The slope of the tangent to the curve y^3x + y^2x^2 = 6 at (2,1)

1 answer:

7 0

Y^3x + y^2x^2 = 6

Using implicit differentiation 

3 y^2.y'.x+y^3+2 y.y'.x^2+2 y^2.x=0

y'(3 y^2.x+2 y. x^2) = -y^3-2 y^2.x 

y' = -[y^2(y+2 x)]/[x y(3 y+2 x)] 

= -[y(y+2 x)]/[x(3 y+2 x)] 

Substituting x =2 and y = 1:

y' = -(1)(1+4)/[2(3+4)] 
 = -5/14 

The slope of the tangent of the curve at (2,1) will be:
m = -5/14

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<h3>Resolution method</h3>

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