(x+s)}{\partial x}$$

I think it's the former, but I have seen sources where the latter is used. Are they both equivalent? If so I can't see the reason.

You should write, since $$f$$ is a function of a single variable $$\frac{\partial f(x+s)}{\partial s}=\frac{\mathrm df}{\mathrm d x}(x+s)\cdot\frac{\partial (x+s)}{\partial s}=\frac{\mathrm df}{\mathrm d x}(x+s).$$ Note the two positions of $$(x+s)$$: on the left side, $$f(x+s)$$ implicitly defines a function of two variables $$g(x,s)$$, whereas in the first factor of the right-hand side, $$\frac{\mathrm df}{\mathrm d x}(x+s)$$ denotes the derivative of the single variable function $$f$$, evaluated at the point $$x+s$$.
Here, you have $$z(y)=f(y)$$ and $$y(x,s)=x+s$$.
Hence $$\frac{dz}{dy}= \frac{df}{dy}$$ and $$\frac{\partial y}{\partial s}= 1$$.
Which leads to $$\frac{\partial f(x+s)}{\partial s}= \frac{df}{dy}(x+s)= f^\prime(x+s).$$