Suppose that h : [itex]\mathbb{R} \rightarrow \mathbb{R}[/itex] is continuous.(adsbygoogle = window.adsbygoogle || []).push({});

Calculate f', if f : [itex]\mathbb{R}^2 \rightarrow \mathbb{R}[/itex] is the function:

[tex]f(x, y) = \int _{\sin (xy)} ^{\cos (xy)} h(t)dt[/tex]

I have:

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

[tex] = \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 ][/tex]

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

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

[tex]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 ][/tex]

Is this right? Thanks.

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Analysis Problem (finding a derivative)

**Physics Forums | Science Articles, Homework Help, Discussion**