an Given f(x,y,z)=x^2-y^2+2mx^2 and a point P(2,1,-1) then a. Find forall f(x,y,x) and er(2.1,-1) b. Find the directional derivative of f(x,y,z) at P(2,1,-1) in the direction of u=langle 1,-1,sqrt (2)rangle

a. To find $\forall f(x,y,z)$, we need to take the partial derivatives of $f(x,y,z)$ with respect to each variable.<br /><br />$\frac{\partial f}{\partial x} = 2x + 4mx$<br />$\frac{\partial f}{\partial y} = -2y$<br />$\frac{\partial f}{\partial z} = 0$<br /><br />Now, we can evaluate these partial derivatives at the point $P(2,1,-1)$:<br /><br />$\frac{\partial f}{\partial x}(2,1,-1) = 2(2) + 4m(2) = 4 + 8m$<br />$\frac{\partial f}{\partial y}(2,1,-1) = -2(1) = -2$<br />$\frac{\partial f}{\partial z}(2,1,-1) = 0$<br /><br />So, $\forall f(x,y,z)$ at $P(2,1,-1)$ is $\langle 4 + 8m, -2, 0 \rangle$.<br /><br />b. The directional derivative of $f(x,y,z)$ at $P(2,1,-1)$ in the direction of $u=\langle 1,-1,\sqrt {2}\rangle$ is given by the off$ at $P$ with the unit vector in the direction of $u$.<br /><br />First, we need to find the unit vector in the direction of $u$:<br /><br />$u = \langle 1,-1,\sqrt {2}\rangle$<br />$|u| = \sqrt{1^2 + (-1)^2 + (\sqrt{2})^2} = \sqrt{ = 2$<br />$u_{\text{unit}} = \frac{u}{|u|} = \frac{1}{2}\langle 1,-1,\sqrt {2}\rangle = \langle \frac{1}{2},-\frac{1}{2},\frac{\sqrt{2}}{2}\rangle$<br /><br />Now, we can find the directional derivative:<br /><br />$\nabla f(P) = \langle 4 + 8m, -2, 0 \rangle$<br />$\frac{\partial f}{\partial u} = \nabla f(P) \cdot u_{\text{unit}} = (4 + 8m) \cdot \frac{1}{2} + (-2) \cdot \left(-\frac{1}{2}\right) + 0 \cdot \frac{\sqrt{2}}{2} = 2 + 4m + 1 = 3 + 4m$<br /><br />So, the directional derivative of $f(x,y,z)$ at $P(2,1,-1)$ in the direction of $u=\langle 1,-1,\sqrt {2}\rangle$ is $3 + 4m$.