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Let $f(x,y) = \int^y_x\frac{ds}{s + \sin(s)}$. What is the derivative of this integral?

I know that the fundamental theorem of calculus says: if $F(x) = \int^x_a\frac{ds}{s + \sin(s)}$ then $F'(x) = \frac{1}{x + \sin(x)}$, but in this case we have $2$ variables so I'm kind of confused.

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    $\begingroup$ What do you mean by "derivative"? In multivariable land, you could mean the total derivative, or partial derivatives. Or there are vector derivatives. $\endgroup$ – Adrian Keister Aug 7 '19 at 2:27
  • $\begingroup$ The total derivative $\endgroup$ – Rainroad Aug 7 '19 at 13:36
  • $\begingroup$ Well, the only independent variables in sight are $x$ and $y$. So, do you mean the total derivative with respect to $x?$ That would be $\dfrac{df}{dx}=\dfrac{\partial f}{\partial x}+\dfrac{\partial f}{\partial y}\,\dfrac{\partial y}{\partial x}.$ Or if you want the total derivative with respect to $y,$ you'd have $\dfrac{df}{dy}=\dfrac{\partial f}{\partial y}+\dfrac{\partial f}{\partial x}\,\dfrac{\partial x}{\partial y}.$ Which one are you after? $\endgroup$ – Adrian Keister Aug 7 '19 at 13:47
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If you take the partial derivative WRT $x$, you can treat $y$ as being constant, and likewise if you take the partial derivative WRT $y$, you can treat $x$ as being constant. Thus, as you stated, you can use the Fundamental Theorem of Calculus to get

$$\frac{\partial f(x,y)}{\partial x} = \frac{-1}{x + \sin(x)} \tag{1}\label{eq1}$$

$$\frac{\partial f(x,y)}{\partial y} = \frac{1}{y + \sin(y)} \tag{2}\label{eq2}$$

The first one is negative due to the $x$ being the lower limit of the integral, so $\int_{x}^{y} \frac{ds}{s + \sin(s)} = -\int_{y}^{x} \frac{ds}{s + \sin(s)}$.

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Your answer would depend on which variable you're taking the derivative with respect to. Here, you can rewrite $f(x,y) = \int^{y}_{0} \frac{1}{s + \sin s} ds - \int^{x}_{0} \frac{1}{s + \sin s} ds$, and apply the Fundamental Theorem of Calculus as you mentioned to take the partial derivative with respect to either $x$ or $y$.

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