Find the slope of the tangent to the curve y^2=x^3/(xy+6) at (6,3)

y^2(xy+6)=x^3

2yy'(xy+6)+y^2xy'=3x^2

y'(2y(xy+6)+y^2x)=3x^2

solve for y' at 6,3

To find the slope of the tangent to the curve at a particular point, we need to take the derivative of the curve equation with respect to x and then evaluate it at the given point.

First, let's find the derivative of the curve equation.

The given curve equation is y^2 = x^3 / (xy + 6).

To find the derivative, we can use the quotient rule. The quotient rule states that for functions u(x) and v(x), the derivative of u(x) / v(x) is given by (u'(x)v(x) - u(x)v'(x)) / (v(x))^2.

Let's apply the quotient rule on the curve equation:

u(x) = x^3, v(x) = (xy + 6),

u'(x) = 3x^2, v'(x) = y + xy'.

Using the quotient rule, the derivative of the curve equation is:

d/dx (y^2) = (3x^2*(xy + 6) - x^3(y + xy')) / ((xy + 6)^2).

Now, let's find the derivative with respect to x only. To do this, we need to use the chain rule, which states that if y is a function of x, then dy/dx = (dy/dt)*(dt/dx), where t is an intermediate variable.

In our case, we have y as a function of x, so we can take d/dx (y^2) as:

d/dx (y^2) = (dy^2 / dy)*(dy / dx).

The derivative of y^2 with respect to y is 2y.

So, applying the chain rule, we have:

d/dx (y^2) = 2y(dy / dx).

Now, let's substitute the equation we found earlier for d/dx (y^2) and solve for dy/dx:

2y(dy / dx) = (3x^2*(xy + 6) - x^3(y + xy')) / ((xy + 6)^2).

Simplifying this equation, we get:

dy / dx = (3x^2*(xy + 6) - x^3(y + xy')) / (2y(xy + 6)^2).

Now, we can evaluate this expression at the given point (6,3) to find the slope of the tangent at that point.

Let's substitute x = 6 and y = 3:

dy / dx = (3(6)^2*(6(3) + 6) - (6)^3(3 + 6(0))) / (2(3)(6(3) + 6)^2).

Calculating this expression, we get:

dy / dx = (3(36)*(18) - (216)(3)) / (2(3)(18)^2).

Simplifying further, we have:

dy / dx = (1944 - 648) / (2(972)).

Calculating this expression, we get:

dy / dx = 1296 / 1944.

Dividing these numbers, we get:

dy / dx = 2/3.

Therefore, the slope of the tangent to the curve at (6,3) is 2/3.