xpand f(xy) = e"sin y in the powers of x and y as far as the terms of third degree

To expand the function f(xy) = e*sin y in the powers of x and y up to the terms of third degree, we can use the Taylor series expansion. The Taylor series expansion of a function f(x) around a point a is given by:

f(x) = f(a) + f'(a)(x-a) + (1/2!)f''(a)(x-a)^2 + (1/3!)f'''(a)(x-a)^3 + ...

In this case, we have f(xy) = e*sin y. To expand it in terms of x and y up to the third degree, we need to find the derivatives of f(xy) with respect to x and y up to the third order.

First, let's find the derivative of f(xy) with respect to x. Since y is treated as a constant in this case, the derivative of e*sin y with respect to x is 0.

Next, let's find the derivative of f(xy) with respect to y. Using the chain rule, we have:

df(xy)/dy = d(esin y)/dy = ecos y

Now, let's find the second derivative of f(xy) with respect to y:

d^2f(xy)/dy^2 = d(ecos y)/dy = -esin y

Finally, let's find the third derivative of f(xy) with respect to y:

d^3f(xy)/dy^3 = d(-esin y)/dy = -ecos y

Now, we can use these derivatives to expand f(xy) in terms of x and y up to the third degree. The expansion is given by:

f(xy) = f(0,0) + (df(xy)/dy)(0,0)y + (1/2!)(d^2f(xy)/dy^2)(0,0)y^2 + (1/3!)(d^3f(xy)/dy^3)(0,0)y^3

Substituting the derivatives we found earlier, we have:

f(xy) = 0 + (ecos 0)y + (1/2!)(-esin 0)y^2 + (1/3!)(-e*cos 0)y^3

Simplifying, we get:

f(xy) = 0 + e*y - (1/2!)ey^2 - (1/3!)ey^3

Therefore, the expansion