Let z=√2x+5y
. Then:
The rate of change in z
at (2,4)
as we change x
but hold y
fixed is
, and
The rate of change in z
at (2,4)
as we change y
but hold x
fixed is
To find the rate of change in z with respect to x, we need to take the partial derivative of z with respect to x while holding y fixed:
∂z/∂x = ∂/∂x (√2x + 5y)
Using the chain rule, we have:
∂z/∂x = ∂/∂x (√2x) = (√2)(1) = √2
So the rate of change in z with respect to x, while holding y fixed, is √2.
To find the rate of change in z with respect to y, we need to take the partial derivative of z with respect to y while holding x fixed:
∂z/∂y = ∂/∂y (√2x + 5y)
Using the chain rule, we have:
∂z/∂y = ∂/∂y (5y) = 5
So the rate of change in z with respect to y, while holding x fixed, is 5.
To find the rate of change in z at (2,4) as we change x but hold y fixed, we can take the partial derivative of z with respect to x.
∂z/∂x = (∂/∂x)√(2x+5y)
To find this derivative, we can use the chain rule:
∂z/∂x = (1/2√(2x+5y)) * (∂/∂x)(2x+5y)
Since we are holding y fixed, the derivative of (2x+5y) with respect to x is simply 2. Plugging this in:
∂z/∂x = (1/2√(2x+5y)) * 2
= 1/√(2x+5y)
Evaluating this derivative at (2,4):
∂z/∂x = 1/√(2(2)+5(4))
= 1/√(4+20)
= 1/√24
= 1/(2√6)
Therefore, the rate of change in z at (2,4) as we change x but hold y fixed is 1/(2√6).
To find the rate of change in z at (2,4) as we change y but hold x fixed, we can take the partial derivative of z with respect to y.
∂z/∂y = (∂/∂y)√(2x+5y)
To find this derivative, we can again use the chain rule:
∂z/∂y = (1/2√(2x+5y)) * (∂/∂y)(2x+5y)
Since we are holding x fixed, the derivative of (2x+5y) with respect to y is simply 5. Plugging this in:
∂z/∂y = (1/2√(2x+5y)) * 5
= 5/√(2x+5y)
Evaluating this derivative at (2,4):
∂z/∂y = 5/√(2(2)+5(4))
= 5/√(4+20)
= 5/√24
= 5/(2√6)
Therefore, the rate of change in z at (2,4) as we change y but hold x fixed is 5/(2√6).