# What is the directional derivative of the function f(x,y,z) = xsin(yz) at P = (1,1,`pi/2` ) in the direction of the vector `<< sqrt(3i)+j-k>>`?

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### 1 Answer

The directional derivative is the dot product of the gradient of the function and the direction of the given vector.

The gradient of the function is the vector with the coordinates

`<(del f)/(del x), (del f)/(del y), (del f)/(del z)>` evaluated at the point P (1, 1, `pi/2` ).

`(del f)/(del x) = sinyz` . At y = 1 and `z = pi/2` , `(del f)/(del x) = 1` .

`(del f)/(del y) = xzcos(yz)` . At x = 1 and y = 1 and `z = pi/2` ,

`(del f)/(del y) = 0` .

`(del f)/(del z) = xycos(yz)` . Similarly to the derivative with respect to y,

`(del f)/(del z) = 0` at point P.

So the gradient vector of this function at point P is <1, 0, 0>.

The dot product of <1, 0, 0> and `<sqrt 3, 1, -1>`

is `1*sqrt(3) + 0 *1 + 0*(-1) = sqrt(3).`

**The directional derivative is** `sqrt(3).`