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Find the slope of the tangent to the curve at the point specified.

x^3+5x^2y+2y^2=4y+11 at (1,2)

So far I simplified it to -(3x^2-10y/5x^2+4y-4) which I think then simplifies to -(8/3). What do I do from here....?

1 answer

I had dy/dx = -(3x^2+10xy/5x^2+4y-4)
plugging in (1,2) we get dy/dx = -(3+20)/(5+8-4)
= -23/9

so now you have a point and the slope, so
y-2 = (-23/9)(x-1) multiply by 9
9y - 18 = -23x + 23
23x + 9y = 41 or y = -(23/9)x + 41/9
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