The partial derivatives of the function $s=\int _u^v\frac{(1-e^t)}{t}dt$

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If $s=\int _u^v\frac{\left(1-e^t\right)}{t}dt$, I want to find $\frac{∂s}{∂v}$ and $\frac{∂s}{∂u}$ and their limits as $u$ and $v$ tend to zero.

first i find for

$\frac{∂s}{∂v}=\frac{∂}{∂v}\int _u^v\frac{1-e^t}{t}dt=\frac{∂}{∂v}\int _u^c\frac{1-e^t}{t}dt+\frac{∂}{∂v}\int _c^v\frac{1-e^t}{t}dt$

i get

$\frac{∂s}{∂v}=\frac{1-e^v}{v}$

and then i find for

$\frac{∂s}{∂u}=\frac{∂}{∂u}\int _u^v\frac{1-e^t}{t}dt=\frac{∂}{∂u}\int _u^c\frac{1-e^t}{t}dt+\frac{∂}{∂u}\int _c^v\frac{1-e^t}{t}dt$

i get

$\frac{∂s}{∂u}=\frac{e^u-1}{u}$

and the limits are $\lim _{v\to 0}\frac{1-e^v}{v}=-1$ and $\lim _{u\to 0}\frac{e^u-1}{u}=1$

this is my answer but i don't know if it's correct or is there an error in my calculation?