题目
Problem
f ( x ) = ( 4 + 3 x ) 2 x , x > 0 \begin{align*}
f(x)=\frac{(4+3\sqrt{x})^2}{x},
\qquad x>0
\end{align*} f ( x ) = x ( 4 + 3 x ) 2 , x > 0 (a) Find f ′ ( x ) f'(x) f ′ ( x )
(5)
(b) Hence find an equation of the tangent to the curve y = f ( x ) y=f(x) y = f ( x ) ( 4 , 25 ) (4,25) ( 4 , 25 ) y = m x + c y=mx+c y = m x + c m m m c c c
(3)
题目中文翻译
已知
f ( x ) = ( 4 + 3 x ) 2 x , x > 0. \begin{align*}
f(x)=\frac{(4+3\sqrt{x})^2}{x},
\qquad x>0.
\end{align*} f ( x ) = x ( 4 + 3 x ) 2 , x > 0. (a) 求 f ′ ( x ) f'(x) f ′ ( x )
(b) 由此求曲线 y = f ( x ) y=f(x) y = f ( x ) ( 4 , 25 ) (4,25) ( 4 , 25 ) y = m x + c y=mx+c y = m x + c m m m c c c
解答
(a)
解法一
思路
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先展开分子，再逐项除以 x x x
答题过程
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First expand the numerator:
( 4 + 3 x ) 2 = 16 + 24 x + 9 x . \begin{align*}
(4+3\sqrt{x})^2
=&\,16+24\sqrt{x}+9x.
\end{align*} ( 4 + 3 x ) 2 = 16 + 24 x + 9 x . Hence
f ( x ) = 16 + 24 x + 9 x x = 16 x − 1 + 24 x − 1 / 2 + 9. \begin{align*}
f(x)
=&\,\frac{16+24\sqrt{x}+9x}{x}\\
=&\,16x^{-1}+24x^{-1/2}+9.
\end{align*} f ( x ) = = x 16 + 24 x + 9 x 16 x − 1 + 24 x − 1/2 + 9. Differentiate term by term:
f ′ ( x ) = − 16 x − 2 + 24 ( − 1 2 ) x − 3 / 2 = − 16 x − 2 − 12 x − 3 / 2 . \begin{align*}
f'(x)
=&\,-16x^{-2}+24\left(-\frac12\right)x^{-3/2}\\
=&\,-16x^{-2}-12x^{-3/2}.
\end{align*} f ′ ( x ) = = − 16 x − 2 + 24 ( − 2 1 ) x − 3/2 − 16 x − 2 − 12 x − 3/2 . Therefore
f ′ ( x ) = − 16 x 2 − 12 x 3 / 2 . \begin{align*}
f'(x)
=-\frac{16}{x^2}-\frac{12}{x^{3/2}}.
\end{align*} f ′ ( x ) = − x 2 16 − x 3/2 12 .
解法二
思路
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保留原来的分式结构，直接使用商法则。先对分子作复合函数求导，再提取公因式，能够很快化到与解法一相同的最简结果。
答题过程
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Let
u = ( 4 + 3 x ) 2 , v = x . \begin{align*}
u=(4+3\sqrt{x})^2,
\qquad v=x.
\end{align*} u = ( 4 + 3 x ) 2 , v = x . Then
u ′ = 2 ( 4 + 3 x ) 3 2 x = 3 ( 4 + 3 x ) x , v ′ = 1. \begin{align*}
u'
=&\,2(4+3\sqrt{x})\frac{3}{2\sqrt{x}}\\
=&\,\frac{3(4+3\sqrt{x})}{\sqrt{x}},
\qquad v'=1.
\end{align*} u ′ = = 2 ( 4 + 3 x ) 2 x 3 x 3 ( 4 + 3 x ) , v ′ = 1. By the quotient rule,
f ′ ( x ) = x 3 ( 4 + 3 x ) x − ( 4 + 3 x ) 2 x 2 = ( 4 + 3 x ) ( 3 x − 4 − 3 x ) x 2 = − 4 ( 4 + 3 x ) x 2 = − 16 x 2 − 12 x 3 / 2 . \begin{align*}
f'(x)
=&\,\frac{x\dfrac{3(4+3\sqrt{x})}{\sqrt{x}}
-(4+3\sqrt{x})^2}{x^2}\\
=&\,\frac{(4+3\sqrt{x})
\bigl(3\sqrt{x}-4-3\sqrt{x}\bigr)}{x^2}\\
=&\,-\frac{4(4+3\sqrt{x})}{x^2}\\
=&\,-\frac{16}{x^2}-\frac{12}{x^{3/2}}.
\end{align*} f ′ ( x ) = = = = x 2 x x 3 ( 4 + 3 x ) − ( 4 + 3 x ) 2 x 2 ( 4 + 3 x ) ( 3 x − 4 − 3 x ) − x 2 4 ( 4 + 3 x ) − x 2 16 − x 3/2 12 .
(b)
解法一
思路
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切线斜率就是该点的导数值。因为题目说 hence，所以直接使用 (a) 的 f ′ ( x ) f'(x) f ′ ( x ) x = 4 x=4 x = 4
答题过程
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At x = 4 x=4 x = 4
f ′ ( 4 ) = − 16 4 2 − 12 4 3 / 2 = − 1 − 12 8 = − 5 2 . \begin{align*}
f'(4)
=&\,-\frac{16}{4^2}-\frac{12}{4^{3/2}}\\
=&\,-1-\frac{12}{8}\\
=&\,-\frac52.
\end{align*} f ′ ( 4 ) = = = − 4 2 16 − 4 3/2 12 − 1 − 8 12 − 2 5 . The tangent passes through ( 4 , 25 ) (4,25) ( 4 , 25 )
y − 25 = − 5 2 ( x − 4 ) y = − 5 2 x + 10 + 25 y = − 5 2 x + 35. \begin{align*}
y-25=&\,-\frac52(x-4)\\
y=&\,-\frac52x+10+25\\
y=&\,-\frac52x+35.
\end{align*} y − 25 = y = y = − 2 5 ( x − 4 ) − 2 5 x + 10 + 25 − 2 5 x + 35. 