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 $x=1$, $y=f(x)=1$.)

$f'(1)$ 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 $x$.

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

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

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

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

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

When $x=y=1,$

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

Hence, $\displaystyle f'(1)=0$.