Question

dervative formula
and example​

Answer 1

Answer:

∂f∂x(x,y)∂f∂x(1,2)∂f∂y(x,y)∂f∂y(1,2)=2x=2=2y=4

So Df(1,2)=[ 2 4 ].

Since both partial derivatives ∂f∂x(x,y) and ∂f∂y(x,y) are continuous functions, we know that f(x,y) is differentiable. Therefore, Df(1,2) is the derivative of f, and the function has a tangent plane there.

To calculate the equation of the tangent plane, the only additional calculation is the value of f at (x,y)=(1,2), which is f(1,2)=12+22=5. The equation for the tangent plane is

z=f(1,2)+∂f∂x(1,2)(x−1)+∂f∂y(1,2)(y−2)=5+2(x−1)+4(y−2)

For a scalar-valued function of two variables such as f(x,y), the tangent plane is the linear approximation. We can write the linear approximation as

L(x,y)=5+2(x−1)+4(y−2).

Example 1'

If looked at the point (2,3), what changes?

Solution: The partial derivatives change, so the derivative becomes

∂f∂x(2,3)∂f∂y(2,3)Df(2,3)=4=6=[ 4 6 ].

The equation for the tangent plane, i.e., the linear approximation, becomes

z=L(x,y)=f(2,3)+∂f∂x(2,3)(x−2)+∂f∂y(2,3)(y−3)=13+4(x−2)+6(y−3)