Here's the solution for question 8.
8. Find dxdy for the relation sin(xy)+cos(x+y)=1.
Step 1: Differentiate both sides of the equation with respect to x.
dxd(sin(xy)+cos(x+y))=dxd(1)
Step 2: Apply the chain rule and product rule to differentiate each term.
For sin(xy):
Using the chain rule, dxd(sinu)=cosu⋅dxdu. Here u=xy.
Using the product rule, dxd(xy)=1⋅y+x⋅dxdy=y+xdxdy.
So, dxd(sin(xy))=cos(xy)(y+xdxdy).
For cos(x+y):
Using the chain rule, dxd(cosv)=−sinv⋅dxdv. Here v=x+y.
dxd(x+y)=1+dxdy.
So, dxd(cos(x+y))=−sin(x+y)(1+dxdy).
For the right side:
dxd(1)=0.
Step 3: Substitute these derivatives back into the equation.
cos(xy)(y+xdxdy)−sin(x+y)(1+dxdy)=0
Step 4: Expand the equation.
ycos(xy)+xcos(xy)dxdy−sin(x+y)−sin(x+y)dxdy=0
Step 5: Group terms containing dxdy on one side and other terms on the other side.
xcos(xy)dxdy−sin(x+y)dxdy=sin(x+y)−ycos(xy)
Step 6: Factor out dxdy.
dxdy(xcos(xy)−sin(x+y))=sin(x+y)−ycos(xy)
Step 7: Solve for dxdy.
dxdy=xcos(xy)−sin(x+y)sin(x+y)−ycos(xy)
The final answer is xcos(xy)−sin(x+y)sin(x+y)−ycos(xy).
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