91Ó°ÊÓ

We start by finding the partial derivatives of the given function with respect to x and y. $$\frac{\partial h}{\partial x}=-e^{-x-y}$$ $$\frac{\partial h}{\partial y}=-e^{-x-y}$$ Now we need to find the gradient at the point \((\ln 2, \ln 3)\). We substitute these values into the partial derivatives. $$\nabla h|_{(\ln(2), \ln(3))} = \left<-\frac{1}{2}\cdot\frac{1}{3},-\frac{1}{2}\cdot\frac{1}{3} \right> = \left<-\frac{1}{6}, -\frac{1}{6}\right>$$

The given direction vector is \(\langle 1,1\rangle\). We need to find the unit vector in this direction, which is found by dividing the direction vector by its magnitude. The magnitude of the given vector is: $$\|\langle 1,1\rangle\|= \sqrt{1^2+1^2} = \sqrt{2}$$ Hence, the unit vector is: $$\frac{\langle 1,1\rangle}{\sqrt{2}}=\left<\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right>$$

Now we have the gradient at point \(P\) and the unit vector in the direction of the given vector. We will find the directional derivative by taking the dot product of the gradient and the unit vector. $$D_{\left<\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right>}h(x,y) = \nabla h|_{(\ln(2), \ln(3))} \cdot \left<\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right>$$ $$D_{\left<\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right>}h(x,y) = \left<-\frac{1}{6}, -\frac{1}{6}\right> \cdot \left<\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right>$$ $$D_{\left<\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right>}h(x,y) = \frac{1}{6\sqrt{2}}+\frac{1}{6\sqrt{2}} = \frac{1}{3\sqrt{2}}$$ The directional derivative of the function \(h(x, y)=e^{-x-y}\) at point \(P(\ln 2, \ln 3)\) in the direction of the vector \(\langle 1,1\rangle\) is \(\frac{1}{3\sqrt{2}}\).