# What is the derivative of this integral?

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.

• What do you mean by "derivative"? In multivariable land, you could mean the total derivative, or partial derivatives. Or there are vector derivatives. – Adrian Keister Aug 7 '19 at 2:27
• The total derivative – Rainroad Aug 7 '19 at 13:36
• 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? – Adrian Keister Aug 7 '19 at 13:47

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 $$\int_{x}^{y} \frac{ds}{s + \sin(s)} = -\int_{y}^{x} \frac{ds}{s + \sin(s)}$$.
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 $$y$$.