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 specific point, we need to take the derivative of the equation and substitute the x-coordinate of the point of interest.
Let's begin by taking the derivative of the equation with respect to x.
Differentiating both sides of the equation, we get:
d/dx (x/y + x^2y^2) = d/dx(6)
To differentiate the left side, we can use the quotient rule.
The quotient rule states that for a function u(x)/v(x), the derivative can be found using the formula:
d/dx (u(x)/v(x)) = (v(x)*u'(x) - u(x)*v'(x)) / [v(x)]^2
Applying the quotient rule to our equation, we have:
[ (y * 1) - (x/y * dy/dx) ] / y^2 + [(2xy^2 * dx/dx) + (x^2 * 2y * dy/dx)]
Simplifying this expression, we get:
[ y - (x/y * dy/dx) ] / y^2 + [2xy^2 + 2x^2y * dy/dx]
Now, let's substitute the values of x and y into the derivative expression.
The point of interest is (2, 1), so x = 2 and y = 1.
Substituting these values, we have:
[1 - (2/1 * dy/dx)] / 1^2 + [2(2)(1)^2 + 2(2)^2(1) * dy/dx]
Simplifying further, we obtain:
[1 - 2(dy/dx)] / 1 + 4 + 8(dy/dx)
Now, we can substitute x = 2 and y = 1 into the equation and solve for dy/dx:
[1 - 2(dy/dx)] / 5 + 8(dy/dx) = 6
Solving this equation will give us the value of dy/dx, which represents the slope of the tangent line at the point (2, 1).