Techniques of Differentiation - Exercise 3

Exercise 3. Solve the equation $\displaystyle \frac{dy}{dx} = 0$ when

\begin{displaymath}y = \frac{x^2 - x + 5}{2x+3}\end{displaymath}

Answer. First we need to find the derivative of y. This will be done via the quotient rule. We have

\begin{displaymath}\frac{dy}{dx} = \frac{(2x-1)(2x+3)- 2(x^2-x+5)}{(2x+3)^2} = \frac{2x^2 + 6x -13}{(2x+3)^2} \;\cdot\end{displaymath}

So to have $\displaystyle \frac{dy}{dx} = 0$, we should have

\begin{displaymath}2x^2 + 6x -13 = 0\;.\end{displaymath}

This is a quadratic equation. The quadratic formulas give

\begin{displaymath}x_1 = \frac{-3 + \sqrt{35}}{2}\;\; \mbox{or}\;\;x_2 = \frac{-3 - \sqrt{35}}{2}\end{displaymath}

which are the two solutions to the equation $\displaystyle \frac{dy}{dx} = 0$.

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