if sin(xy)=x, then dy/dx=
for derivative:
cos(xy)(xdy/dx + y) = 1
expand
xdy/dx(cos(xy)) + ycos(xy) = 1
dy/dx = (1 - ycos(xy)/xcos(xy))
(1-ycos(xy))/(xcos(xy))
Well, well, well, looks like we have a mathematical conundrum! Let's see if we can make sense of it, shall we?
If sin(xy) = x, then we can try to differentiate both sides with respect to x.
But before we dive into that, may I remind you that I'm a clown bot, not a math professor? So, grab your popcorn and let's give it a shot!
If we differentiate sin(xy) with respect to x, we get:
d/dx(sin(xy)) = d/dx(x)
Now, we need to take a deep breath and remember our chain rule.
The derivative of sin(u) with respect to u is cos(u). So, using the chain rule, we have:
cos(xy) * (dy/dx * y) = 1
Okay, now let's solve for dy/dx by isolating it:
dy/dx = 1 / (cos(xy) * y)
And voila! There's your answer. But remember, take it with a grain of salt – I'm a clown bot, after all!
To find dy/dx, we first need to differentiate both sides of the equation sin(xy) = x with respect to x.
Using the chain rule, we can differentiate sin(xy) with respect to x as follows:
d/dx(sin(xy)) = d/dx(x)
To differentiate sin(xy) with respect to x, we consider y as a function of x. Therefore, we apply the chain rule:
cos(xy) * (d(xy)/dx) = 1
Since y is implicitly a function of x, we can write d(xy)/dx as y + x * dy/dx.
Substituting into the equation, we have:
cos(xy) * (y + x * dy/dx) = 1
Expanding and rearranging the equation, we get:
cos(xy) * y + x * cos(xy) * dy/dx = 1
To isolate dy/dx, we move the terms involving dy/dx to one side and the other terms to the other side:
x * cos(xy) * dy/dx = 1 - cos(xy) * y
Finally, we divide both sides by x * cos(xy) to solve for dy/dx:
dy/dx = (1 - cos(xy) * y) / (x * cos(xy))
To find dy/dx given that sin(xy) = x, we need to differentiate both sides of the equation with respect to x.
Starting with the left-hand side (LHS):
LHS = sin(xy)
To differentiate sin(xy) with respect to x, we can use the chain rule. Let's denote u = xy, then we have:
LHS = sin(u)
Now differentiating sin(u) with respect to x using the chain rule, the derivative of sin(u) with respect to x is given by:
d/dx[sin(u)] = cos(u) * du/dx
To find du/dx, we differentiate u = xy with respect to x:
du/dx = d/dx (xy) = y(dx/dx) + x(dy/dx) = y + x(dy/dx)
Now substituting du/dx back into the equation for d/dx[sin(u)], we have:
LHS = cos(u) * du/dx = cos(u) * (y + x(dy/dx))
Moving to the right-hand side (RHS):
RHS = x
Now equating the LHS and the RHS, we can write:
cos(u) * (y + x(dy/dx)) = x
Expanding the equation, we get:
cos(u)*y + cos(u)*x(dy/dx) = x
Rearranging the equation, we have:
cos(u)*x(dy/dx) = x - cos(u)*y
Now isolating dy/dx, we can divide both sides of the equation by cos(u)*x:
dy/dx = (x - cos(u)*y) / (cos(u)*x)
Finally, substituting u = xy back into the equation, we obtain:
dy/dx = (x - cos(xy)*y) / (cos(xy)*x)
Therefore, the derivative dy/dx is given by (x - cos(xy)*y) / (cos(xy)*x).