Hello I trying to do a problem on partial derivative. I know the partial derivative of 3x+2y with respect to x would be 3 and with respect to y would be 2. But wut if it is squarroot of 3x+2y, wut is the partial derivative of it with respect to x and y
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
k tnx for ur help
-5(3x+2)
To find the partial derivatives of the function √(3x + 2y) with respect to x and y, we can use the chain rule.
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) = √u (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:
∂/∂x [f(g(x))] = ∂/∂x [√(3x + 2y)] = df/du * du/dx
df/du represents the derivative of √u 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, ∂/∂x [√(3x + 2y)], is given by:
(3x + 2y)^(-1/2) / 2 * 3 = 3 / (2√(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:
∂/∂y [f(g(x))] = ∂/∂y [√(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, ∂/∂y [√(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 / (2√(3x + 2y)) and the partial derivative with respect to y is (3x + 2y)^(-1/2).