\(\phi = \mathrm x^2+2\mathrm xy \) at (1, –1, 3)
directional vector \(\vec p = \hat i+2\hat j+2\hat k\)
We know that direction derivative \(= \nabla\phi.\hat p = \nabla \phi\frac{\vec p}{|\vec p|}\)
Now, \(\nabla \phi = \nabla(\mathrm x^2+2\mathrm xy)\)
\(=\hat i \frac{\partial}{\partial \mathrm x}(\mathrm x^2+2\mathrm xy)\) \(+\hat j\frac{\partial}{\partial \mathrm x}(\mathrm x^2+2\mathrm xy)\) \(+\hat k\frac{\partial}{\partial z}(\mathrm x^2+2\mathrm xy)\)
\(=(2\mathrm x+2y)\hat i + 2\mathrm x\hat j\)
\((\nabla \phi)_{(1, -1, 3)}=(2-2)\hat i + 2\hat j=2\hat j\)
And \(\hat p = \frac{\vec p}{| \vec p|}\) \(=\frac{\hat i+2\hat j+2\hat k}{\sqrt{1^2+2^2+2^2}}\) \(=\frac{1}{3}(\hat i+2\hat j+2\hat k)\)
Now, directional derivative of \(\phi\) at (1, –1, 3) in the direction of vector p
\(=(\nabla \phi)_{(1, -1, 3)}\hat p \) \(=2\hat j.\left(\frac{1}{3}(\hat i+2\hat j+2\hat k)\right)\)
\(=\frac{4}{3}\) \(\Big(\hat i.\hat j=\hat j.\hat k = 0\) but \(\hat j.\hat j =1\Big)\)
