题目

Problem

The curve $C$ has equation

\begin{align*} y=\frac{5x^3-8}{2x^2},\qquad x>0. \end{align*}

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

(3)

The point $P(2,4)$ lies on $C$ .

(b) Find an equation of the tangent to $C$ at $P$ , giving your answer in the form $ax+by+c=0$ , where $a$ , $b$ and $c$ are integers to be found.

(3)

解答

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先把分式拆成幂函数，再求导会更直接。分母是单项式，所以可以分别除以 $2x^2$ 。

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Rewrite $y$ as powers of $x$ :

\begin{align*} y =&\,\frac{5x^3}{2x^2}-\frac{8}{2x^2}\\ =&\,\frac52x-4x^{-2}. \end{align*}

Differentiate term by term:

\begin{align*} \frac{dy}{dx} =&\,\frac52+8x^{-3}\\ =&\,\frac52+\frac{8}{x^3}. \end{align*}

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切线斜率就是该点处的导数值。求出 $x=2$ 时的斜率后，用点斜式写直线，再整理成题目要求的整数形式。

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At $x=2$ ,

\begin{align*} \frac{dy}{dx} =&\,\frac52+\frac{8}{2^3}\\ =&\,\frac52+1\\ =&\,\frac72. \end{align*}

So the tangent has gradient $\frac72$ and passes through $P(2,4)$ :

\begin{align*} y-4=&\,\frac72(x-2)\\ 2y-8=&\,7x-14\\ 7x-2y-6=&\,0. \end{align*}