Question 1008865
<pre>
{{{x^2+y^2=10y}}}

{{{2x+2y*expr((dy)/(dx))=10expr((dy)/(dx))}}}

Divide through by 2

{{{x+y*expr((dy)/(dx))=5expr((dy)/(dx))}}}

{{{y*expr((dy)/(dx))-5expr((dy)/(dx))=-x}}}

{{{expr((dy)/(dx))(y-5)=-x}}}

{{{expr((dy)/(dx))=-x/(y-5)}}}

{{{expr((dy)/(dx))=-x/(-5+y)}}}

{{{expr((dy)/(dx))=-x/(-(5-y))}}}

{{{expr((dy)/(dx))=x/(5-y)}}}

Yes you were right about the derivative.

The derivative is a formula for the slope of the 
tangent line at the point (x,y).

So the slope m of the tangent line at (3,1) is found by
substituting (x,y) = (3,1) in the derivative:

{{{m=3/(5-1)}}}

{{{m=3/4}}}

Then use the point-slope formula for a line

{{{y-y[1]=m(x-x[1])}}} where (x<sub>1</sub>, y<sub>1</sub>) = (3,1)

{{{y-1 = expr(3/4)(x-3)}}}

{{{y-1 = expr(3/4)x-9/4}}}

{{{y = expr(3/4)x-9/4+1}}}

{{{y = expr(3/4)x-9/4+4/4}}}

{{{y = expr(3/4)x-5/4}}}

{{{drawing(320,400,-6,6,-4,11, 

circle(3,1,0.15),circle(3,1,0.13),circle(3,1,0.11),circle(3,1,0.09),circle(3,1,0.07),circle(3,1,0.05),circle(3,1,0.03),circle(3,1,0.01),

locate(3,1,"(3,1)"),



graph(320,400,-6,6,-4,11), circle(0,5,5),line(-13,-11,11,7)  )}}}

Edwin</pre>