Finding the partial derivatives of an integral given by $f(x,y) = \int_{y}^{x} \cos (t^2) \ dt$

Question

I am asked to find the partial derivatives of an integral given by $f(x,y) = \int_{y}^{x} \cos (t^2) \ dt$.

What I did was

\begin{align} f(x,y) &= \int_{y}^{x} \cos (t^2) \ dt\\ \\ \frac{\partial f}{\partial x} &= \frac{\partial}{\partial x} \int_{y}^{x} \cos (t^2) \ dt = \cos (x^2)\\ \frac{\partial f}{\partial y} &= - \frac{\partial}{\partial y} \int_{x}^{y} \cos (t^2) \ dt = - \cos (y^2)\\ \end{align}

It seems very similar to what we do on single variable calculus (Fundamental Theorem of Calculus part 1), that is, substituting the variable and aplying the chain rule (which in this case will lead to a multiplication by $1$ in both cases). Can someone confirm my answer?

Answer 1

Yes, your answer is correct. It's as simple as that!

Answer 2

Yes, your calculations are correct. In general, note: $$\int g(t)dt =G(t)+C \Rightarrow G'(t)=g(t).$$ Hence: $$f(x,y)=\int_y^x g(t)dt =G(t)\bigg{|}_y^x=G(x)-G(y) \Rightarrow f'_x=\left[G(x)-G(y)\right]'_x=G'(x)=g(x).$$ See here for more: http://www.mathmistakes.info/facts/CalculusFacts/learn/doi/doi.html