Question

what is the slope of the line tangent to the curve $y^{3}-xy^{2}+x^{3}=5$ at the point $(1,2)$?
a $\frac{1}{10}$
b $\frac{1}{8}$
c $\frac{5}{12}$
d $\frac{11}{4}$

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