How do you take the partial derivative of y = (1200z1z2)^1/2 in respect to z1. Please note z1 and z2 are variables.

f'z1 = (1200^1/2)(z1^1/2)(z2^1/2)
f'z1 = ??

the answer is...

f'z1 = ((1200^1/2)(z2^1/2))/((2(z1^1/2))

but I do not know how to get this answer. Please help me by providing full steps and explanations.

nvrm I figured it out myself

To find the partial derivative of y = (1200z1z2)^(1/2) with respect to z1, we can use the chain rule. The chain rule states that if we have a function g(u) raised to the power n, and u is a function of another variable x, then the derivative of g(u)^(n) with respect to x is given by:

(d/dx) [g(u)^n] = n * g(u)^(n-1) * (du/dx)

In this case, g(u) = u^(1/2) and u = 1200z1z2.

1. Derivative of u with respect to z1:
To find du/dz1, we treat z2 as a constant since we're differentiating with respect to z1. So, we have:
du/dz1 = 1200z2 * (d/dz1) [z1] = 1200z2

2. Replace g(u) and du/dz1 in the chain rule formula:
(d/dz1) [(1200z1z2)^(1/2)] = (1/2) * (1200z1z2)^(-1/2) * (1200z2)

Simplifying,
= (1/2) * (1200z2) / (1200z1z2)^(1/2)
= (1200z2/2) / (1200z1z2)^(1/2)
= 600z2 / (1200z1z2)^(1/2)

Further simplifying,
= 600z2 / (2 * (z1z2)^(1/2))
= 600z2 / (2z1^(1/2) * (z2)^(1/2))
= (600z2 / 2) * (z1^(-1/2)) * (z2^(-1/2))
= 300z2 / (z1^(-1/2) * z2^(-1/2))

We can simplify it further by multiplying the term inside the parentheses by (z1^(1/2) * z2^(1/2))/(z1^(1/2) * z2^(1/2)):
= (300z2 / (z1^(-1/2) * z2^(-1/2))) * (z1^(1/2) * z2^(1/2))/(z1^(1/2) * z2^(1/2))
= (300z2z1^(1/2) * z2^(1/2)) / (z1 * z2)
= (300z2z1^(1/2) * z2^(1/2)) / (z1z2)
= (300(z1z2)^(1/2))

So, the partial derivative of y = (1200z1z2)^(1/2) with respect to z1 is:
f'z1 = (300(z1z2)^(1/2))