题目

Problem

The curve $C$ has equation

\begin{align*} y=\frac{x^3}{8}-\frac{24}{\sqrt{x}}+1. \end{align*}

(a) Find $\dfrac{dy}{dx}$ , giving the answer in its simplest form.

(3)

The point $P(4,-3)$ lies on $C$ .

(b) Find the equation of the tangent to $C$ at the point $P$ . Write your answer in the form $y=mx+c$ , where $m$ and $c$ are constants to be found.

(3)

解答

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先把根号分母写成负指数，再逐项求导。

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\begin{align*} y=&\,\frac18x^3-24x^{-1/2}+1,\\ \frac{dy}{dx} =&\,\frac38x^2+12x^{-3/2}\\ =&\,\frac38x^2+\frac{12}{x^{3/2}}. \end{align*}

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切线斜率是导数在 $x=4$ 时的值。求出斜率后，用点 $P(4,-3)$ 写直线。

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At $x=4$ ,

\begin{align*} \frac{dy}{dx} =&\,\frac38(4)^2+\frac{12}{4^{3/2}}\\ =&\,6+\frac{12}{8}\\ =&\,\frac{15}{2}. \end{align*}

So the tangent is

\begin{align*} y+3=&\,\frac{15}{2}(x-4)\\ y=&\,\frac{15}{2}x-33. \end{align*}