Find differential. Is it the same as finding dy?

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These questions told me to find the differential at a number when given a specific dx. Wanted to check if my work is right:

1. $$x = 0, dx = 0.1$$

$$y = e^{\frac{x}{10}}$$ $$y' = e^{\frac{x}{10}} \cdot \frac{1}{10} $$

so

$$dy = e^{\frac{x}{10}} \cdot \frac{1}{10} \cdot 0.1 = \frac{1}{100}$$

Is that right?

2.

$$x = 2, dx = 0.05$$ $$y = \frac{x+1}{x-1}$$ $$ y' = \frac{(x-1) - (x+1)}{(x-1)(x-1)}$$ $$ = \frac{-2}{(x-1)^2}$$

so

$$dy = \frac{-2}{1} \cdot 0.05 = -0.1$$

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Yes. The definition of a differential, at least according to Stewart's calculus text, is

$$dy= f'(x)dx,$$ where $dx$ and $x$ are independent variables. What this is providing is a best linear approximation you can get, namely $$f(a+dx)\approx f(a)+dy.$$