Question

32: find the slope of the tangent line to the curve y = y(x) given implicitly by the relation y² - xy = 10 at the point (-1,1). (a) -\frac{2}{3} (b) -\frac{1}{3} (c) 1 (d) \frac{2}{3} (e) \frac{1}{3}

Explanation:

Step1: Differentiate implicitly

Differentiate $y^{2}-xy = 10$ with respect to $x$. Using the chain - rule for $y^{2}$ (gives $2y\frac{dy}{dx}$) and the product - rule for $xy$ (gives $y + x\frac{dy}{dx}$). So, $2y\frac{dy}{dx}-(y + x\frac{dy}{dx})=0$.

Step2: Rearrange terms

Expand and collect the terms with $\frac{dy}{dx}$: $2y\frac{dy}{dx}-y - x\frac{dy}{dx}=0$. Then, $(2y - x)\frac{dy}{dx}=y$. So, $\frac{dy}{dx}=\frac{y}{2y - x}$.

Step3: Substitute the point

Substitute $x=-1$ and $y = 1$ into $\frac{dy}{dx}$. We get $\frac{dy}{dx}=\frac{1}{2\times1-(-1)}=\frac{1}{3}$.

Answer:

e. $\frac{1}{3}$