## What is the slope of the line tangent to the curve y3−xy2+x3=5 at the point (1,2) ?

### To find the slope of the line tangent to the curve at a specific point, we need to take the derivative of the equation with respect to the independent variable (in this case, x) and then substitute the x-coordinate of the given point to get the slope

To find the slope of the line tangent to the curve at a specific point, we need to take the derivative of the equation with respect to the independent variable (in this case, x) and then substitute the x-coordinate of the given point to get the slope.

First, let’s find the derivative of the equation with respect to x. We can do this by differentiating each term with respect to x using the power rule.

Taking the derivative, we have:

d/dx(y^3 – xy^2 + x^3) = d/dx(5)

Applying the power rule, we get:

3y^2 * dy/dx – y^2 – 2xy * dy/dx + 3x^2 = 0

Simplifying, we have:

dy/dx * (3y^2 – 2xy) = y^2 – 3x^2

Now, we can substitute the x-coordinate of the given point (1,2) into the equation.

At (x, y) = (1,2), we have:

dy/dx * (3(2)^2 – 2(1)(2)) = (2)^2 – 3(1)^2

Simplifying further, we get:

dy/dx * (12 – 4) = 4 – 3

dy/dx * 8 = 1

Solving for dy/dx (the slope), we have:

dy/dx = 1/8

Therefore, the slope of the line tangent to the curve y^3 – xy^2 + x^3 = 5 at the point (1,2) is 1/8.

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