课程： 7.3 指数函数和对数函数的衍生物

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选择节指数函数和对数函数的衍生物

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指数函数和对数函数的衍生物
Using the limit definition of the derivative, $\begin{align*}f^{\prime}(x) = \lim\limits_{h\rightarrow 0} \frac{f(x+h) - f(x)}{h}\end{align*}$ , it is possible to determine the of the exponential function $\begin{align*}f(x)=b^x\end{align*}$ , and the logarithm function $\begin{align*}f(x) = \log_b x\end{align*}$ . Before you proceed, see if you can apply the definition to obtain these derivatives; then compare your results with the derivations in the Guidance section. Try letting the $\begin{align*}b = e\end{align*}$ to simplify the work. Your results should show that the rate of change of the exponential varies directly with the exponential; and the rate of change of the logarithm varies inversely with $\begin{align*}x\end{align*}$ .
::使用衍生物的限值定义 fä( x) =limh0f( x+h) - f( x) h, 有可能确定指数函数 f( x) =bx 和对数函数 f( x) =logb* x 。在您继续之前, 查看您能否应用该定义获取这些衍生物; 然后比较您的结果和“ 指导” 一节中的衍生物。尝试让 b=e 来简化工作。您的结果应该显示指数变化率与指数直接不同; 对数变化率与 x 反差。

Derivatives of Exponential and Logarithmic Functions
::指数函数和对数函数的衍生物

We will look first at the derivative of , and then the derivative of the exponential function.
::我们首先研究... 的衍生物,然后研究指数函数的衍生物。

The Derivative of a Logarithmic Function
::对数函数的衍生要素

Given the logarithmic function $\begin{align*}f(x) = \log_b x\end{align*}$ , where $\begin{align*}x>0\end{align*}$ , and the base $\begin{align*}b>0\end{align*}$ and $\begin{align*}b \neq 1\end{align*}$ , then:
::根据对数函数 f(x) =logbx, 其中 x>0, 以及基数 b>0 和 b1, 然后 :

$\begin{aligned} \frac{d}{d x} [\log_{b} x] = \frac{1}{x \log_{e} b} = \frac{1}{x \ln b} \end{aligned}$
$\begin{align*}\frac{d}{dx}[\log_b x] = \frac{1}{x \log_e b} = \frac{1}{x \ln b}\end{align*}$
::ddx [logbx] =1xlogeb=1xlnb

where $\begin{align*}e\end{align*}$ is the base of the natural logarithm.
::e 是自然对数的基数。

Given the logarithm function $\begin{align*}f(x) = \log_b u\end{align*}$ , where $\begin{align*}u = g(x) > 0\end{align*}$ and $\begin{align*}g(x)\end{align*}$ is a differentiable function, then by the :
::鉴于对数函数 f( x) =logbu, 即 u=g( x) > 0 和 g( x) 是可区别的函数, 则由 :

$\begin{aligned} \frac{d}{d x} [\log_{b} u] = \frac{1}{u \log_{e} b} \cdot \frac{d u}{d x} = \frac{1}{u \ln b} \cdot \frac{d u}{d x} \end{aligned}$
$\begin{align*}\frac{d}{dx}[\log_b u] = \frac{1}{u \log_e b} \cdot \frac{du}{dx} = \frac{1}{u \ln b} \cdot \frac{du}{dx}\end{align*}$
::ddx [logbu] = 1lugebdudx = 1ulnbdudx

Note that $\begin{align*}\frac{d}{dx} [\ln u] = \frac{1}{u}\frac{du}{dx}\end{align*}$
::注意 ddx [lnu] = 1ududdx

Apply the equations from above and find the derivative of each of the functions:
::应用上面的方程式并找到每个函数的衍生物 :

$\begin{align*}y = x^3 \log_5 2x\end{align*}$
::y=x3log5=%2x

For $\begin{align*}y=x^3 \log_5 2x\end{align*}$ , we use the Product Rule along with the derivative formula
::y=x3log52x,我们使用产品规则以及衍生公式

$\begin{aligned} \frac{d}{d x} [\log_{b} u] & = \frac{1}{u \ln b} \frac{d u}{d x} \\ \frac{d}{d x} [x^{3} \log_{5} 2 x] & = x^{3} \cdot \frac{d}{d x} [\log_{5} 2 x] + \frac{d}{d x} [x^{3}] \cdot \log_{5} 2 x \\ = x^{3} \cdot \frac{1}{x \ln 5} + 3 x^{2} \cdot \log_{5} 2 x \\ = \frac{x^{2}}{\ln 5} + 3 x^{2} \log_{5} 2 x . \end{aligned}$
$\begin{align*}\frac{d}{dx} [\log_b u] & = \frac{1}{u \ln b} \frac{du}{dx}\\ \frac{d}{dx} [x^3 \log_5 2x] & = x^3 \cdot \frac{d}{dx} [\log_5 2x] + \frac{d}{dx} [x^3] \cdot \log_5 2x\\ & = x^3 \cdot \frac{1}{x \ln 5} + 3x^2 \cdot \log_5 2x\\ & = \frac{x^2}{\ln 5} + 3x^2 \log_5 2x.\end{align*}$
::dx[logbu] = 1ulnbdudxdx[x3log5}2x] =x3{ddx[log5}2x] +ddx[x3] }}}}[x3}log5}2x=x3}1x_1xln}5+3x2x2}log5}2x=x2ln}5+3x2x2log5}#2x

$\begin{align*}y = \ln (2x^2 -4x+3)\end{align*}$
::y=ln( 2x2 - 4x+3)

For $\begin{align*}y = \ln (2x^2-4x+3)\end{align*}$ , let $\begin{align*}u = 2x^2-4x+3\end{align*}$ .
::y=ln( 2x2 - 4x+3) , let u= 2x2 - 4x+3 。

$\begin{aligned} \frac{d}{d x} [\ln u] & = \frac{1}{u} \frac{d u}{d x} \\ \frac{d u}{d x} & = 4 x - 4 \\ \frac{d y}{d x} & = \frac{1}{2 x^{2} - 4 x + 3} \frac{d}{d x} [2 x^{2} - 4 x + 3] \\ = \frac{1}{2 x^{2} - 4 x + 3} (4 x - 4) \\ = \frac{4 (x - 1)}{2 x^{2} - 4 x + 3} . \end{aligned}$
$\begin{align*}\frac{d}{dx} [\ln u] & = \frac{1}{u}\frac{du}{dx}\\ \frac{du}{dx} & = 4x-4\\ \frac{dy}{dx} & = \frac{1}{2x^2-4x+3} \frac{d}{dx}[2x^2-4x+3]\\ & = \frac{1}{2x^2-4x+3}(4x-4)\\ & = \frac{4(x-1)}{2x^2-4x+3}.\end{align*}$
::dx[lnu] = 1ududxduddudx=4x4-4uddx=12x2-4x+3ddx[2x2-4x+3] =12x2-4x+3] =12x2-4x+3(4x-4)=4(x-1)2x2-4x+3。

Derivation of the Derivative of a Logarithmic Function
::对对数函数衍生的衍生结果

The above results for the derivative of a logarithmic function are obtained by using the limit definition of the derivative that you already studied:
::对数函数衍生物的上述结果,通过使用您已经研究过的对数函数衍生物的限值定义获得:

$\begin{align*}f^{\prime} (x) = \lim\limits_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h} = \lim\limits_{w \rightarrow x} \frac{f(w) - f(x)}{w-x}\end{align*}$ .
::f_(x) = limh_0f(x+h) - f(x) = limw_xf(w) - f(x) w-x。

To get the derivative of $\begin{align*}y = \log_b x\end{align*}$ , use the limit definition of the derivative and the rules of logarithms:
::要获取 y=logbx 的衍生物, 请使用衍生物的限值定义和对数规则 :

$\begin{aligned} \frac{d}{d x} [\log_{b} x] & = lim_{w \to x} \frac{\log_{b} w - \log_{b} x}{w - x} \\ = lim_{w \to x} \frac{\log_{b} (\frac{w}{x})}{w - x} \\ = lim_{w \to x} [\frac{1}{w - x} \log_{b} (\frac{w}{x})] \\ = lim_{w \to x} [\frac{1}{w - x} \log_{b} (\frac{x + (w - x)}{x})] \\ = lim_{w \to x} [\frac{1}{w - x} \log_{b} (1 + \frac{w - x}{x})] \\ = lim_{w \to x} [\frac{1}{x (w - x)} \log_{b} (1 + \frac{w - x}{x})] \\ = lim_{w \to x} [\frac{x}{x (w - x)} \log_{b} (1 + \frac{w - x}{x})] . \end{aligned}$
$\begin{align*}\frac{d}{dx} [\log_b x] & = \lim_{w \rightarrow x} \frac{\log_b w - \log_b x}{w-x}\\ & = \lim_{w \rightarrow x} \frac{\log_b \left ( \frac{w}{x} \right )}{w-x}\\ & = \lim_{w \rightarrow x} \left [ \frac{1}{w-x} \log_b \left ( \frac{w}{x} \right )\right ]\\ & = \lim_{w \rightarrow x} \left [ \frac{1}{w-x} \log_b \left ( \frac{x+(w-x)}{x} \right )\right ]\\ & = \lim_{w \rightarrow x} \left [ \frac{1}{w-x} \log_b \left ( 1+ \frac{w-x}{x} \right )\right ]\\ & = \lim_{w \rightarrow x} \left [ \frac{1}{x(w-x)} \log_b \left ( 1+ \frac{w-x}{x} \right )\right ]\\ & = \lim_{w \rightarrow x} \left [ \frac{x}{x(w-x)} \log_b \left ( 1+ \frac{w-x}{x} \right )\right ]. \end{align*}$
::dx[logbx] = limww}xlogb}xw}xw_x=limw}xlogb}(wx) w -x=limw}x[1w-xlogb}(wx)] = limw}x[1w-x}(xx) =limw}x[1w-xlogb}(1+w-xx) = limw}x[1+x)xlogb}(1+w_x)] =limw}x}[1x(wx)x)x logb}(1+w_x)]

At this stage, let $\begin{align*}a=\frac{(w-x)}{(x)}\end{align*}$ , so that the limit of $\begin{align*}w \rightarrow x\end{align*}$ then becomes $\begin{align*}a \rightarrow 0\end{align*}$ .
::在现阶段,让我们a=(w-x)(x),这样w__x的极限就变成a_0。

Substituting, we get
::替代,我们得到

$\begin{aligned} = lim_{a \to 0} [\frac{1}{x} \frac{1}{a} \log_{b} (1 + a)] \\ = \frac{1}{x} lim_{a \to 0} [\frac{1}{a} \log_{b} (1 + a)] \\ = \frac{1}{x} lim_{a \to 0} [\log_{b} (1 + a)^{\frac{1}{a}}] . \end{aligned}$
$\begin{align*}& = \lim_{a\rightarrow 0} \left [ \frac{1}{x} \frac{1}{a} \log_b (1+a) \right ]\\ & = \frac{1}{x} \lim_{a\rightarrow 0} \left [ \frac{1}{a} \log_b (1+a) \right ]\\ & = \frac{1}{x} \lim_{a\rightarrow 0} \left [ \log_b (1+a)^{\frac{1}{a}} \right ].\end{align*}$
::=lima%0[1x1logb(1+a)]=1xlima0[1alogb(1+a)]=1xlima0[1+a)1a]。

Inserting the limit,
::插入限制,

$\begin{align*}= \frac{1}{x} \log_b \left [ \lim\limits_{a\rightarrow 0} (1+a)^{\frac{1}{a}} \right ]\end{align*}$ .
::=1xlogb[lima%0(1+a)1a]。

But by the definition $\begin{align*}e = \lim\limits_{a \rightarrow 0} (1+a)^{\frac{1}{a}}\end{align*}$ ,
::但根据e=lima%0(1+a)1a的定义

$\begin{align*}\frac{d}{dx} [\log_b x] = \frac{1}{x} \log_b e\end{align*}$ .
::ddx[logbx]=1xlogbe。

We can express $\begin{align*}\log_b e\end{align*}$ in terms of natural logarithm by the using the formula: $\begin{align*}\log_b w = \frac{\ln w}{\ln b}\end{align*}$ .
::我们可以使用公式(logbw=lnwlnb)来表示自然对数的对数对数。

Then
::然后

$\begin{align*}\log_b e = \frac{\ln e}{\ln b} = \frac{1}{\ln b}\end{align*}$ .
::logbe=lnelnb=1lnb。

Thus we conclude
::因此,我们就此结束结束

$\begin{align*}\frac{d}{dx} [\log_b x] = \frac{1}{x \ln b}\end{align*}$ .
::dx[logbx]=1xlnb。

In the special case where $\begin{align*}b = e\end{align*}$ ,
::在特别情况下,b=e,

$\begin{align*}\frac{d}{dx} [\ln x] = \frac{1}{x}\end{align*}$ .
::ddx[lnx]=1x。

The Chain Rule can be used to obtain the generalized derivative rule for logarithmic functions, if $\begin{align*}u\end{align*}$ is a differentiable function of $\begin{align*}x\end{align*}$ and if $\begin{align*}u(x)>0\end{align*}$ .
::链规则可用于获取对数函数的通用衍生规则,如果u是x的不同函数,如果u(x)>0。

Students often wonder why the constant $\begin{align*}e\end{align*}$ is defined the way it is. The answer is in the derivative of $\begin{align*}f(x) = \ln x\end{align*}$ . With any other base, the derivative of $\begin{align*}f(x) = \log_b x\end{align*}$ would be equal to the more complicated expression $\begin{align*}f^{\prime} (x) = \frac{1}{x \ln b} \end{align*}$ , rather than $\begin{align*}\frac{1}{x}\end{align*}$ . This is similar to the situation where the derivative of $\begin{align*}f(x) = \sin x\end{align*}$ is the simple expression $\begin{align*}f^{\prime} (x) = \cos (x)\end{align*}$ only if $\begin{align*}x\end{align*}$ is in radians. In degrees, $\begin{align*}f^{\prime} (x) = \frac{\pi}{180} \cos (x)\end{align*}$ , which is more cumbersome and harder to remember.
::学生往往想知道为什么对常数 e 进行定义。答案在 f( x) = lnx 的衍生物中。在任何其他基数中, f( x) =logbx 的衍生物等于更复杂的表达式 fä( x) = 1xlnb, 而不是 1x。这类似于 f( x) = sinx 的衍生物只有 x 以弧度表示才为简单表达式 f\( x) = cos( x) 的情况。在度上, f\( x) = 180cos( x) 更麻烦, 更难记住。

Derivatives of Exponential Functions
::指数函数的衍生因素

Given the exponential function $\begin{align*}f(x) = b^x\end{align*}$ , where the base $\begin{align*}b\end{align*}$ is a positive, real number, then:
::根据指数函数 f( x) =bx, b b 是正数, 实际数为正数, 那么 :

$\begin{aligned} \frac{d}{d x} [b^{x}] = \ln b \cdot b^{x} \end{aligned}$
$\begin{align*}\frac{d}{dx} [b^x] = \ln b \cdot b^x\end{align*}$
::ddx [bx] =ln bx

Given the exponential function $\begin{align*}f(x) = b^u \end{align*}$ , where $\begin{align*}u=g(x)\end{align*}$ and $\begin{align*}g(x)\end{align*}$ is a differentiable function, then:
::鉴于指数函数 f( x) =bu, 即 u=g( x) 和 g( x) 是可区别函数, 则 :

$\begin{aligned} \frac{d}{d x} [b^{u}] = (\ln b \cdot b^{u}) \frac{d u}{d x} \end{aligned}$
$\begin{align*}\frac{d}{dx} [b^u] = (\ln b \cdot b^u) \frac{du}{dx}\end{align*}$
::ddx[bu] = (ln_bbu) dudx

Using the equations from above, find the derivative of the following functions
::使用上面的方程,找到下列函数的衍生函数

$\begin{align*}y=2^{x^2}\end{align*}$
::y=2x2 y=2x2

For $\begin{align*}y=2^{x^2}\end{align*}$ , let $\begin{align*}u = x^2\end{align*}$ ,
::y=2x2,让u=x2,

$\begin{align*}\frac{d}{dx} [b^u] = (\ln b \cdot b^u) \frac{du}{dx}\end{align*}$
::ddx[bu] = (ln_bbu) dudx

$\begin{aligned} y^{'} & = (2 x) 2^{x^{2}} \ln 2 \\ = 2^{x^{2} + 1} \cdot x \cdot \ln 2. \end{aligned}$
$\begin{align*}y^\prime & = (2x) 2^{x^2} \ln 2\\ & = 2^{x^2 + 1} \cdot x \cdot \ln 2.\end{align*}$
::y( 2x) 2x2ln2=2x2+1xxxln2。

$\begin{align*}y = e^{x^2}\end{align*}$
::y=ex2 y=ex2

For $\begin{align*}y=e^{x^2}\end{align*}$ , let $\begin{align*}u = x^2\end{align*}$ .
::y=ex2,让我们u=x2。

$\begin{aligned} \frac{d}{d x} [e^{u}] & = u^{'} e^{u}, \\ y^{'} & = 2 x e^{x^{2}} . \end{aligned}$
$\begin{align*}\frac{d}{dx} [e^u] & = u^\prime e^u,\\ y^\prime & = 2xe^{x^2}.\end{align*}$
::ddx[eu]=ueu,y2xex2。

Derivation of the Derivative of an Exponential Function
::指向函数衍生的衍生要素

We have discussed above that the exponential function is simply the inverse function of the logarithmic function. To obtain a derivative formula for the exponential function with base $\begin{align*}b\end{align*}$ we rewrite $\begin{align*}y=b^x\end{align*}$ as
::我们在上文中讨论过,指数函数仅仅是对数函数的反函数。要获得指数函数的衍生公式,要使用 b 的基 b 我们重写 y=bx 作为

$\begin{align*}x = \log_b y\end{align*}$ .
::x=logby. =logby。 =logby。 =logby。 =logby。 =logby。

Differentiating implicitly,
::隐含地区别对待,

$\begin{align*}1 = \frac{1}{y \ln b} \cdot \frac{dy}{dx}\end{align*}$ .
::1=1ylnbdex。

Solving for $\begin{align*}\frac{dy}{dx}\end{align*}$ and replacing $\begin{align*}y\end{align*}$ with $\begin{align*}b^x\end{align*}$ ,
::用 bx 替换 y 和 dydx 的溶解方法,

$\begin{align*}\frac{dy}{dx} = y \ln b = b^x \ln b\end{align*}$ .
::丁丁二烯b=bxlnb。

Thus the derivative of an exponential function is
::因此指数函数的衍生物是

$\begin{align*}\frac{d}{dx} [b^x] = b^x \ln b\end{align*}$ .
::dx[bx]=bxlnb。

In the special case where the base is $\begin{align*}b^x = e^x\end{align*}$ , since $\begin{align*}\ln e=1\end{align*}$ the derivative rule becomes
::在基准为bx=ex的特殊情况中,由于Ine=1,衍生规则成为

$\begin{align*}\frac{d}{dx} [e^x] = e^x\end{align*}$ .
::dx[ex]=ex。

To generalize, if $\begin{align*}u\end{align*}$ is a differentiable function of $\begin{align*}x\end{align*}$ , with the use of the Chain Rule the above derivatives take the general form
::如果u是x的不同函数,则通过使用链条规则,上述衍生物具有一般形式。

$\begin{align*}\frac{d}{dx} [b^u] = b^u \cdot \ln b \cdot \frac{du}{dx}\end{align*}$ .
::ddx[bu] = bunbdudx。

And if $\begin{align*}b=e\end{align*}$ ,
::如果b=e,

$\begin{align*}\frac{d}{dx} [e^u] = e^u \cdot \frac{du}{dx}\end{align*}$ .
::ddx[eu]=eududx。

Examples
::实例

Example 1
::例1

Write and equation for the tangent line of the function $\begin{align*}f(x) = e^{x^3} \ln \left ( \frac{1}{x+1} \right )\end{align*}$ at $\begin{align*}x=1\end{align*}$ .
::x=1 的函数 f( x) =ex3ln( 1x+1) 的正切线的写法和方程式。

For $\begin{align*}x=1\end{align*}$ , $\begin{align*}f(1) = e \ln \left ( \frac{1}{2} \right ) = -e \ln 2\end{align*}$ . The tangent line will be at the point $\begin{align*}(1, -e \ln 2)\end{align*}$ .
::x=1, f(1)=eln( 12) eln2. 切线将在点(1, -eln2)。

The derivative of the function is:
::函数的衍生物为:

$\begin{aligned} \frac{d}{d x} [e^{x^{3}} \ln (\frac{1}{x + 1})] & = 3 x^{2} e^{x^{3}} \ln (\frac{1}{x + 1}) + e^{x^{3}} (x + 1) \frac{- 1}{(x + 1)^{2}} \\ = [3 x^{2} \ln (\frac{1}{x + 1}) - \frac{1}{(x + 1)}] e^{x^{3}} \end{aligned}$
$\begin{align*}\frac{d}{dx} \left [ e^{x^3} \ln \left ( \frac{1}{x+1} \right ) \right ] & = 3x^2 e^{x^3} \ln \left ( \frac{1}{x+1} \right ) + e^{x^3} (x+1) \frac{-1}{(x+1)^2}\\ & = \left [ 3x^2 \ln \left ( \frac{1}{x+1} \right ) - \frac{1}{(x+1)} \right ] e^{x^3}\end{align*}$
::dx[ ex3ln( 1x+1)] = 3x2ex3ln( 1x+1) +ex3( x+1) - 1( x+1) 2= [3x2ln( 1x+1) - 1x+1)]ex3

Therefore $\begin{align*}f^{\prime} (1) = -e \left ( \frac{6 \ln 2+1}{2} \right ) \end{align*}$ .
::因此,fä(1) e(6ln) 2+12。

The equation of the tangent line at $\begin{align*}(1, -e \ln 2)\end{align*}$ can be expressed in point-slope form as:
::相切线(1,-eln2)的等式可以用点斜体表示如下:

Point-slope form: $\begin{align*}y + e \ln 2 = -e \left ( \frac{6 \ln 2+1}{2} \right ) (x-1)\end{align*}$ .
::point- slope 窗体: y+eln2e( 6ln2+12) (x-1) 。

Slope-intercept form: $\begin{align*}y = -e \left ( \frac{6 \ln 2+1}{2} \right ) x + e \left ( \frac{4 \ln 2+1}{2} \right )\end{align*}$
::斜坡- 截取窗体: ye( 6ln2+12) x+e( 4ln2+12)

Example 2
::例2

Find $\begin{align*}f^{\prime} (x)\end{align*}$ , for the function $\begin{align*}f(x) = \frac{1}{\sqrt{\pi \sigma}} e^{-\alpha k(x-x_0)^2}\end{align*}$ , where $\begin{align*}\sigma, \alpha, x_0,\end{align*}$ and $\begin{align*}k\end{align*}$ are constants and $\begin{align*}\sigma \neq 0\end{align*}$ .
::对于函数 f( x) =1\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\F\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

The function $\begin{align*}f(x) = \frac{1}{\sqrt{\pi \sigma}} e^{-\alpha k(x-x_0)^2}\end{align*}$ , can be thought of as $\begin{align*}f(x) = \frac{1}{\sqrt{\pi \sigma}} e^{u(x)}\end{align*}$ , where $\begin{align*}u(x) = - \alpha k(x-x_0)^2\end{align*}$ .
::函数 f( x) = 1 @ e @ k( x- x0) 2 , 也可以视为 f( x) = 1 @ eu( x) , 其中 u( x) @ k( x- x0) 2 。

We apply the exponential derivative and the Chain Rule:
::我们运用指数衍生物和链规则:

$\begin{aligned} f^{'} (x) & = \frac{d}{d x} [\frac{1}{\sqrt{π σ}} e^{u (x)}] \\ = \frac{1}{\sqrt{π σ}} e^{u (x)} \frac{d}{d x} u (x) \\ = \frac{1}{\sqrt{π σ}} e^{- α k (x - x_{0})^{2}} \frac{d}{d x} [- α k (x - x_{0})^{2}] \\ = \frac{- 2 α k (x - x_{0})^{2}}{\sqrt{π σ}} e^{- α k (x - x_{0})^{2}} \end{aligned}$
$\begin{align*}f^{\prime}(x) & = \frac{d}{dx} \left [ \frac{1}{\sqrt{\pi \sigma}} e^{u(x)} \right ]\\ & = \frac{1}{\sqrt{\pi \sigma}} e^{u(x)} \frac{d}{dx} u(x)\\ & = \frac{1}{\sqrt{\pi \sigma}} e^{-\alpha k (x-x_0)^2} \frac{d}{dx} \left [ -\alpha k (x-x_0)^2 \right ]\\ & = \frac{-2\alpha k(x-x_0)^2}{\sqrt{\pi \sigma}} e^{-\alpha k (x-x_0)^2} \end{align*}$
::f(x) = ddxx[1] (x) = 1(x) ddxu(x) = 1(x) (x-x0) 2dx[*k(x-x0) 2) 2α(x-x0) 2(e) (k) (x-x0) 2

Review
::回顾

For #1-8, find the derivative and give the applicable domain if necessary.
::对于 #1-8, 找到衍生物, 并在必要时给出适用的域。

$\begin{align*}y=\log_7 (2x+5)\end{align*}$
::y=log7( 2x+5)

$\begin{align*}y=\frac{1}{\log x}\end{align*}$
::y=1logx

$\begin{align*}y = \frac{5}{\log (x+4)}\end{align*}$
::y=5log(x+4)

$\begin{align*}y = \log_2 x^2 \cdot \log_5 x^3\end{align*}$
::y=log2\\\ x2\log5\\ x3 y=log2\\ x2\ log5\\ x3

$\begin{align*}y = \ln (\sin x)\end{align*}$
::y= = = = = (sin=x) y= = = = = (sin=x)

$\begin{align*}y = \ln (\cos 5x)^3\end{align*}$
::y= = = = (cos=5x)3

$\begin{align*}y=\ln \frac{x}{x+1}\end{align*}$
::y=xxx+1

$\begin{align*}y = \ln (\sin (\ln x))\end{align*}$
::y= = = = = = = = = (sin = ) (xx) = = = = = = = = = = = = = = = = = = = = (x) = = = = = = = = = = = = = = = = = = (x) = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =

For #9-15, find the derivative.
::915号找到衍生物

$\begin{align*}y=e^{6x}\end{align*}$
::y=e6x y=e6x

$\begin{align*}y=e^{3x^3-2x^2 +6}\end{align*}$
::y= e3x3 - 2x2+6 y= e3x3 - 2x2+6

$\begin{align*}y=x^2 e^{-3x}\end{align*}$ ; find the extrema and use the 2 ^nd derivative to identify maxima/minima.
::y=x2e- 3x; 找到extrema, 并使用第二衍生物来识别最大值/最小值。

$\begin{align*}y=\frac{e^x - e^{-x}}{e^x + e^{-x}}\end{align*}$
::y=ex- ex- ex- xex+e- x

$\begin{align*}y=\csc (e^x)\end{align*}$
::y=csc( ex)

For the function $\begin{align*}y=\frac{\log (x)}{e^x}\end{align*}$ :

Find the expression for the derivative
::查找衍生物的表达式

Determine the location(s) of the extrema.
::确定extrema 的位置。

::函数 y=log(x)ex: 查找衍生物的表达式确定 extrema 的位置。

$\begin{align*}y = e^{x^2} \cdot \ln \left ( \frac{1}{x} \right )\end{align*}$
::y=ex2 ( 1x)

Review (Answers)
::回顾(答复)

Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键。

章节大纲

常规

指数函数和对数函数的衍生物

Derivatives of Exponential and Logarithmic Functions ::指数函数和对数函数的衍生物

The Derivative of a Logarithmic Function ::对数函数的衍生要素

Derivation of the Derivative of a Logarithmic Function ::对对数函数衍生的衍生结果

Derivatives of Exponential Functions ::指数函数的衍生因素

Derivation of the Derivative of an Exponential Function ::指向函数衍生的衍生要素

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Review ::回顾

Review (Answers) ::回顾(答复)

Derivatives of Exponential and Logarithmic Functions
::指数函数和对数函数的衍生物

The Derivative of a Logarithmic Function
::对数函数的衍生要素

Derivation of the Derivative of a Logarithmic Function
::对对数函数衍生的衍生结果

Derivatives of Exponential Functions
::指数函数的衍生因素

Derivation of the Derivative of an Exponential Function
::指向函数衍生的衍生要素

Examples
::实例

Example 1
::例1

Example 2
::例2

Review
::回顾

Review (Answers)
::回顾(答复)