what is the slope of the line tangent to the curve $y^{3}-xy^{2}+x^{3}=5$ at the point $(1,2)$?\na…

Explanation:

Step1: Differentiate both sides with respect to (x)

Using implicit differentiation: For (y^{3}), the derivative is (3y^{2}y') (by the chain - rule (\frac{d}{dx}(u^{n})=nu^{n - 1}u'), where (u = y)). For (-xy^{2}), use the product - rule ((uv)'=u'v + uv'). Here (u=-x), (u'=-1) and (v = y^{2}), (v'=2yy'). So the derivative of (-xy^{2}) is (-y^{2}-2xyy'). For (x^{3}), the derivative is (3x^{2}). The derivative of the constant (5) is (0). So, (\frac{d}{dx}(y^{3}-xy^{2}+x^{3})=\frac{d}{dx}(5)) gives (3y^{2}y'-y^{2}-2xyy'+3x^{2}=0).

Step2: Solve for (y')

Group the terms with (y'): (y'(3y^{2}-2xy)=y^{2}-3x^{2}). Then (y'=\frac{y^{2}-3x^{2}}{3y^{2}-2xy}).

Step3: Substitute (x = 1) and (y = 2)

(y'=\frac{2^{2}-3\times1^{2}}{3\times2^{2}-2\times1\times2}=\frac{4 - 3}{12-4}=\frac{1}{8}).

Answer:

B. (\frac{1}{8})