题目

Problem

The curve $C$ has equation

\begin{align*} y=6x^2+3\sqrt{x}+\frac58, \qquad x>0 \end{align*}

(a) Find $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ writing the answer in simplest form.

(3)

The point $P\left(\dfrac14,\dfrac52\right)$ lies on $C$ .

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

(3)

解答

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先把 $\sqrt{x}$ 写成 $x^{1/2}$ ，再逐项求导。常数项求导后为 $0$ 。

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Rewrite $y$ as

\begin{align*} y=6x^2+3x^{1/2}+\frac58. \end{align*}

Differentiate term by term:

\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,12x+3\left(\frac12\right)x^{-1/2}+0\\ =&\,12x+\frac32x^{-1/2}. \end{align*}

Therefore

\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =12x+\frac{3}{2\sqrt{x}}. \end{align*}

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切线斜率是该点的导数值。代入 $x=\frac14$ 后得到斜率，再用点斜式写出切线方程。

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At $x=\dfrac14$ ,

\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,12\left(\frac14\right) +\frac{3}{2\sqrt{1/4}}\\ =&\,3+\frac{3}{2(1/2)}\\ =&\,3+3\\ =&\,6. \end{align*}

The tangent passes through $P\left(\dfrac14,\dfrac52\right)$ , so

\begin{align*} y-\frac52=&\,6\left(x-\frac14\right)\\ y-\frac52=&\,6x-\frac32\\ y=&\,6x+1. \end{align*}