题目

Problem

A curve $C$ has equation

\begin{align*} y=2+10x^{\frac12}-2x^{\frac32}, \qquad x>0. \end{align*}

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

(3)

(b) Hence find the exact value of the gradient of the tangent to $C$ at the point where $x=2$ giving your answer in simplest form.

(Solutions relying on calculator technology are not acceptable.)

(2)

解答

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这里直接使用幂函数求导。常数项求导后为 $0$ ，每个 $x^n$ 项求导时变成 $nx^{n-1}$ 。

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Differentiate term by term:

\begin{align*} \frac{dy}{dx} =&\,0+10\cdot \frac12 x^{-\frac12} -2\cdot \frac32 x^{\frac12}\\ =&\,5x^{-\frac12}-3x^{\frac12}. \end{align*}

Therefore

\begin{align*} \frac{dy}{dx}=5x^{-\frac12}-3x^{\frac12}. \end{align*}

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切线的斜率就是该点的 $\frac{dy}{dx}$ 。题目要求 exact value，所以不要用小数，把 $\sqrt2$ 形式化简到最简。

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At $x=2$ ,

\begin{align*} \frac{dy}{dx} =&\,5(2)^{-\frac12}-3(2)^{\frac12}\\ =&\,\frac5{\sqrt2}-3\sqrt2. \end{align*}

Rationalise the first term:

\begin{align*} \frac5{\sqrt2}-3\sqrt2 =&\,\frac{5\sqrt2}{2}-3\sqrt2\\ =&\,\frac{5\sqrt2}{2}-\frac{6\sqrt2}{2}\\ =&\,-\frac{\sqrt2}{2}. \end{align*}

So the exact gradient is

\begin{align*} -\frac{\sqrt2}{2}. \end{align*}