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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.
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.
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