Question

Find the gradient and directional derivative of f at P in the direction of v:
f(x,y)=ln(x^2 + y^2 +1)+ e^(2xy)
P(0,2)
v=5i-12j

Answer 1

The gradient ∇f
= 2x/(x^2+y^2+1) + 2ye^(2xy)<b>i</b>
+ 2y/(x^2+y^2+1) + 2xe^(2xy)<b>j</b>

The directional derivative along <b>v</v> is ∇f•<b>v</v>

So, at (0,2)
Fx = 4
Fy = 4/5

and

∇f•<b>v</v> = <4,4/5>•<5,-12> = <20,-48/5></b></b></b>

Answer 2

Sorry about the boldface.

The gradient ∇f
= 2x/(x^2+y^2+1) + 2ye^(2xy)<b>i</b>
+ 2y/(x^2+y^2+1) + 2xe^(2xy)<b>j</b>

The directional derivative along <b>v</v> is ∇f•<b>v</b>

So, at (0,2)
Fx = 4
Fy = 4/5

and

∇f•<b>v</b> = <4,4/5>•<5,-12> = <20,-48/5></b>