Compute the directional derivative of the following function at the given point P in the direction of the given vector. Be sure to use a unit vector for the direction vector. f(x, y)= ln(2 +3x^2 +3y^2); P(1,-2); (3,1)

The directional derivate is given by:

Step-by-step explanation:

The directional derivative at point (x,y) is given by:

In which a is the x component of the unit vector and b is the y component of the unit vector.

Vector:

We are given the following vector:

Its modulus is given by:

The unit vector is given by each component divided by it's modulus. So

This means that

Partial derivatives:

So

Directional derivative:

The directional derivate is given by:

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.