Question

Find the derivative of the function at P 0 in the direction of A. ​f(x,y,z) = 3 e^x cos(yz)​, P0 (0, 0, 0), A = - i + 2 j + 3k

Answer 1

The derivative of $f(x,y,z)$ at a point $p_0=(x_0,y_0,z_0)$ in the direction of a vector $\vec a=a_x\,\vec\imath+a_y\,\vec\jmath+a_z\,\vec k$ is

$\nabla f(x_0,y_0,z_0)\cdot\dfrac{\vec a}{\|\vec a\|}$

We have

$f(x,y,z)=3e^x\cos(yz)\implies\nabla f(x,y,z)=3e^x\cos(yz)\,\vec\imath-3ze^x\sin(yz)\,\vec\jmath-3ye^x\sin(yz)\,\vec k$

and

$\vec a=-\vec\imath+2\,\vec\jmath+3\,\vec k\implies\|\vec a\|=\sqrt{(-1)^2+2^2+3^2}=\sqrt{14}$

Then the derivative at $p_0$ in the direction of $\vec a$ is

$3\,\vec\imath\cdot\dfrac{-\vec\imath+2\,\vec\jmath+3\,\vec k}{\sqrt{14}}=-\dfrac3{\sqrt{14}}$