The partial derivative with respect to y for z=e^y sin(xy) is:
zy= xe^y cos(xy) + e^y sin(xy)
How do i get to this answer? Thanks.
d z/dy= d/dy e^y sin(xy)
= sin xy * e^y + cos(xy)*d(xy)/dy *e^y
but d(xy)/dy=x
so there it is.
But why wouldnt the answer be just
xe^y cos(xy)? if the partial derivative with respect to x is just ye^y cos(xy).
To find the partial derivative of a function with respect to a specific variable, you need to differentiate the function with respect to that variable while treating all others as constants. In this case, we want to find the partial derivative of z with respect to y.
Given the function z = e^y * sin(xy), we follow these steps:
Step 1: Treat x as a constant. We are only interested in the derivatives with respect to y.
∂/∂y [z] = ∂/∂y [e^y * sin(xy)]
Step 2: Differentiate each term separately using the product and chain rules.
Differentiating e^y:
d/∂y [e^y] = e^y
Differentiating sin(xy):
d/∂y [sin(xy)] = cos(xy) * d/∂y [xy]
Applying the chain rule:
d/∂y [xy] = x * d/∂y [y]
Step 3: Simplify the results and combine the terms.
The derivative of y with respect to y is simply 1.
So, substituting the derivative of e^y and sin(xy) back into the original equation, we get:
∂/∂y [z] = e^y * cos(xy) * x * 1
Simplifying further:
∂z/∂y = xe^y * cos(xy)
Therefore, the partial derivative of z with respect to y is:
∂z/∂y = xe^y * cos(xy)
This is the correct answer.
Note: It's important to note that the use of "zy" in the proposed answer seems to be a typo because the partial derivative is denoted as ∂z/∂y.