Science:Math Exam Resources/Courses/MATH104/December 2012/Question 06 (a)/Solution 1

$f(1)=1$

(Explanation: We are given that the curve $x^2+y^3-2xy=0$ passes through the point $(1,1)$. Near this point, the curve is the graph of a function, $y=f(x)$. When $x=1$, $y=f(x)=1$.)

$f^{\prime }(1)$ is the same thing as $\frac{dy}{dx}$ evaluated at the point $(x,y)=(1,1)$. In order to find this, we use implicit differentiate the equation $x^2+y^3-2xy=0$ with respect to $x$.

$\begin{array}{r}2x+3y^2\frac{dy}{dx}-2y-2x\frac{dy}{dx}=0\end{array}$

In order to solve for $\frac{dy}{dx}$, we first factorize,

$\begin{array}{r}2x-2y+(3y^2-2x)\frac{dy}{dx}=0.\end{array}$

Then subtract $2x-2y$ from both sides and divide by $(3y^2-2x)$ to obtain

$\begin{array}{r}\frac{dy}{dx}=\frac{2y-2x}{3y^2-2x}.\end{array}$

When $x=y=1,$

$\begin{array}{r}\frac{dy}{dx}=\frac{2(1)-2(1)}{3(1{)}^2-2(1)}=0.\end{array}$

Hence, $f^{\prime }(1)=0$.