The slope of the tangent to the curve y^3x + y^2x^2 = 6 at (2,1)
1 answer:
Y^3x + y^2x^2 = 6
<span>Using implicit differentiation </span>
<span>
3 y^2.y'.x+y^3+2 y.y'.x^2+2 y^2.x=0
</span>
<span>y'(3 y^2.x+2 y. x^2) = -y^3-2 y^2.x </span>
<span>
y' = -[y^2(y+2 x)]/[x y(3 y+2 x)] </span>
<span>
= -[y(y+2 x)]/[x(3 y+2 x)] </span>
<span>
Substituting x =2 and y = 1:
</span>
<span>y' = -(1)(1+4)/[2(3+4)] </span>
<span> = -5/14 </span>
<span>The slope of the tangent of the curve at (2,1) will be:
m = -5/14</span>
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<em><u>Hope this helps.</u></em>
