What is the derivative of y=x+sin(xy) in terms of x and y at any point on the curve
So I got to y'\1+cos(xy)*....
I dont know what to do next
Sorry that is suppose to say y'=
y=x+sin(xy)
You will need the product rule for the sine term
dy/dx = 1 + cos(xy)(xdy/dx + y)
the second dy/dx is imbedded in a product so we have to expand
I will use y' for dy/dx
y' = 1 + xcos(xy)*y' + ycos(xy)
y' - xcos(xy) y' = 1 + ycos(xy)
y' (1 - xcos(xy) ) = 1 + ycos(xy)
y' = (1 + ycos(xy))/(1 - xcos(xy) )
check: https://www.wolframalpha.com/input/?i=derivative+of+y%3Dx%2Bsin(xy)
To find the derivative of y = x + sin(xy) with respect to x and y, you can use the partial derivative method.
First, let's calculate the partial derivative with respect to x (denoted as ∂/∂x):
To find ∂/∂x, treat y as a constant and differentiate x with respect to x:
∂/∂x (x) = 1
Next, we'll calculate the partial derivative with respect to y (∂/∂y):
To find ∂/∂y, treat x as a constant and differentiate sin(xy) with respect to y:
Using the chain rule, we have:
∂/∂y (sin(xy)) = cos(xy) * x
Now, we can combine both partial derivatives to find the total derivative:
y' = ∂/∂x + ∂/∂y
Substituting the partial derivatives we found:
y' = 1 + cos(xy) * x
Therefore, the derivative of y = x + sin(xy) with respect to x and y at any point on the curve is given by y' = 1 + cos(xy) * x.