题目

Problem

In this question you must show all stages of your working.

Solutions relying on calculator technology are not acceptable.

\begin{align*} y=x^3+96\sqrt{x}+5,\qquad x>0 \end{align*}

(a) Find $\dfrac{dy}{dx}$ , giving each term in simplest form.

(3)

(b) Find the solution of the equation

\begin{align*} \frac{d^2y}{dx^2}=0 \end{align*}

writing the answer in the form $2^k$ where $k$ is a constant.

(3)

解答

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把 $\sqrt{x}$ 写成 $x^{1/2}$ ，再逐项求导。最后把 $48x^{-1/2}$ 写回分式根式形式。

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\begin{align*} y=x^3+96x^{1/2}+5. \end{align*}

Differentiate term by term:

\begin{align*} \frac{dy}{dx} =&\,3x^2+96\cdot\frac12x^{-1/2}+0\\ =&\,3x^2+48x^{-1/2}\\ =&\,3x^2+\frac{48}{\sqrt{x}}. \end{align*}

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先对 (a) 的导数再求一次导。令二阶导数等于 $0$ 后，整理成 $x^{5/2}=4$ ，再把 $4$ 写成 $2^2$ ，用指数法则得到 $x=2^{4/5}$ 。

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From part (a),

\begin{align*} \frac{dy}{dx}=3x^2+48x^{-1/2}. \end{align*}

Differentiate again:

\begin{align*} \frac{d^2y}{dx^2} =&\,6x+48\left(-\frac12\right)x^{-3/2}\\ =&\,6x-24x^{-3/2}\\ =&\,6x-\frac{24}{x^{3/2}}. \end{align*}

Set this equal to zero:

\begin{align*} 6x-\frac{24}{x^{3/2}}=&\,0\\ 6x=&\,\frac{24}{x^{3/2}}\\ x=&\,\frac{4}{x^{3/2}}. \end{align*}

Since $x>0$ , multiply by $x^{3/2}$ :

\begin{align*} x^{5/2}=&\,4\\ x=&\,4^{2/5}\\ =&\,(2^2)^{2/5}\\ =&\,2^{4/5}. \end{align*}

Therefore

\begin{align*} x=2^{4/5}. \end{align*}