The slope of tangent to the curve y³x + y²x² = 6at (2, 1) is &_____?
- -5/14
- -1
- 3/2
- None of these
Explanation
To find the slope of the tangent to the curve y³x + y²x² = 6 at (2, 1), we'll use implicit differentiation:
Given: y³x + y²x² = 6
Differentiating both sides with respect to x:
y³ + x_3y²(dy/dx) + y²_2x + x²*2y(dy/dx) = 0
Rearranging terms to isolate dy/dx:
(3xy² + 2x²y)(dy/dx) = -y³ - 2xy²
dy/dx = (-y³ - 2xy²) / (3xy² + 2x²y)
Now, evaluating dy/dx at (2, 1):
dy/dx = (-(1)³ - 2_2_(1)²) / (3_2_(1)² + 2*(2)²*1)
= (-1 - 4) / (6 + 8)
= -5 / 14
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