How do you differentiate #xy = cot(xy)#?

1 Answer
Aldrich
Sep 6, 2015

#dy/dx= ( ycsc^2(xy) - y)/(x - xcsc^2(xy))#

Explanation:

Start off by differentiating both sides.

#d/dx (xy) = d/dx (cot(xy))#

First, let's discuss how to differentiate the left side.
We can apply the product rule.

#x(d/dx y) + y(d/dxx) = d/dx (cot(xy))#

#d/dx y = dy/dx# due to implicit differentiation and the chain rule.

So:

#x(d/dx y) + y(d/dxx) = d/dx (cot(xy))#

#xdy/dx + y = d/dx (cot(xy))#

Next, we can differentiate the right side using the chain rule.
In short terms, the chain rule is:
1) The derivative of the outer function, with the inner function plugged in...
2) ...multiplied by the derivative of the inner function.

The outer function is #cot(x)#.
The inner function is #xy#.

The derivative of the outer function is #csc^2(x)#.
Plug in the inner function and we get #csc^2(xy)#.
Then we multiply this by the derivative of the inner function.

So #xdy/dx + y = d/dx (cot(xy))# becomes:

#xdy/dx + y = csc^2(xy) * d/dx (xy)#

We've already solved #d/dx (xy)# on the left side, so we can just copy it:

#xdy/dx + y = csc^2(xy) * [xdy/dx + y]#

#xdy/dx + y =xcsc^2(xy) dy/dx + ycsc^2(xy) #

Then we need to isolate #dy/dx# so we bring all #dy/dx# terms to one side:

#xdy/dx - xcsc^2(xy) dy/dx = ycsc^2(xy) - y#

Then we factor out #dy/dx# to isolate it:

#dy/dx(x - xcsc^2(xy))= ycsc^2(xy) - y#

#dy/dx= ( ycsc^2(xy) - y)/(x - xcsc^2(xy))#

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