6.[判断题]判断题假定f(x,y)=(x+sin y)(y-cos x)且h(y)=f((pi)/(3),y),则(partial f)/(partial y)(pi/3,-pi/2)=(dh)/(dy)(-pi/2)1.对2.错

为了判断给定的陈述是否正确，我们需要计算$\frac{\partial f}{\partial y}(\pi/3, -\pi/2)$和$\frac{dh}{dy}(-\pi/2)$，然后比较它们的值。 首先，我们从函数$f(x, y) = (x + \sin y)(y - \cos x)$开始。我们需要找到偏导数$\frac{\partial f}{\partial y}$。 使用乘积法则，我们得到： \[ \frac{\partial f}{\partial y} = \frac{\partial}{\partial y}[(x + \sin y)(y - \cos x)] = (x + \sin y) \frac{\partial}{\partial y}[y - \cos x] + (y - \cos x) \frac{\partial}{\partial y}[x + \sin y] \] 由于$\frac{\partial}{\partial y}[y - \cos x] = 1$和$\frac{\partial}{\partial y}[x + \sin y] = \cos y$，我们有： \[ \frac{\partial f}{\partial y} = (x + \sin y) \cdot 1 + (y - \cos x) \cdot \cos y = x + \sin y + (y - \cos x) \cos y \] 现在，我们需要在$(\pi/3, -\pi/2)$处评估这个偏导数： \[ \frac{\partial f}{\partial y}\left(\frac{\pi}{3}, -\frac{\pi}{2}\right) = \frac{\pi}{3} + \sin\left(-\frac{\pi}{2}\right) + \left(-\frac{\pi}{2} - \cos\left(\frac{\pi}{3}\right)\right) \cos\left(-\frac{\pi}{2}\right) \] 由于$\sin\left(-\frac{\pi}{2}\right) = -1$和$\cos\left(-\frac{\pi}{2}\right) = 0$，我们得到： \[ \frac{\partial f}{\partial y}\left(\frac{\pi}{3}, -\frac{\pi}{2}\right) = \frac{\pi}{3} - 1 + \left(-\frac{\pi}{2} - \frac{1}{2}\right) \cdot 0 = \frac{\pi}{3} - 1 \] 接下来，我们考虑函数$h(y) = f\left(\frac{\pi}{3}, y\right)$。将$x = \frac{\pi}{3}$代入函数$f$，我们得到： \[ h(y) = \left(\frac{\pi}{3} + \sin y\right)\left(y - \cos\left(\frac{\pi}{3}\right)\right) = \left(\frac{\pi}{3} + \sin y\right)\left(y - \frac{1}{2}\right) \] 我们需要找到导数$\frac{dh}{dy}$： \[ \frac{dh}{dy} = \frac{d}{dy}\left[\left(\frac{\pi}{3} + \sin y\right)\left(y - \frac{1}{2}\right)\right] = \left(\frac{\pi}{3} + \sin y\right) \frac{d}{dy}\left[y - \frac{1}{2}\right] + \left(y - \frac{1}{2}\right) \frac{d}{dy}\left[\frac{\pi}{3} + \sin y\right] \] 由于$\frac{d}{dy}\left[y - \frac{1}{2}\right] = 1$和$\frac{d}{dy}\left[\frac{\pi}{3} + \sin y\right] = \cos y$，我们有： \[ \frac{dh}{dy} = \left(\frac{\pi}{3} + \sin y\right) \cdot 1 + \left(y - \frac{1}{2}\right) \cdot \cos y = \frac{\pi}{3} + \sin y + \left(y - \frac{1}{2}\right) \cos y \] 现在，我们需要在$y = -\frac{\pi}{2}$处评估这个导数： \[ \frac{dh}{dy}\left(-\frac{\pi}{2}\right) = \frac{\pi}{3} + \sin\left(-\frac{\pi}{2}\right) + \left(-\frac{\pi}{2} - \frac{1}{2}\right) \cos\left(-\frac{\pi}{2}\right) \] 由于$\sin\left(-\frac{\pi}{2}\right) = -1$和$\cos\left(-\frac{\pi}{2}\right) = 0$，我们得到： \[ \frac{dh}{dy}\left(-\frac{\pi}{2}\right) = \frac{\pi}{3} - 1 + \left(-\frac{\pi}{2} - \frac{1}{2}\right) \cdot 0 = \frac{\pi}{3} - 1 \] 由于$\frac{\partial f}{\partial y}\left(\frac{\pi}{3}, -\frac{\pi}{2}\right) = \frac{dh}{dy}\left(-\frac{\pi}{2}\right)$，给定的陈述是正确的。 答案是$\boxed{1}$。