5.2.1-5.2.2 导数的运算
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基础知识
基础初等函数的导数
基本初等函数的导数公式
原函数 | 导函数 | 原函数 | 导函数 |
---|---|---|---|
\(f(x)=c(c是常数)\) | \(f' (x)=0\) | \(f(x)=x^n\) \((n∈Q^*,且n\neq 0)\) |
\(f' (x)=nx^{n-1}\) |
\(f(x)=\sin x\) | \(f' (x)=\cos x\) | \(f(x)=\cos x\) | \(f' (x)=-\sin x\) |
\(f(x)=a^x (a>0)\) \((a>0且a≠1)\) |
\(f' (x)=a^x\ln a\) | \(f(x)=e^x\) | \(f' (x)=e^x\) |
\(f(x)=\log _ax\) \((a>0且a≠1)\) |
\(f^{\prime}(x)=\dfrac{1}{x \ln a}\) | \(f(x)=\ln x\) | \(f^{\prime}(x)=\dfrac{1}{x}\) |
根据导数的定义求函数\(y=f(x)\)的导数,就是求出当\(Δx→0\)时, \(\dfrac{\Delta y}{\Delta x}\)无限趋近的那个定值.
下面求几个常用函数的导数.
【例1】函数\(y=x^2\)的导数.
解 因为 \(\dfrac{\Delta y}{\Delta x}=\dfrac{f(x+\Delta x)-f(x)}{\Delta x}=\dfrac{(x+\Delta x)^2-x^2}{\Delta x}\)\(=\dfrac{x^2+2 x \cdot \Delta x+(\Delta x)^2-x^2}{\Delta x}=2 x+\Delta x\),
所以\(y^{\prime}=\lim\limits _{\Delta x \rightarrow 0} \dfrac{\Delta y}{\Delta x}=\lim\limits _{\Delta x \rightarrow 0}(2 x+\Delta x)=2 x\).
【例2】函数\(y= \dfrac{1}{x}\)的导数.
解 因为\(\dfrac{\Delta y}{\Delta x}=\dfrac{f(x+\Delta x)-f(x)}{\Delta x}=\dfrac{\dfrac{1}{x+\Delta x}-\dfrac{1}{x}}{\Delta x}=\dfrac{x-(x+\Delta x)}{x(x+\Delta x) \Delta x}=-\dfrac{1}{x^2+x \cdot \Delta x}\),
所以\(y^{\prime}=\lim\limits _{\Delta x \rightarrow 0} \dfrac{\Delta y}{\Delta x}=\lim\limits _{\Delta x \rightarrow 0}\left(-\dfrac{1}{x^2+x \cdot \Delta x}\right)=-\dfrac{1}{x^2}\) .
导数运算法则
(1)\([f(x)± g(x)]'=f' (x)±g' (x)\);
拓展 :\([f(x)± g(x)±⋯± u(x)]'=f' (x)±g' (x)±⋯± u'(x)\);
记忆:函数的和差的导数等于函数导数的和差;
(2)\([f(x)⋅g(x)]'=f' (x) g(x)+f(x) g' (x)\);
记忆 :两函数积的导数等于“前导后不导+后导前不导”;
特别 :\([C\cdot f(x)]'=C\cdot f' (x)\),\(C\)为常数;
证明 \([cf(x)]'=c' f(x)+cf' (x)=cf' (x)\);
(3) \(\left[\dfrac{f(x)}{g(x)}\right]^{\prime}=\dfrac{f^{\prime}(x) g(x)-f(x) g^{\prime}(x)}{[g(x)]^2}(g(x) \neq 0)\).
记忆 :两函数商的导数等于“分母平分,子导母不导减母导子不导”.
【例】 求下列函数的导数
(1)\(f(x)=2x^3-x^2-4\) \(\qquad \qquad\) (2) \(f(x)=x \cdot \ln x\) \(\qquad \qquad\) (3) \(f(x)=\dfrac{e^x}{x}\)
解 (1)\(f' (x)=(2x^3 )'-(x^2 )'-4'=6x^2-2x\);
(2)\(f' (x)=x'\cdot \ln x+x\cdot (\ln x )'=\ln x+x\cdot \dfrac{1}{x}=\ln x+1\);
(3) \(f^{\prime}(x)=\dfrac{\left(e^x\right)^{\prime} \cdot x-x^{\prime} \cdot e^x}{x^2}=\dfrac{x \cdot e^x-e^x}{x^2}\) .
导数的几何意义
函数\(y=f(x)\)在点\(x=x_0\)处的导数的几何意义是曲线\(y=f(x)\)在点\(P(x_0 ,f(x_0))\)处的切线的斜率,
即:曲线\(y=f(x)\)在点\(P(x_0 ,f(x_0))\)处的切线\(l\)的斜率\(k=f'(x_0)\),
切线\(l\)的方程为\(y-f(x_0)=f'(x_0)(x-x_0)\).
解释
"过点\(x=x_0\)”与"在点\(x=x_0\)处"的区别
曲线\(C:y=f(x)\)在点\(P(x_0 ,y_0)\)处的切线指的是\(P\)为切点的切线,如图一;
过点\(P(x_0 ,y_0)\)的切线是指切线过点\(P\),点\(P\)是否切点均可,切线可多条,如图二.
\(\qquad \qquad\)
【例】 求曲线\(f(x)=x^2\)在点\((1,1)\)处的切线方程.
解 \(\because f' (x)=2x\),\(\therefore\) 切线方程的斜率为\(k=f' (1)=2\),
\(\therefore\)切线方程为\(y-1=2(x-1)\),化简为\(2x-y-1=0\).
基本方法
【题型1】 基本初等函数的导数
【典题1】 求下列函数的导数:
(1) \(y=(2x^2+3)(3x-1)\); \(\qquad \qquad\) (2)\(y=x^3 e^x\) \(\qquad \qquad\) (3) \(y=\dfrac{3 \sin x}{1-2 \sin ^2 \frac{x}{2}}\);
解析 (1) 方法一 \(\because y=(2x^2+3)(3x-1)\),
\(\therefore y'=(2x^2+3)' (3x-1)+(2x^2+3) (3x-1)'\)
\(=4x(3x-1)+3(2x^2+3)=18x^2-4x+9\);
方法二 \(\because y=(2x^2+3)(3x-1)=6x^3-2x^2+9x-3\),
\(\therefore y'=18x^2-4x+9\).
(2) \(y'=(x^3 )'\cdot e^x+(e^x )'\cdot x^3=3x^2\cdot e^x+e^x\cdot x^3=x^2\cdot e^x (x+3)\);
(3) \(\because y=\dfrac{3 \sin x}{1-2 \sin ^2 \dfrac{x}{2}}=\dfrac{3 \sin x}{\cos x}\),
\(\therefore y^{\prime}=3 \cdot \dfrac{(\sin x)^{\prime} \cdot \cos x-(\cos x)^{\prime} \cdot \sin x}{\cos ^2 x}=3 \cdot \dfrac{\cos ^2 x+\sin ^2 x}{\cos ^2 x}=\dfrac{3}{\cos ^2 x}\).
点拨 求导先要明确函数的结构,是函数“加减形式”还是“乘除形式”,若可以化简先化简,再选择简单形式求导.
【典题2】一物体做直线运动,其位移\(s\)(单位:\(m\))与时间\(t\)(单位:\(s\))的关系是 \(s(t)=\dfrac{s^{\prime}(1)}{3} \cdot t^2+4 t\),则该物体在\(t=4s\)时的瞬时速度是\(\underline{\quad \quad}\) .
解析 \(\because s(t)=\dfrac{s^{\prime}(1)}{3} \cdot t^2+4 t\),
\(\therefore s^{\prime}(t)=\dfrac{2}{3} \cdot s^{\prime}(1) \cdot t+4\),
令\(t=1\),得 \(s^{\prime}(1)=\dfrac{2}{3} \cdot s^{\prime}(1)+4\),解得\(s' (1)=12\),
\(\therefore s' (t)=8t+4\),
\(\therefore\) 该物体在\(t=4s\)时的瞬时速度是\(s' (4)=36\).
【巩固练习】
1.求下列函数的导函数
(1) \(y=x^2 \sin x\);\(\qquad \qquad\) (2) \(y=\dfrac{e^x+1}{e^x-1}\).\(\qquad \qquad\) (3)\(y=3x^2+x\sin x\)
2.设函数\(f(x)\)的导函数是\(f'(x)\),若 \(f(x)=f^{\prime}\left(\dfrac{\pi}{2}\right) \cdot \cos x-\sin x\),则 \(f^{\prime}\left(\dfrac{\pi}{3}\right)=\) \(\underline{\quad \quad}\).
3.有一机器人的运动方程为 \(s=t^2+\dfrac{3}{t}\)(\(t\)是时间,\(s\)是位移),则该机器人在时刻\(t=2\)时的瞬时速度为\(\underline{\quad \quad}\).
4.血液在血管中的流速满足关系式\(v(r)=k(R^2-r^2 )\)其中\(k\)为常数,\(R\)和\(r\)分别为血管的外径和内径(单位:\(cm\)). 现假定\(k=1000\),\(R=0.2cm\),求\(v(0.1)\)及\(v' (0.1)\),并对所得结果作出解释.
5.通过某导体横截面的电量\(q\)(单位: \(C\)) 关于时间\(t\)(单位: \(s\)) 的函数关系式为\(q(t)=2t^3+3t\).
(1)求当\(t\)从\(1s\)变到\(5s\)时,电量\(q\)关于时间\(t\)的平均变化率,并解释它的实际意义;
(2)求\(q'(5)\),并解释它的实际意义.
参考答案
-
答案 (1)\(y'=2x\sin x+x^2 \cos x\);(2) \(y^{\prime}=\dfrac{-2 e^x}{\left(e^x-1\right)^2}\) ;(3)\(y'=6x+\sin x+x\cos x\).
解析 (1)\(y'=(x^2 \sin x)'=2x\sin x+x^2 \cos x\)
(2) \(y^{\prime}=\left(\dfrac{e^x+1}{e^x-1}\right)^{\prime}=\dfrac{\left(e^x+1\right)^{\prime}\left(e^x-1\right)-\left(e^x+1\right)\left(e^x-1\right)^{\prime}}{\left(e^x-1\right)^2}=\dfrac{-2 e^x}{\left(e^x-1\right)^2}\).
(3)\(∵y=3x^2+x\sin x\),\(\therefore y'=6x+\sin x+x\cos x\). -
答案 \(-\dfrac{1}{2}\)
解析 \(f(x)=f^{\prime}\left(\dfrac{\pi}{2}\right) \cdot \cos x-\sin x\),
则\(f^{\prime}(x)=-f^{\prime}\left(\dfrac{\pi}{2}\right) \sin x-\cos x\),
\(\therefore f^{\prime}\left(\dfrac{\pi}{2}\right)=-f^{\prime}\left(\dfrac{\pi}{2}\right) \sin \dfrac{\pi}{2}-\cos \dfrac{\pi}{2}\), \(\therefore f^{\prime}\left(\dfrac{\pi}{2}\right)=0\),
\(\therefore f^{\prime}(x)=-\cos x\), \(\therefore f^{\prime}\left(\dfrac{\pi}{3}\right)=-\dfrac{1}{2}\). -
答案 \(\dfrac{13}{4}\)
解析 机器人的运动方程为 \(s=t^2+\dfrac{3}{t}\),所以\(s^{\prime}=2 t-\dfrac{3}{t^2}\),
\(t=2\)时,\(s^{\prime}=2 \times 2-\dfrac{3}{4}=\dfrac{13}{4}\),
则该机器人在时刻\(t=2\)时的瞬时速度为\(\dfrac{13}{4}\).
故答案为:\(\dfrac{13}{4}\). -
答案 \(v(0.1)=30\),\(v' (0.1)=-200\),当血管内径为\(0.1cm\)时,血液流速为\(30\),加速度为\(-200\).
解析 由已知可得\(v(r)=1000(0.04-r^2 )=40-1000r^2\),
则\(v' (r)=-2000r\),
所以\(v(0.1)=40-1000×0.1^2=30\), \(v' (0.1)=-2000×0.1=-200\).
意义为: 当血管内径为\(0.1cm\)时,血液流速为\(30\),加速度为\(-200\). -
答案 (1) \(65(C/s)\), 它表示从\(1s\)到\(5s\)这段时间内,平均每秒通过该导体横截面的电量为\(65C\);
(2) \(153(C/s)\), 电流强度为\(153C/s\).
解析 (1)当\(t\)从\(1s\)变到\(5s\)时,电量\(q\)从\(5C\)变到\(265C\),
此时电量\(q\)关于时间\(t\)的平均变化率为\(\dfrac{q(5)-q(1)}{5-1}=\dfrac{265-5}{5-1}=65(\mathrm{C} / \mathrm{s})\),
它表示从\(1s\)到\(5s\)这段时间内,平均每秒通过该导体横截面的电量为\(65C\),
(2)\(\because q' (t)=6t^2+3\), \(\therefore q' (5)=6×5^2+3=153(C/s)\)
它表示在\(t=5s\)时,每秒通过该导体横截面的电量为\(153C\),即电流强度为\(153C/s\).
【题型2】 导数的几何意义
【典题1】已知曲线\(y=2x^3+4\).
(1)求曲线在点\(P(-1,2)\)处的切线方程;
(2)求曲线过点\(P(-1,2)\)的切线方程;
(3)求斜率为\(24\)的切线方程.
解析 (1) \(y=2x^3+4\)的导数\(y'=6x^2\),则切线的斜率为\(6\),
则切线方程为:\(y-2=6(x+1)\),
即为\(y=6x+8\);
(2)若点\(P\)是切点,由题意可知切线方程为\(y=6x+8\);
若点\(P\)不是切点,令切点\((m,n) (m≠-1)\),
则切线的斜率为\(6m^2\),由两点的斜率公式得\(\dfrac{n-2}{m+1}=6 m^2\),①
又\(n=2m^3+4\)②,由①②解得,\(m=\dfrac{1}{2}\),则切线的斜率为\(\dfrac{3}{2}\),
即切线方程为\(y-2=\dfrac{3}{2}(x+1)\),即\(3x-2y+7=0\),
故所求的切线方程为:\(y=6x+8\)或\(x-2y+7=0\);
(3) \(y=2x^3+4\)的导数\(y'=6x^2\),
令\(y'=24\),则\(x=±2\),
将\(x=2\)代入曲线方程,得\(y=20\);将\(x=-2\)代入曲线方程,得\(y=-12\).
即切点为\((2,20)\),或\((-2,-12)\).
则切线方程为:\(y-20=24(x-2)\)或\(y+12=24(x+2)\),
即有\(y=24x-28\)或\(y=24x+36\).
点拨 注意"过点\(x=x_0\)”与"在点\(x=x_0\)处"的区别,确定点\(P\)是否切点.
【典题2】若函数\(f(x)=2\ln x+4x^2+bx+5\)的图象上的任意一点的切线斜率都大于\(0\),则\(b\)的取值范围是 \(\underline{\quad \quad}\) .
解析 根据题意,函数\(f(x)=2\ln x+4x^2+bx+5\),其定义域为\((0,+∞)\),
其导数 \(f^{\prime}(x)=\dfrac{2}{x}+8 x+b\),
若函数\(f(x)\)的图象上的任意一点的切线斜率都大于\(0\),
则有\(f^{\prime}(x)=\dfrac{2}{x}+8 x+b>0\)在\((0,+∞)\)上恒成立,
变形可得 \(b>-\left(\dfrac{2}{x}+8 x\right)\)在\((0,+∞)\)上恒成立,
又由 \(\dfrac{2}{x}+8 x \geq 2 \times \sqrt{\dfrac{2}{x} \times 8 x}=8\),
当且仅当\(x=\dfrac{1}{2}\)时等号成立,即 \(\dfrac{2}{x}+8 x\)有最小值\(8\),
所以 \(-\left(\dfrac{2}{x}+8 x\right)\)在\((0,+∞)\)上最大值\(-8\),
若 \(b>-\left(\dfrac{2}{x}+8 x\right)\)在\((0,+∞)\)上恒成立,必有\(b>-8\),
即\(b\)的取值范围为\((-8,+∞)\).
【巩固练习】
1.曲线\(y=x^3+\ln x+1\)在点\((1,2)\)处的切线方程为( )
A.\(3x-y-1=0\) \(\qquad \qquad\) B.\(4x-y-2=0\) \(\qquad \qquad\) C.\(4x+y-6=0\)\(\qquad \qquad\) D.\(3x+y-5=0\)
2.已知直线\(y=kx\)是曲线\(y=e^x\)的切线,则实数\(k\)的值为( )
A. \(\dfrac{1}{e}\) \(\qquad \qquad \qquad \qquad\) B. \(-\dfrac{1}{e}\)\(\qquad \qquad \qquad \qquad\) C.\(-e\) \(\qquad \qquad \qquad \qquad\) D.\(e\)
3.设曲线\(y=e^x\)在点\((0,1)\)处的切线与曲线\(y= \dfrac{1}{x}(x>0)\)上点\(P\)的切线垂直,则\(P\)的坐标为\(\underline{\quad \quad}\).
4.若曲线\(f(x)=ax^2+\ln x\)存在平行于\(x\)轴的切线,则实数\(a\)的取值范围是\(\underline{\quad \quad}\).
5.若直线\(y=3x+1\)是曲线\(y=x^3-a\)的一条切线,求实数\(a\)的值.
参考答案
-
答案 \(B\)
解析 由\(y=x^3+\ln x+1\),得\(y'=3x^2+ \dfrac{1}{x}\),
\(\therefore\)曲线在\((1,2)\)处的斜率 \(k=\left.y^{\prime}\right|_{x=1}=4\),
\(\therefore\)曲线在点\((1,2)\)处的切线方程为\(y-2=4(x-1)\),
即\(4x-y-2=0\).
故选:\(B\). -
答案 \(D\)
解析 曲线\(y=e^x\)的导数为\(y'=e^x\),设切点为\(P(x_0,e^{x_0} )\),
则过\(P\)的切线方程为\(y-e^{x_0}=e^{x_0}(x-x_0 )\)
代入\((0,0)\)点得\(x_0=1\),\(\therefore P(1,e)\),\(\therefore k=e\),
故选:\(D\). -
答案 \((1,1)\)
解析 \(\because f'(x)=e^x\),\(\therefore f'(0)=e^0=1\).
\(\because y=e^x\)在\((0,1)\)处的切线与\(y= \dfrac{1}{x}(x>0)\)上点\(P\)的切线垂直
\(\therefore\)点\(P\)处的切线斜率为\(-1\).
又\(y'=- \dfrac{1}{x^2 }\),设点\(P(x_0,y_0 )\)
\(\therefore-\dfrac{1}{x_0{ }^2}=-1\),\(\therefore x_0=±1\),
\(\because x>0\),\(\therefore x_0=1\),
\(\therefore y_0=1\),\(\therefore 点P(1,1)\). -
答案 \((-∞,0)\)
解析 函数的定义域为\((0,+∞)\),
若\(f(x)=ax^2+\ln x\)存在平行于\(x\)轴的切线,则等价为\(f'(x)=0\)有解,
即\(f'(x)=2ax+ \dfrac{1}{x}=0\)在\((0,+∞)\)上有解,即\(2a=- \dfrac{1}{x^2}\) ,
\(\because x>0\),\(\therefore - \dfrac{1}{x^2 }<0\),
则\(2a<0\),则\(a<0\),
故答案为:\((-∞,0)\). -
答案 \(-3\)或\(1\)
解析 设切点为\(P(x_0,y_0 )\),
对\(y=x^3-a\)求导数是\(y'=3x^2\),\(\therefore 3x_0^2=3\),\(\therefore x_0=±1\).
(1)当\(x=1\)时,
\(\because P(x_0,y_0 )\)在\(y=3x+1\)上,\(\therefore y_0=3×1+1=4\),即\(P(1,4)\).
又\(P(1,4)\)也在\(y=x^3-a\)上,\(\therefore 4=1-a\),\(\therefore a=-3\).
(2)当\(x=-1\)时,
\(\because P(x_0,y_0 )\)在\(y=3x+1\)上,
\(\therefore y_0=3×(-1)+1=-2\),即\(P(-1,-2)\).
又\(P(-1,-2)\)也在\(y=x^3-a\)上,
\(\therefore -2=(-1)^3-a\),\(\therefore a=1\).
综上可知,实数\(a\)的值为\(-3\)或\(1\).
分层练习
【A组---基础题】
1.已知函数 \(f(x)=e^x-\dfrac{e}{2} x^2\),\(f'(x)\)为\(f(x)\)的导函数,若\(f'(a)=f(a)\),则\(a=\)( )
A.\(0\) \(\qquad \qquad \qquad \qquad\) B.\(-1\) \(\qquad \qquad \qquad \qquad\) C.\(2\) \(\qquad \qquad \qquad \qquad\) D.\(0\)或\(2\)
2.已知函数\(f(x)=2x+3f'(0)\cdot e^x\),则\(f'(1)=\)( )
A.\(\dfrac{3}{2} e\) \(\qquad \qquad \qquad \qquad\) B.\(3-2e\) \(\qquad \qquad \qquad \qquad\) C.\(2-3e\) \(\qquad \qquad \qquad \qquad\) D.\(2+3e\)
3.一个质量\(m=5kg\)的物体作直线运动,设运动距离\(s\)(单位:\(m\))与时间\(t\)(单位:\(s\))的关系可用函数\(s(t)=1+t^2\)表示,并且物体的动能\(E_k=\dfrac{1}{2} mv^2\),则物体开始运动后第\(4s\)时的动能是( )
A.\(160J\) \(\qquad \qquad \qquad \qquad\) B.\(165J\) \(\qquad \qquad \qquad \qquad\) C.\(170J\) \(\qquad \qquad \qquad \qquad\) D.\(175J\)
4.已知函数 \(f(\theta)=\dfrac{\sin \theta}{2+\cos \theta}\),则\(f'(0)=\) \(\underline{\quad \quad}\).
5.曲线\(y=x^3+x\)在点\((1,2)\)处的切线方程为\(\underline{\quad \quad}\).
6.现有一个圆柱形空杯子,盛液体部分的底面半径为\(2cm\),高为\(8cm\),用一个注液器向杯中注入溶液,已知注液器向杯中注入的溶液的容积\(V\)(单位: \(ml\)) 关于时间(单位: \(s\)) 的函数解析式为\(V=πt^3+3πt^2 (t≥0)\),不考虑注液过程中溶液的流失,则当\(t=2\)时,杯中溶液上升高度的瞬时变化率为\(\underline{\quad \quad}\) .
7.曲线\(y=\ln x- \dfrac{1}{x}\)在\(x=1\)处的切线的倾斜角为\(α\),则\(\sin 2α=\)\(\underline{\quad \quad}\).
8.已知函数\(y=e^x\)的图象在点\((a_k,e^{a_k})\)处的切线与\(x\)轴的交点的横坐标为\(a_{k+1}\),其中\(k∈N^*\),\(a_1=0\),则\(a_1+a_3+a_5=\) \(\underline{\quad \quad}\).
9.如图,煤场的煤堆形如圆锥,设圆锥母线与底面所成角为\(α\).
(1) 高\(h\)与半径\(r\)有什么关系?
(2) 传输带以\(0.3m^3/min\)送煤,当半径\(r=1.7m\)时,求\(r\)对时间\(t\)的变化率.
(参考数据:\(π\)取\(3.14\),\(1.7^2=2.89\),\(1.7^3≈4.91\),为计算方便可取\(3.14×2.89≈9\),\(3.14×4.91≈15\))
10.已知函数\(f(x)=x^3-4x^2+5x-4\),求经过点\(A(2,-2)\)的曲线\(f(x)\)的切线方程.
参考答案
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答案 \(D\)
解析 因为\(f'(x)=e^x-ex\),根据条件得 \(e^a-\dfrac{e}{2} a^2=e^a-e a\),
解得\(a=0\)或\(2\).故选:\(D\). -
答案 \(C\)
解析 \(f' (x)=2+3f' (0)\cdot e^x\),
\(\therefore f'(0)=2+3f'(0)\),解得\(f' (0)=-1\),
\(\therefore f'(x)=2-3e^x\),\(\therefore f' (1)=2-3e\).
故选:\(C\). -
答案 \(A\)
解析 根据题意,物体的运动距离\(s\)与时间\(t\)的关系式为\(s(t)=1+t^2\),
则有\(s'(t)=2t\),
物体开始运动后第\(4s\)时速度\(v=s'(4)=8\),
物体开始运动后第\(4s\)时的动能\(E=\dfrac{1}{2}×mv^2=\dfrac{1}{2}×5×64=160J\);
故选:\(A\). -
答案 \(\dfrac{1}{3}\)
解析 函数\(f(\theta)=\dfrac{\sin \theta}{2+\cos \theta}\),
则\(f^{\prime}(\theta)=\dfrac{\cos \theta(2+\cos \theta)-\sin \theta \times(-\sin \theta)}{(2+\cos \theta)^2}=\dfrac{1+2 \cos \theta}{(2+\cos \theta)^2}\),
所以\(f^{\prime}(0)=\dfrac{1+2}{(2+1)^2}=\dfrac{1}{3}\). -
答案 \(4x-y-2=0\)
解析 \(y=x^3+x\)的导数为\(y'=3x^2+1\),
可得曲线\(y=x^3+x\)在点\((1,2)\)处的切线斜率为\(k=3+1=4\),
则切线的方程为\(y-2=4(x-1)\),
即为\(4x-y-2=0\). -
答案 \(6cm/s\)
解析 由题设,杯子底面积为\(S=4π\),
则溶液上升高度 \(h=\dfrac{V}{S}=\dfrac{t^3+3 t^2}{4}\),
所以\(h^{\prime}=\dfrac{3 t^2+6 t}{4}\),则 \(h^{\prime} |_{t=2}=6 \mathrm{~cm} / \mathrm{s}\). -
答案 \(\dfrac{4}{5}\)
解析 由\(y=\ln x- \dfrac{1}{x}\),得\(y'= \dfrac{1}{x}+ \dfrac{1}{x^2 }\),
\(\therefore\)曲线\(y=\ln x- \dfrac{1}{x}\)在\(x=1\)处的切线斜率\(k=2\),
\(\because\) 曲线\(y=\ln x- \dfrac{1}{x}\)在\(x=1\)处的切线的倾斜角为\(α\),
\(\therefore \tan α=2\), \(\therefore \sin 2 \alpha=2 \sin \alpha \cos \alpha=\dfrac{2 \tan \alpha}{1+\tan ^2 \alpha}=\dfrac{4}{5}\).
故答案为: \(\dfrac{4}{5}\). -
答案 \(-6\)
解析 \(\because y=e^x\),\(\therefore y'=e^x\),
\(\therefore y=e^x\)在点\((a_k,e^{a_k})\)处的切线方程是 \(y-e^{a_k}=e^{a_k}\left(x-a_k\right)\),
整理,得 \(e^{a_k} x-y-a_k e^{a_k}+e^{a_k}=0\),
\(\because\)切线与\(x\)轴交点的横坐标为\(a_{k+1}\),
\(\therefore a_{k+1}=a_k-1\),
\(\therefore\left\{a_n\right\}\)是首项为\(a_1=0\),公差\(d=-1\)的等差数列,
\(\therefore a_1+a_3+a_5=0-2-4=-6\).
故答案为:\(-6\). -
答案 (1)\(h=r\tan α\);(2)\(\dfrac{0.033}{\tan \alpha}\) .
解析 (1)由题意知, \(\tan \alpha=\dfrac{h}{r}\),\(\therefore h=r\tan α\).
(2)记\(t min\)时煤堆的体积为\(V\),
则 \(V=\dfrac{1}{3} \pi r^2 h=\dfrac{1}{3} \pi r^3 \tan \alpha=0.3 t\)①
\(\therefore r=\sqrt[3]{\dfrac{0.9}{\pi \tan \alpha}} t^{\frac{1}{3}}\),②
两边对\(t\)求导, 得 \(r^{\prime}(t)=\dfrac{1}{3} \sqrt[3]{\dfrac{0.9}{\pi \tan \alpha}} t^{-\frac{2}{3}}\)③
设\(r=1.7m\)时对应的时刻为\(t_0\), 由①得 \(t_0=\dfrac{\pi \tan \alpha}{0.9} \times 1.7^3\)
\(\therefore t_0^{-\frac{2}{3}}=\left(\dfrac{\pi \tan \alpha}{0.9}\right)^{-\frac{2}{3}} \times 1.7^{-2}\),
代入③式得\(r\left(t_0\right)=\dfrac{1}{3} \sqrt[3]{\dfrac{0.9}{\pi \tan \alpha}} t_0^{-\frac{2}{3}}=\dfrac{1}{3} \sqrt[3]{\dfrac{0.9}{\pi \tan \alpha}} \times\left(\dfrac{\pi \tan \alpha}{0.9}\right)^{-\frac{2}{3}} \times 1.7^{-2}\)
\(=\dfrac{0.3}{\pi \tan \alpha} \times 1.7^{-2}\approx \dfrac{0.3}{9 \tan \alpha}=\dfrac{0.033}{\tan \alpha}(\mathrm{m} / \mathrm{min})\) . -
答案 \(y+2=0\)或\(x-y-4=0\).
解析 设切点坐标为\(P(a,a^3-4a^2+5a-4)\),
\(\because f(x)=x^3-4x^2+5x-4\),\(\therefore f' (x)=3x^2-8x+5\),
\(\therefore\)切线的斜率为\(f' (a)=3a^2-8a+5\),
由点斜式可得切线方程为\(y-(a^3-4a^2+5a-4)=(3a^2-8a+5)(x-a)\),①
又根据已知,切线方程过点\(A(2,-2)\),
\(\therefore -2-(a^3-4a^2+5a-4)=(3a^2-8a+5)(2-a)\),
即\(a^3-5a^2+8a-4=0\),
\(\therefore (a-1)(a^2-4a+4)=0\),即\((a-1)(a-2)^2=0\),
解得\(a=1\)或\(a=2\),
将\(a=1\)或\(a=2\)代入①可得,切线方程为\(y+2=0\)或\(x-y-4=0\),
故经过点\(A(2,-2)\)的曲线\(f(x)\)的切线方程为\(y+2=0\)或\(x-y-4=0\).
【B组---提高题】
1.已知函数 \(f(x)=\dfrac{3}{e^x+1}+x^3\),其导函数为\(f' (x)\),则\(f(2020)+f(-2020)+f' (2019)-
f'(-2019)\)的值为( )
A.\(1\) \(\qquad \qquad \qquad \qquad\) B.\(2\) \(\qquad \qquad \qquad \qquad\) C.\(3\) \(\qquad \qquad \qquad \qquad\) D.\(4\)
2.已知函数\(f(x)=ax^3+bx^2+cx+d(a≠0)\)的对称中心为\(M(x_0,y_0)\),且点\(M\)在函数\(y=f(x)\)图象上,记函数\(f(x)\)的导函数为\(f'(x)\),\(f'(x)\)的导函数为\(f''(x)\),则有\(f''(x_0)=0\).若函数\(f(x)=x^3-3x^2\),则可求得 \(f\left(\dfrac{1}{2020}\right)+f\left(\dfrac{2}{2020}\right)+\cdots+f\left(\dfrac{4038}{2020}\right)+f\left(\dfrac{4039}{2020}\right)=\)( )
A.\(4039\) \(\qquad \qquad \qquad \qquad\) B.\(-4039\) \(\qquad \qquad \qquad \qquad\) C.\(8078\) \(\qquad \qquad \qquad \qquad\) D.\(-8078\)
3.已知\(M(1,0)\),\(N\)是曲线\(y=e^x\)上一点,则\(|MN|\)的最小值为( )
A.\(1\) \(\qquad \qquad \qquad \qquad\) B. \(\sqrt{2}\) \(\qquad \qquad \qquad \qquad\) C.\(e\) \(\qquad \qquad \qquad \qquad\) D. \(\sqrt{e^4+1}\)
参考答案
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答案 \(C\)
解析 \(f^{\prime}(x)=\dfrac{-3 e^x}{\left(e^x+1\right)^2}+3 x^2\), \(f^{\prime}(-x)=\dfrac{-3 e^x}{\left(e^x+1\right)^2}+3 x^2\)
\(\therefore f'(x)\)为偶函数,\(f' (2019)-f'(-2019)=0\),
因为 \(f(-x)+f(x)=\dfrac{3}{1+e^x}+x^3+\dfrac{3}{1+e^{-x}}-x^3=\dfrac{3}{1+e^x}+\dfrac{3 e^x}{1+e^x}=3\)
所以\(f(2020)+f(-2020)+f' (2019)-f'(-2019)=3\).
故选:\(C\). -
答案 \(D\)
解析 由题意\(f(x)=x^3-3x^2\),则\(f' (x)=3x^2-6x\),\(f″(x)=6x-6\),
由\(f'' (x_0 )=0\),得\(x_0=1\),而\(f(1)=-2\),
故函数\(f(x)=x^3-3x^2\)关于点\((1,-2)\)对称,
即\(f(x)+f(2-x)=-4\).
\(\therefore f\left(\dfrac{1}{2020}\right)+f\left(\dfrac{2}{2020}\right)+\cdots+f\left(\dfrac{4038}{2020}\right)+f\left(\dfrac{4039}{2020}\right)\)
\(=\left[f\left(\dfrac{1}{2020}\right)+f\left(\dfrac{4039}{2020}\right)\right]+\left[f\left(\dfrac{2}{2020}\right)+f\left(\dfrac{4038}{2020}\right)\right]+\cdots\) \(+\left[f\left(\dfrac{2019}{2020}\right)+f\left(\dfrac{2021}{2020}\right)\right]+f\left(\dfrac{2020}{2020}\right)\)
\(=-4×2019+(-2)=-8078\),
故选:\(D\). -
答案 \(B\)
解析 \(y=e^x\)的导数为\(y'=e^x\).设\(N(m,e^m)\),可得过\(N\)的切线的斜率为\(e^m\).
当\(MN\)垂直于切线时,\(|MN|\)取得最小值,
可得 \(\dfrac{e^m}{m-1}=-\dfrac{1}{e^m}\) ,则\(e^{2m}+m=1\).
因为\(f(x)=e^{2x}+x\)单调递增,且\(f(0)=1\),所以\(m=0\).
所以\(|MN|\)的最小值为 \(\sqrt{1^2+1^2}=\sqrt{2}\).
故选:\(B\).
【C组---拓展题】
1.设\(f(x)=x(x-1)(x-2)(x-3)……(x-99)\),则\(f'(0)=\) \(\underline{\quad \quad}\).
2.设曲线\(f(x)=e^x+2x\)(\(e\)为自然对数的底数)上任意一点处的切线为\(l_1\),总存在曲线\(g(x)=-ax+\sin x\)上某点处的切线\(l_2\),使得\(l_1⊥l_2\),则实数\(a\)的取值范围为\(\underline{\quad \quad}\) .
3.若直线\(y=kx+b\)是曲线\(f(x)=e^x\)与曲线\(g(x)=\ln x+2\)的公切线,求\(k\),\(b\).
4.求 \(\lim\limits _{x \rightarrow 0} \dfrac{\sin x}{x}\).
参考答案
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答案 \(-99!\)
解析 等式两边取对数得:\(\ln f(x)=\ln x+\ln (x-1)+\ln (x-1)+⋯+\ln (x-99)\),
求导得: \(f^{\prime}(x)=f(x)\left[\dfrac{1}{x}+\dfrac{1}{x-1}+\dfrac{1}{x-2}+\cdots+\dfrac{1}{x-99}\right]\)
\(=[(x-1)(x-2)…(x-99)+x(x-2)…(x-99)+x(x-1)(x-2)…(x-98)]\)
代入\(0\),\(f' (0)=-1\cdot (-2)\cdot (-3)\cdot \cdot \cdot (-99)=-99!\) -
答案 \(\left[-\dfrac{1}{2},1\right]\)
解析 \(f(x)=e^x+2x\)的导数为\(f' (x)=e^x+2\),
设\((x_1,y_1 )\)为\(f(x)\)上的任一点,
则过\((x_1,y_1 )\)处的切线\(l_1\)的斜率为 \(k_1=e^{x_1}+2\),
\(g(x)=-ax+\sin x\)的导数为\(g' (x)=\cos x-a\),
过\(g(x)\)图象上一点\((x_2,y_2 )\)处的切线\(l_2\)的斜率为\(k_2=-a+\cos x_2\).
由\(l_1⊥l_2\),可得 \(\left(e^{x_1}+2\right) \cdot\left(-a+\cos x_2\right)=-1\),
即\(-a+\cos x_2=-\dfrac{1}{e^{x_1+2}}\),
任意的\(x_1∈R\),总存在\(x_2∈R\)使等式成立.
则有\(y_1=-a+\cos x_2\)的值域为\(A=[-a-1,-a+1]\).
\(\dfrac{1}{e^{x_1+2}}\)的值域为\(B=\left(-\dfrac{1}{2}, 0\right)\),
有\(B⊆A\),即 \(\left(-\dfrac{1}{2}, 0\right) \subseteq[-a-1,-a+1]\),
即 \(\left\{\begin{array}{l} -a-1 \leq-\dfrac{1}{2} \\ 1-a \geq 0 \end{array}\right.\),解得:\(\left[-\dfrac{1}{2},1\right]\). -
答案 \(\left\{\begin{array}{l} k=e \\ b=0 \end{array}\right.\)或 \(\left\{\begin{array}{l} k=1 \\ b=1 \end{array}\right.\).
解析 对函数\(y=e^x\)求导,得\(y'=e^x\),对函数\(y=\ln x+2\)求导,得\(y'= \dfrac{1}{x}\).
设直线\(y=kx+b\)与\(y=e^x\)切于点 \(P\left(x_1, e^{x_1}\right)\),与\(y=\ln x+2\)切于\(Q(x_2,\ln x_2+2)\).
则在点\(P\)处的切线方程为: \(y-e^{x_1}=e^{x_1}\left(x-x_1\right)\),
即 \(y=e^{x_1} x+\left(1-x_1\right) e^{x_1}\).
在点\(Q\)处的切线方程为: \(y-\ln x_2-2=\dfrac{1}{x_2}\left(x-x_2\right)\),
即\(y= \dfrac{1}{x}_2 x+\ln x_2+1\).
这两条直线为同一条直线,所以有 \(\left\{\begin{array}{l} e^{x_1}=\dfrac{1}{x_2} \\ \left(1-x_1\right) e^{x_1}=\ln x_2+1 \end{array}\right.\)
故 \(\dfrac{\left(1-x_1\right)\left(1-x_2\right)}{x_2}=0\),则\(x_1=1\)或\(x_2=1\).
当\(x_1=1\)时,切线方程为\(y=ex\),所以 \(\left\{\begin{array}{l} k=e \\ b=0 \end{array}\right.\).
当\(x_2=1\)时,切线方程为\(y=x+1\),所以 \(\left\{\begin{array}{l} k=1 \\ b=1 \end{array}\right.\). -
答案 \(1\)
解析 设\(f(x)=\sin x\),由求导公式可知\(f' (x)=\cos x\),
\(\lim\limits _{x \rightarrow 0} \dfrac{\sin x}{x}=\lim\limits _{x \rightarrow 0} \dfrac{\sin x-\sin 0}{x-0}\)(构造出导数的形式)
\(=\lim\limits _{x \rightarrow 0} \dfrac{f(x)-f(0)}{x-0}=f^{\prime}(0)\) (由导数的概念可知)
\(=1\).