For d/dx(partial)of (3x + 2y)^1/2, just treat y as a constant and use the usual rules. In this case, use the chain rule and let u = 3x + 2y
df/dx = df/du*du/dx
d/dx(partial)of (3x + 2y)^1/2
=(1/2)(3x + 2y)^-1/2 * 3
= (3/2)(3x + 2y)^-1/2
df/dx = df/du*du/dx
d/dx(partial)of (3x + 2y)^1/2
=(1/2)(3x + 2y)^-1/2 * 3
= (3/2)(3x + 2y)^-1/2
The chain rule states that if we have a function of the form f(g(x)), then the derivative of f(g(x)) with respect to x is given by d(f(g(x)) / dx = f'(g(x)) * g'(x), where f'(g(x)) represents the derivative of f with respect to g(x), and g'(x) represents the derivative of g with respect to x.
In this case, we have f(u) = āu (where u = 3x + 2y) and g(x) = 3x + 2y.
To find the partial derivative with respect to x, we differentiate f(g(x)) with respect to x while treating y as a constant:
ā/āx [f(g(x))] = ā/āx [ā(3x + 2y)] = df/du * du/dx
df/du represents the derivative of āu with respect to u, and du/dx represents the derivative of u with respect to x.
So, df/du = (1/2) * (3x + 2y)^(-1/2) = (3x + 2y)^(-1/2) / 2
And, du/dx = 3
Thus, the partial derivative with respect to x, ā/āx [ā(3x + 2y)], is given by:
(3x + 2y)^(-1/2) / 2 * 3 = 3 / (2ā(3x + 2y))
Similarly, to find the partial derivative with respect to y, we treat x as a constant and differentiate f(g(x)) with respect to y:
ā/āy [f(g(x))] = ā/āy [ā(3x + 2y)] = df/du * du/dy
df/du = (1/2) * (3x + 2y)^(-1/2) = (3x + 2y)^(-1/2) / 2 (same as the previous derivative)
And, du/dy = 2
Therefore, the partial derivative with respect to y, ā/āy [ā(3x + 2y)], is given by:
(3x + 2y)^(-1/2) / 2 * 2 = (3x + 2y)^(-1/2)
So, the partial derivative with respect to x is 3 / (2ā(3x + 2y)) and the partial derivative with respect to y is (3x + 2y)^(-1/2).