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 \(y^{3}\), by the chain - rule \(\frac{d}{dx}(y^{3}) = 3y^{2}\frac{dy}{dx}\)
For \(-xy^{2}\), use the product rule \(\frac{d}{dx}(-xy^{2})=-y^{2}-2xy\frac{dy}{dx}\)
For \(x^{3}\), \(\frac{d}{dx}(x^{3}) = 3x^{2}\)
Differentiating \(y^{3}-xy^{2}+x^{3}=5\) gives:
\(3y^{2}\frac{dy}{dx}-y^{2}-2xy\frac{dy}{dx}+3x^{2}=0\)

Step2: Solve for \(\frac{dy}{dx}\)

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

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

\(\frac{dy}{dx}\mid_{(x = 1,y = 2)}=\frac{2^{2}-3\times1^{2}}{3\times2^{2}-2\times1\times2}=\frac{4 - 3}{12-4}=\frac{1}{8}\)

Answer:

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