# Analysis Problem (finding a derivative)

1. Oct 26, 2004

### AKG

Suppose that h : $\mathbb{R} \rightarrow \mathbb{R}$ is continuous.
Calculate f', if f : $\mathbb{R}^2 \rightarrow \mathbb{R}$ is the function:

$$f(x, y) = \int _{\sin (xy)} ^{\cos (xy)} h(t)dt$$

I have:

$$f' = \left [D_1f(x, y) \ \ \ \ D_2f(x, y)\right ] = \left [\frac{\partial f}{\partial x} \ \ \ \ \frac{\partial f}{\partial y}\right ]$$

$$= \left [\frac{\partial}{\partial x}\int_{\sin (xy)} ^{\cos (xy)} h(t)dt \ \ \ \ \frac{\partial}{\partial y}\int_{\sin (xy)} ^{\cos (xy)} h(t)dt\right ]$$

Let H be the antiderivative of h (can I assume it exists from the continuity of h? I would think so), then:

$$f' = \left [\frac{\partial}{\partial x}H(\cos xy) - H(\sin xy) \ \ \ \ \frac{\partial}{\partial y}H(\cos xy) - H(\sin xy)\right ]$$

$$f' = -\left [y(\sin (xy)h(\cos (xy)) + \cos (xy)h(\sin (xy))) \ \ \ \ x(\sin (xy)h(\cos (xy)) + \cos (xy)h(\sin (xy)))\right ]$$

Is this right? Thanks.

Last edited: Oct 26, 2004
2. Oct 26, 2004

### HallsofIvy

Staff Emeritus
Yes.
(and other stuff to make more than 10 characters)

3. Oct 26, 2004

Thanks......