HELP MEe g: R² → R a differentiable function at (0, 0), with g (x, y) = 0 only at the point (x, y) = (0, 0). Consider (a) Calculate as partial derivatives of f in (0, 0), in terms of the partial derivatives of g.

(b) Show that f is differentiable at (0, 0).

(a) This follows from the definition for the partial derivative, with the help of some limit properties and a well-known limit.

• Recall that for , we have the partial derivative with respect to defined as

The derivative at (0, 0) is then

• By definition of , , so

• Expanding the tangent in terms of sine and cosine gives

• Introduce a factor of in the numerator, then distribute the limit over the resulting product as

• The first limit is 1; recall that for , we have

The second limit is also 1, which should be obvious.

• In the remaining limit, we end up with

and this is exactly the partial derivative of with respect to .

For the same reasons shown above,

(b) To show that is differentiable at (0, 0), we first need to show that is continuous.

• By definition of continuity, we need to show that

is very small, and that as we move the point closer to the origin, converges to .

We have

The first expression in the product is bounded above by 1, since for all . Then as approaches the origin,

So, is continuous at the origin.

• Now that we have continuity established, we need to show that the derivative exists at (0, 0), which amounts to showing that the rate at which changes as we move the point closer to the origin, given by

approaches 0.

Just like before,

and this converges to , since differentiability of means

So, is differentiable at (0, 0).

dc=6 square root of 3 and ac is 12 square root 3

c and d

step-by-step explanation:

square root of 3/2 = 18/c

cross multiply to get 18*2=√3*c

36/√3 * √3/√3 = c

c= 36*√3/3

c=12√3