Find the slope of the tangent line to the curve at (2,1)
x/y + x^2y^2 = 6
To find the slope of the tangent line to the curve at a given point, you need to take the derivative of the function with respect to x and evaluate it at that point.
First, let's rewrite the equation of the curve in a more convenient form:
x/y + x^2y^2 = 6
Multiply both sides of the equation by y to eliminate the fraction:
x + x^2y^3 = 6y
Next, differentiate both sides of the equation with respect to x using the product rule:
1 + (2xy^3 + 3x^2y^2(dy/dx)) = 6(dy/dx)
Now, let's find the derivative dy/dx:
2xy^3 + 3x^2y^2(dy/dx) = 6(dy/dx) - 1
Rearrange the equation to solve for dy/dx:
(3x^2y^2 - 6)dy/dx = 1 - 2xy^3
dy/dx = (1 - 2xy^3) / (3x^2y^2 - 6)
Now, substitute the coordinates of the point (2,1) into the equation to find the slope of the tangent line at that point:
dy/dx = (1 - 2(2)(1)^3) / (3(2)^2(1)^2 - 6)
= (1 - 4) / (12 - 6)
= -3/6
= -1/2
So, the slope of the tangent line to the curve at the point (2,1) is -1/2.