The gradient ∇f
= 2x/(x^2+y^2+1) + 2ye^(2xy)i
+ 2y/(x^2+y^2+1) + 2xe^(2xy)j
The directional derivative along v</v> is ∇f•v</v>
So, at (0,2)
Fx = 4
Fy = 4/5
and
∇f•v</v> = <4,4/5>•<5,-12> = <20,-48/5>
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
2 answers
Sorry about the boldface.
The gradient ∇f
= 2x/(x^2+y^2+1) + 2ye^(2xy)i
+ 2y/(x^2+y^2+1) + 2xe^(2xy)j
The directional derivative along v</v> is ∇f•v
So, at (0,2)
Fx = 4
Fy = 4/5
and
∇f•v = <4,4/5>•<5,-12> = <20,-48/5>
The gradient ∇f
= 2x/(x^2+y^2+1) + 2ye^(2xy)i
+ 2y/(x^2+y^2+1) + 2xe^(2xy)j
The directional derivative along v</v> is ∇f•v
So, at (0,2)
Fx = 4
Fy = 4/5
and
∇f•v = <4,4/5>•<5,-12> = <20,-48/5>
1 answer