『学习笔记』导数

导数：函数上某一点切线的斜率。

$y=f(x)$ 在 $x_0$ 的某个邻域内有定义，$x$ 在 $x_0$ 取增量 $\Delta x$， $\Delta y=f(x_0+\Delta x)-f(x_0)$。

$f’(x_0)=\lim\limits_{\Delta x\to 0}\frac{\Delta y}{\Delta x}=\lim\limits_{\Delta x\to 0} \frac{f(x_0+\Delta x)-f(x_0)}{\Delta x}$

符号：$y’\big|_{x=x_0}$，$\frac{dy}{dx}\big|_{x=x_0}$，$\frac{df(x)}{dx}\big|_{x=x_0}$

定义一：$\lim\limits_{\Delta x\to 0} \frac{f(x_0+\Delta x)-f(x_0)}{\Delta x} \Leftrightarrow \lim\limits_{h\to 0} \frac{f(x_0+h)-f(x_0)}{h}$

定义二：$\lim\limits_{x\to x_0} \frac{f(x)-f(x_0)}{x-x_0}$

导函数：在开区间 $(a,b)$ 上每个点都可导。

符号：$y’$，$f’(x)$，$\frac{dy}{dx}$，$\frac{df(x)}{dx}$

==Ⅰ：$f(x) = C$==

$f’(x) = \lim\limits_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x} = \lim\limits_{\Delta x \to 0} \frac{C - C}{\Delta x} = 0$

==Ⅱ：$f(x) = x^n$，$n \in \mathbb{N}_+$==

前置知识：二项式定理 $(a + b)^n = \sum_{k=0}^{n} \mathrm{C}_n^k \, a^{n-k} b^k$

  • 当 $n = 1$ 时： $f’(x) = \lim\limits_{h \to 0} \frac{f(x + h) - f(x)}{h} = \lim\limits_{h \to 0} \frac{(x + h) - x}{h} = \lim\limits_{h \to 0} \frac{h}{h} = 1$

  • 当 $n > 1$ 时： $f’(x) = \lim\limits_{h \to 0} \frac{f(x + h) - f(x)}{h} = \lim\limits_{h \to 0} \frac{(x + h)^n - x^n}{h}=\lim\limits_{h \to 0} \frac{\mathrm{C}_n^0 x^n + \mathrm{C}_n^1 x^{n - 1}h + \mathrm{C}_n^2 x^{n - 2}h^2 + \cdots + h^n - x^n}{h}$

    化简得：$f’(x) = \lim\limits_{h \to 0} \left( n x^{n - 1} + \mathrm{C}_n^2 x^{n - 2}h + \cdots + h^{n - 1} \right) = n x^{n - 1}$

可以推广到 $n\in\mathbb R$，留给读者自证。

==Ⅲ：$(\sin x)’=\cos x,(\cos x)’=-\sin x$==

$
\begin{align}
(\sin x)’ &= \lim\limits_{h \to 0} \frac{\sin(x + h) - \sin x}{h} = \lim\limits_{h \to 0} \frac{\sin x \cos h + \cos x \sin h - \sin x}{h} \
&= \lim\limits_{h \to 0} \left( \sin x \cdot \frac{\cos h - 1}{h} + \cos x \cdot \frac{\sin h}{h} \right) = \sin x \cdot \lim\limits_{h \to 0} \frac{\cos h - 1}{h} + \cos x \cdot \lim\limits_{h \to 0} \frac{\sin h}{h} \
&= \sin x \cdot 0 + \cos x \cdot 1 \ &= \cos x \
\end{align}
$

$
\begin{align}
(\cos x)’ &= \lim\limits_{h \to 0} \frac{\cos(x + h) - \cos x}{h} = \lim\limits_{h \to 0} \frac{\cos x \cos h - \sin x \sin h - \cos x}{h} \
&= \lim\limits_{h \to 0} \left( \cos x \cdot \frac{\cos h - 1}{h} - \sin x \cdot \frac{\sin h}{h} \right) = \cos x \cdot \lim\limits_{h \to 0} \frac{\cos h - 1}{h} - \sin x \cdot \lim\limits_{h \to 0} \frac{\sin h}{h} \
&= \cos x \cdot 0 - \sin x \cdot 1 \ &= -\sin x
\end{align}
$

==Ⅳ：$(a^x)’=a^x\ln a$==

$
\begin{align}
(a^x)’ &= \lim\limits_{h \to 0} \frac{a^{x+h} - a^x}{h} = \lim\limits_{h \to 0} \frac{a^x \cdot a^h - a^x}{h} = a^x \cdot \lim\limits_{h \to 0} \frac{a^h - 1}{h} \
&令\ t = a^h - 1，则\ a^h = t + 1，h = \log_a(t + 1) = \frac{\ln(t + 1)}{\ln a}，当\ h \to 0\ 时，t \to 0 \
&= a^x \cdot \lim\limits_{t \to 0} \frac{t}{\frac{\ln(t + 1)}{\ln a}} = a^x \cdot \ln a \cdot \lim\limits_{t \to 0} \frac{t}{\ln(t + 1)} = a^x \cdot \ln a \cdot \lim\limits_{t \to 0} \frac{1}{\frac{1}{t}\ln(t + 1)} \
&= a^x \cdot \ln a \cdot \lim\limits_{t \to 0} \frac{1}{\ln(1 + t)^{\frac{1}{t}}} = a^x \cdot \ln a \cdot \frac{1}{\ln e} \
&= a^x \cdot \ln a
\end{align}
$

不难发现：$(e^x)’=e^x$。

==Ⅴ：$(\log_a^x)’=\frac{1}{x\ln a}$==

$
\begin{align}
(\log_a x)’ &= \lim\limits_{h \to 0} \frac{\log_a (x + h) - \log_a x}{h} = \lim\limits_{h \to 0} \frac{\log_a \frac{x + h}{x}}{h} = \lim\limits_{h \to 0} \frac{1}{h} \log_a \left(1 + \frac{h}{x}\right) \
&= \lim\limits_{h \to 0} \frac{1}{x} \cdot \frac{x}{h} \log_a \left(1 + \frac{h}{x}\right) = \frac{1}{x} \lim\limits_{h \to 0} \log_a \left(1 + \frac{h}{x}\right)^{\frac{x}{h}} \
&令\ t = \frac{h}{x}，则\ h = xt，当\ h \to 0\ 时，t \to 0 \
&= \frac{1}{x} \lim\limits_{t \to 0} \log_a \left(1 + t\right)^{\frac{1}{t}} = \frac{1}{x} \log_a \left[ \lim\limits_{t \to 0} \left(1 + t\right)^{\frac{1}{t}} \right] = \frac{1}{x} \log_a e = \frac{1}{x} \cdot \frac{\ln e}{\ln a} \
&= \frac{1}{x \ln a}
\end{align}
$

不难发现：$\ln x=\frac{1}{x}$。

以上为一些简单函数的导函数。

$\text{Problem}$：$f(x) = |x|$，在 $x = 0$ 处的导数。

$\lim\limits_{h \to 0} \frac{f(0 + h) - f(0)}{h} = \lim\limits_{h \to 0} \frac{|h|}{h}$

  • 当 $h \to 0^+$ 时：$\lim\limits_{h \to 0^+} \frac{h}{h} = 1$ 当 $h \to 0^-$ 时：$\lim\limits_{h \to 0^-} \frac{-h}{h} = -1$

所以 $f(x) = |x|$，在 $x = 0$ 处不可导。结论：函数在某一点可导需要满足连续且光滑。

单侧导数：

右导数：$f’_{+}(x_0) = \lim\limits_{\Delta x \to 0^+} \frac{f(x_0 + \Delta x) - f(x_0)}{\Delta x}$

左导数：$f’_{-}(x_0) = \lim\limits_{\Delta x \to 0^-} \frac{f(x_0 + \Delta x) - f(x_0)}{\Delta x}$

综上：$f(x)$ 在 $x_0$ 处可导 $\Leftrightarrow$ $f’_{+}(x_0)$ 和 $f’_{-}(x_0)$ 存在 且 $f’_{+}(x_0)=f’_{-}(x_0)$。

$f(x)$ 在 $[a,b]$ 上可导 $\Leftrightarrow$ $f(x)$ 在 $(a,b)$ 上可导，并且 $a$ 点有右导数，$b$ 点有左导数。

导数的几何意义：

切线斜率：$f’(x_0)$，切线方程：$y-y_0=f’(x_0)(x-x_0)$，法线方程：$y-y_0=-\frac{1}{f’(x_0)}\left ( x-x_0 \right )$

$\text{Problem}$：求 $y=x^{\frac 3 2}$ 过 $(0,-4)$ 的切线方程。

将 $(0,-4)$ 带入，发现不在函数上，故设切点坐标 $(x_0,y_0)$，所以 $y_0=x_0^{\frac 3 2}$。

对函数求导，得：$y’|_{x=x_0} = \frac 3 2 x_0^{\frac 1 2}$。所以切线方程：$y-x_0^{\frac 3 2} = \frac 3 2 x_0^{\frac 1 2}(x-x_0)$，将点 $(0,-4)$ 带入，得：$-4-x_0^{\frac 3 2}=-\frac 3 2x_0^{\frac 3 2}$，解得 $x_0=4,y_0=8$ ，所以切线方程为 $3x-y-4=0$。

（一元函数）可导与连续的关系：

Ⅰ：可导必连续 Ⅱ：连续未必可导

可导：$f’(x) = \lim\limits_{\Delta x \to 0} \frac{\Delta y}{\Delta x}$ 存在。一定满足 $\Delta y\to 0$。

  • 证明：$\frac{\Delta y}{\Delta x} = f’(x) + \alpha$，$\Delta y = f’(x)\Delta x + \alpha \Delta x$ ，$\lim\limits_{\Delta x \to 0} \Delta y = \lim\limits_{\Delta x \to 0} \left( f’(x)\Delta x + \alpha \Delta x \right) = 0$

连续：①：$\lim\limits_{x \to x_0} f(x) = f(x_0)$ ②：$\lim\limits_{\Delta x \to 0} \Delta y = 0$

结论Ⅰ证毕！结论Ⅱ：可以发现 $y=|x|$ 等函数满足连续未必可导。

求导公式：

  • ==$(u\pm v)’ = u’\pm v’$==

    $
    \begin{align}
    (u \pm v)’ &= \lim\limits_{\Delta x \to 0} \frac{[u(x + \Delta x) \pm v(x + \Delta x)] - [u(x) \pm v(x)]}{\Delta x} \
    &= \lim\limits_{\Delta x \to 0} \frac{[u(x + \Delta x) - u(x)] \pm [v(x + \Delta x) - v(x)]}{\Delta x} \
    &= \lim\limits_{\Delta x \to 0} \left( \frac{u(x + \Delta x) - u(x)}{\Delta x} \pm \frac{v(x + \Delta x) - v(x)}{\Delta x} \right) \
    &= \lim\limits_{\Delta x \to 0} \frac{u(x + \Delta x) - u(x)}{\Delta x} \pm \lim\limits_{\Delta x \to 0} \frac{v(x + \Delta x) - v(x)}{\Delta x} \
    &= u’(x) \pm v’(x)
    \end{align}
    $

  • ==$(u\times v)’ =u’\cdot v+u\cdot v’$==

    $
    \begin{align}
    (u \times v)’ &= \lim\limits_{\Delta x \to 0} \frac{u(x + \Delta x)v(x + \Delta x) - u(x)v(x)}{\Delta x} \
    &= \lim\limits_{\Delta x \to 0} \frac{u(x + \Delta x)v(x + \Delta x) - u(x)v(x + \Delta x) + u(x)v(x + \Delta x) - u(x)v(x)}{\Delta x} \
    &= \lim\limits_{\Delta x \to 0} \left[ \frac{u(x + \Delta x) - u(x)}{\Delta x} \cdot v(x + \Delta x) + u(x) \cdot \frac{v(x + \Delta x) - v(x)}{\Delta x} \right] \
    &= \lim\limits_{\Delta x \to 0} \frac{u(x + \Delta x) - u(x)}{\Delta x} \cdot \lim\limits_{\Delta x \to 0} v(x + \Delta x) + u(x) \cdot \lim\limits_{\Delta x \to 0} \frac{v(x + \Delta x) - v(x)}{\Delta x} \
    &= u’(x) \cdot v(x) + u(x) \cdot v’(x)
    \end{align}
    $

  • ==$\large{(\frac u v)’ = \frac{u’\cdot v-u\cdot v’}{v^2}}$==

    $
    \begin{align}
    \left( \frac{u}{v} \right)’ &= \lim\limits_{\Delta x \to 0} \frac{\frac{u(x + \Delta x)}{v(x + \Delta x)} - \frac{u(x)}{v(x)}}{\Delta x} \
    &= \lim\limits_{\Delta x \to 0} \frac{u(x + \Delta x)v(x) - u(x)v(x + \Delta x)}{\Delta x \cdot v(x)v(x + \Delta x)} \
    &= \lim\limits_{\Delta x \to 0} \frac{u(x + \Delta x)v(x) - u(x)v(x) + u(x)v(x) - u(x)v(x + \Delta x)}{\Delta x \cdot v(x)v(x + \Delta x)} \
    &= \lim\limits_{\Delta x \to 0} \frac{1}{v(x)v(x + \Delta x)} \left[ v(x) \cdot \frac{u(x + \Delta x) - u(x)}{\Delta x} - u(x) \cdot \frac{v(x + \Delta x) - v(x)}{\Delta x} \right] \
    &= \frac{u’(x)v(x) - u(x)v’(x)}{[v(x)]^2}
    \end{align}
    $

  • ==$(C\cdot u)’=C\cdot u’$==

    $
    \begin{align}
    (C \cdot u)’ &= \lim\limits_{\Delta x \to 0} \frac{C \cdot u(x + \Delta x) - C \cdot u(x)}{\Delta x} \
    &= C \cdot \lim\limits_{\Delta x \to 0} \frac{u(x + \Delta x) - u(x)}{\Delta x} \
    &= C \cdot u’(x)
    \end{align}
    $

拓展：

  • $(u+v+w)’=u’+v’+w’$
  • $(u\cdot v\cdot w)’= u’\cdot v\cdot w+u\cdot v’\cdot w+u\cdot v\cdot w’$
  • $\left( \frac{u}{v} \right)’ = \left( u \cdot \frac{1}{v} \right)’ = u’ \cdot \frac{1}{v} + u \cdot \left( \frac{1}{v} \right)’ = \frac{u’}{v} - u \cdot \frac{v’}{v^2} = \frac{u’v - uv’}{v^2}$（利用乘法求导证明除法求导）
  • $(\tan x)’ =\sec^2 x$
  • $(\cot x)’ =-\csc^2 x$
  • $(\sec x)’ =\sec x\tan x$
  • $(\csc x)’ =-\csc x\cot x$

正确性显然，留给读者自证。

反函数求导

$x = f(y)$ 在 $I_y$ 单调、可导，且 $f’(y) \neq 0$，则 $y = f^{-1}(x)$ 在 $I_x$ 也可导，且： $\left[ f^{-1}(x) \right]’ = \frac{1}{f’(y)}$ ，或 $\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}}$

  • ==$(\arcsin x)’ = \frac{1}{\sqrt{1 - x^2}}$==

    $
    \begin{align}
    \text{设 } y = \arcsin x &\implies x = \sin y,\ y \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \
    \frac{dx}{dy} &= (\sin y)’ = \cos y \
    (\arcsin x)’ = \frac{dy}{dx} &= \frac{1}{\cos y} = \frac{1}{\sqrt{1 - \sin^2 y}} = \frac{1}{\sqrt{1 - x^2}}
    \end{align    }
    $

  • ==$(\arccos x)’ = -\frac{1}{\sqrt{1 - x^2}}$==

    $
    \begin{align}
    \text{设 } y = \arccos x &\implies x = \cos y,\ y \in (0, \pi) \
    \frac{dx}{dy} &= (\cos y)’ = -\sin y \
    (\arccos x)’ = \frac{dy}{dx} &= \frac{1}{-\sin y} = -\frac{1}{\sqrt{1 - \cos^2 y}} = -\frac{1}{\sqrt{1 - x^2}}
    \end{align    }
    $

  • ==$(\arctan x)’ = \frac{1}{1 + x^2}$==

    $
    \begin{align}
    \text{设 } y = \arctan x &\implies x = \tan y,\ y \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \
    \frac{dx}{dy} &= (\tan y)’ = \sec^2 y = 1 + \tan^2 y \
    (\arctan x)’ = \frac{dy}{dx} &= \frac{1}{1 + \tan^2 y} = \frac{1}{1 + x^2}
    \end{align    }
    $

  • ==$(\text{arccot}\, x)’ = -\frac{1}{1 + x^2}$==

    $
    \begin{align}
    \text{设 } y = \text{arccot}\, x &\implies x = \cot y,\ y \in (0, \pi) \
    \frac{dx}{dy} &= (\cot y)’ = -\csc^2 y = -(1 + \cot^2 y) \
    (\text{arccot}\, x)’ = \frac{dy}{dx} &= \frac{1}{-(1 + \cot^2 y)} = -\frac{1}{1 + x^2}
    \end{align    }
    $

复合函数求导法则（链式法则 Chain rule）

定理：设 $ y = f(u) $，$ u = g(x) $， $ y = f(g(x)) $。 若 $ u = g(x) $ 在 $ x $ 处可导，且 $ y = f(u) $ 在 $ g(x) $ 处可导，那么 $ y = f(g(x)) $ 在 $ x $ 处可导，且有：$\frac{dy}{dx} = f’(u) \cdot g’(x) $ 或：$ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} $

推广：若 $ y = f(u) $，$ u = g(t) $，$ t = h(x) $，则复合函数 $ y = f(g(h(x))) $ 的导数为： $ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dt} \cdot \frac{dt}{dx} $

高阶导数

符号：$y’’ = (y’)’ $，$ \frac{d}{dx}\left( \frac{dy}{dx} \right) = \frac{d\left( \frac{dy}{dx} \right)}{dx} = \frac{d^2 y}{dx^2} $ 。$y’’’ = (y’’)’ $，当阶数 $>3$ 时，记作 $y^{(n)}$ 表示 $y$ 的 $n$ 阶导，同理 $\frac{d^ny}{dx^n}$。

  1. $(\sin x)^{(n)}=\sin\left( x + n \cdot \frac{\pi}{2} \right)$

    数学归纳法证明：$(\sin x)’ = \cos x = \sin\left(x + \frac{\pi}{2}\right)$

    $(\sin x)’’ = \left[\sin\left(x + \frac{\pi}{2}\right)\right]’ = \cos\left(x + \frac{\pi}{2}\right) = \sin\left(x + 2 \cdot \frac{\pi}{2}\right)$

    $(\sin x)’’’ = \left[\sin\left(x + 2 \cdot \frac{\pi}{2}\right)\right]’ = \cos\left(x + 2 \cdot \frac{\pi}{2}\right) = \sin\left(x + 3 \cdot \frac{\pi}{2}\right)$

    $(\sin x)^{(n)} = \sin\left(x + n \cdot \frac{\pi}{2}\right)$

  2. $(\cos x)^{(n)} = \cos\left( x + n \cdot \frac{\pi}{2} \right)$

    数学归纳法证明：$(\cos x)’ = -\sin x = \cos\left(x + \frac{\pi}{2}\right)$

    $(\cos x)’’ = \left[ \cos\left(x + \frac{\pi}{2}\right) \right]’ = -\sin\left(x + \frac{\pi}{2}\right) = \cos\left(x + 2 \cdot \frac{\pi}{2}\right)$

    $(\cos x)’’’ = \left[ \cos\left(x + 2 \cdot \frac{\pi}{2}\right) \right]’ = -\sin\left(x + 2 \cdot \frac{\pi}{2}\right) = \cos\left(x + 3 \cdot \frac{\pi}{2}\right)$

    $(\cos x)^{(n)} = \cos\left(x + n \cdot \frac{\pi}{2}\right)$

  3. $[\ln(x+1)]^{(n)} = (-1)^{n-1} \cdot \frac{(n-1)!}{(x+1)^n}$

    数学归纳法证明：$[\ln(x+1)]’ = \frac{1}{x+1} = (-1)^{1-1} \cdot \frac{(1-1)!}{(x+1)^1}$

    $[\ln(x+1)]’’ = \left( \frac{1}{x+1} \right)’ = -\frac{1}{(x+1)^2} = (-1)^{2-1} \cdot \frac{(2-1)!}{(x+1)^2}$

    $[\ln(x+1)]’’’ = \left( -\frac{1}{(x+1)^2} \right)’ = \frac{2}{(x+1)^3} = (-1)^{3-1} \cdot \frac{(3-1)!}{(x+1)^3}$

    $[\ln(x+1)]^{(n)} = (-1)^{n-1} \cdot \frac{(n-1)!}{(x+1)^n}$

高阶求导法则

  • $(u\pm v)^{(n)}=u^{(n)}\pm v^{(n)}$

  • 莱布尼茨公式：$(uv)^{(n)} = \mathrm{C}_n^0 u^{(n)}v^{(0)} + \mathrm{C}_n^1 u^{(n - 1)}v^{(1)} + \cdots + \mathrm{C}_n^n u^{(0)}v^{(n)}= \sum\limits_{k=0}^{n} \mathrm{C}_n^k u^{(n-k)} v^{(k)} $

  • $\displaystyle \left( \frac{u}{v} \right)^{(n)} = \sum_{k=0}^{n} \binom{n}{k} u^{(n-k)} \cdot \left[ (-1)^k \cdot k! \cdot \frac{\sum_{i=0}^{k} (-1)^i \binom{k}{i} v^{(k-i)} \cdot 0^i}{v^{k+1}} \right]$

隐函数求导

对于隐函数 $F(x,y)=0$ 两边分别求导，提取 $y’$。

$\text{Problem}$：$e^y + xy = e$，求 $y’’$。

两边对 $x$ 求导：$e^y \cdot y’ + y + x \cdot y’ = 0$，$y’(e^y + x) = -y \implies y’ = -\frac{y}{e^y + x}$

对 $y’ = -\frac{y}{e^y + x}$ 两边求导：$y’’ = -\frac{y’(e^y + x) - y \cdot (e^y \cdot y’ + 1)}{(e^y + x)^2}= -\frac{y’ \cdot (-\frac{y}{y’}) - y \cdot (e^y \cdot y’ + 1)}{(e^y + x)^2}$ 整理，$y’’ = \frac{y e^y \cdot y’}{(e^y + x)^2} = \frac{y e^y \cdot (-\frac{y}{e^y + x})}{(e^y + x)^2} = -\frac{y^2 e^y}{(e^y + x)^3}$

对数求导

幂指函数：$y=u^{v}$，$u,v$ 为关于 $x$ 的表达式，且 $u>0$。

  • 两边同时取 $\ln$，或两边同时作为 $e$ 的指数。

$\text{Problem}$：求 $y = (1 + \cos x)^{\sin x}$（定义域：$1 + \cos x > 0$，即 $x \neq (2k+1)\pi, k \in \mathbb{Z}$）的导数。

两边取自然对数：$\ln y = \ln (1 + \cos x)^{\sin x} = \sin x \cdot \ln(1 + \cos x)$

两边对 $x$ 求导：$\frac{1}{y} \cdot y’ = \cos x \cdot \ln(1 + \cos x) - \frac{\sin^2 x}{1 + \cos x}$

解出 $y’ = y \cdot \left[ \cos x \cdot \ln(1 + \cos x) - \frac{\sin^2 x}{1 + \cos x} \right]= (1 + \cos x)^{\sin x} \cdot \left[ \cos x \cdot \ln(1 + \cos x) - \frac{\sin^2 x}{1 + \cos x} \right]$

参数方程求导

函数 $y = f(x)$，$\begin{cases} x = \varphi(t) \ y = \psi(t) \end{cases}$ 对 $x$ 求导，得：$\dfrac{dy}{dx} = \dfrac{dy}{dt} \cdot \dfrac{dt}{dx} = \dfrac{dy}{dt} \cdot \dfrac{1}{\dfrac{dx}{dt}} = \dfrac{\psi’(t)}{\varphi’(t)}$

$\dfrac{d^2y}{dx^2}= \dfrac{d\left( \dfrac{dy}{dx} \right)}{dx}=\dfrac{\dfrac {d\left( \dfrac{dy}{dx} \right)}{dt}}{\dfrac{dx}{dt}}= \dfrac{\psi’’(t)\varphi’(t) - \psi’(t)\varphi’’(t)}{\left( \varphi’(t) \right)^2}\cdot \dfrac{1}{\varphi’(t)}= \dfrac{\psi’’(t)\varphi’(t) - \psi’(t)\varphi’’(t)}{\left( \varphi’(t) \right)^3}$