Derivative and Differential


What is a derivative

velocity and tangent question

velocity question

Average velocity:

    \begin{equation*} \bar{v}=\frac{s-s_{0}}{t-t_{0}}=\frac{f(t)-f\left(t_{0}\right)}{t-t_{0}}\end{equation*}

Instantaneous velocity:

    \begin{equation*} v=\lim _{t \rightarrow t_{0}} \frac{f(t)-f\left(t_{0}\right)}{t-t_{0}}\end{equation*}

tangent question

secant:

    \begin{equation*} k = \tan \varphi=\frac{y-y_{0}}{x-x_{0}}=\frac{f(x)-f\left(x_{0}\right)}{x-x_{0}}\end{equation*}

 

 

tangent:

    \begin{equation*} k = \tan \alpha=\lim _{x \rightarrow x_{0}} \frac{f(x)-f\left(x_{0}\right)}{x-x_{0}}\end{equation*}

derivative definition

The derivative at some point

    \begin{equation*} f^{\prime}\left(x_{0}\right)=\lim _{\Delta x \rightarrow 0} \frac{\Delta y}{\Delta x}=\lim _{\Delta x \rightarrow 0} \frac{f\left(x_{0}+\Delta x\right)-f\left(x_{0}\right)}{\Delta x}\end{equation*}

Or as:

    \begin{equation*} \left.y^{\prime}\right|_{x=x_{0}},\left.\frac{\mathrm{d} y}{\mathrm{d} x}\right|_{x=x_{0}},\left.\frac{\mathrm{d} f(x)}{\mathrm{d} x}\right|_{x=x_{0}}\end{equation*}

Or as:

    \begin{equation*} f^{\prime}\left(x_{0}\right)=\lim _{h \rightarrow 0} \frac{f\left(x_{0}+h\right)-f\left(x_{0}\right)}{h}\end{equation*}

derived function

    \begin{equation*} y^{\prime}=\lim _{\Delta x \rightarrow 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}\end{equation*}

Or as:

    \begin{equation*} f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\end{equation*}

Or as:

    \begin{equation*} y^{\prime},f^{\prime}(x),\frac{d y}{d x},\frac{\mathrm{d} f(x)}{\mathrm{d} x}\end{equation*}

Ordinary derivative

1.Derivative of f(x)=C(C \text{\quad is constant} ):

    \begin{equation*} f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=\lim _{h \rightarrow 0} \frac{C-C}{h}=0\end{equation*}

2.Derivative of f(x)=x^{\mu}(\mu \in \mathbf{R}):

    \begin{equation*} \begin{aligned}f^{\prime}(x) &=\lim _{h \rightarrow 0}\frac{f(x+h)-f(x)}{h} =\lim _{h \rightarrow 0}\frac{(x+h)^{\mu}-x^{\mu}}{h}\\&=\lim _{h \rightarrow 0}x^{\mu-1} \cdot \frac{\left(1+\frac{h}{x}\right)^{\mu}-1}{\frac{h}{x}}=\mu x^{\mu-1}\end{aligned}\end{equation*}

3.Derivative of f(x)=\sin x:

    \begin{equation*} \begin{aligned} f^{\prime}(x) &=\lim _{h \rightarrow 0} f(x+h)-f(x)= \lim _{h \rightarrow 0} \frac{1}{h} \cdot 2 \cos \left(x+\frac{h}{2}\right) \sin \frac{h}{2} \\ &=\lim _{h \rightarrow 0} \cos\left(x+\frac{h}{2}\right) \cdot \frac{\sin \frac{h}{2}}{\frac{h}{2}} =\cos x \end{aligned}\end{equation*}

and in a similar way:

    \begin{equation*} (\cos x)^{\prime}=-\sin x\end{equation*}

4.Derivative of f(x)=a^{x}(a>0, a \neq 1):

    \begin{equation*} \begin{aligned} f^{\prime}(x)&=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=\lim _{h \rightarrow 0} \frac{a^{x+h}-a^{x}}{h}\\&=a^{x} \lim _{h \rightarrow 0} \frac{a^{h}-1}{h}=a^{x} \ln a.\end{aligned}\end{equation*}

specifically:\left(e^{x}\right)^{\prime}=e^{x}.

5.Derivative of f(x)=\log _{a} x \quad(a>0, a \neq 1)

    \begin{equation*} \begin{aligned} f^{\prime}(x) &=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=\lim _{h \rightarrow 0} \frac{\log _{a}(x+h)-\log _{a} x}{h} \\ &=\lim _{h \rightarrow 0} \frac{1}{h} \log _{a} \frac{x+h}{x}=\lim _{h \rightarrow 0} \frac{1}{x} \cdot \frac{x}{h} \log _{a}\left(1+\frac{h}{x}\right) \\ &=\frac{1}{x} \lim _{h \rightarrow 0} \frac{\log _{a}\left(1+\frac{h}{x}\right)}{\frac{h}{x}}=\frac{1}{x \ln a} \end{aligned}\end{equation*}

specifically:(\ln x)^{\prime}=\dfrac{1}{x}.

Unilateral derivative

    \begin{equation*} {f_{-}^{\prime}\left(x_{0}\right)=\lim _{h \rightarrow 0^{-}} \dfrac{f\left(x_{0}+h\right)-f\left(x_{0}\right)}{h}} \end{equation*}

    \begin{equation*}{f_{+}^{\prime}\left(x_{0}\right)=\lim _{h \rightarrow 0^{+}} \dfrac{f\left(x_{0}+h\right)-f\left(x_{0}\right)}{h}}\end{equation*}

The geometric meaning of the derivative

The equation of the tangent line of a curve

    \begin{equation*} y-y_{0}=f^{\prime}\left(x_{0}\right)\left(x-x_{0}\right)\end{equation*}

The normal equation of the curve

    \begin{equation*} y-y_{0}=-\frac{1}{f^{\prime}\left(x_{0}\right)}\left(x-x_{0}\right)\end{equation*}

Differentiable and continuous

Derivative must be continuous, continuous may not be derivative.

Derivative method

sum, difference, product, quotient method

    \begin{equation*} \begin{aligned} &(1)[u(x) \pm v(x)]^{\prime}=u^{\prime}(x) \pm v^{\prime}(x) \\ &(2)[u(x) v(x)]^{\prime}=u^{\prime}(x) v(x)+u(x) v^{\prime}(x) \\&(3)\left[\dfrac{u(x)}{v(x)}\right]^{\prime}=\dfrac{u^{\prime}(x) v(x)-u(x) v^{\prime}(x)}{v^{2}(x)}(v(x) \neq 0)\end{aligned}\end{equation*}

How to demonstrate (3)

    \begin{equation*} \begin{aligned}\left[\frac{u(x)}{v(x)}\right]^{\prime} &=\lim _{\Delta x \rightarrow 0} \dfrac{\dfrac{u(x+\Delta x)}{v(x+\Delta x)}-\dfrac{u(x)}{v(x)}}{\Delta x} \\ &=\lim _{\Delta x \rightarrow 0} \frac{u(x+\Delta x) v(x)-u(x) v(x+\Delta x)}{v(x+\Delta x) v(x) \Delta x} \\ &=\lim _{\Delta x \rightarrow 0} \frac{[u(x+\Delta x)-u(x)] v(x)-u(x)[v(x+\Delta x)-v(x)]}{v(x+\Delta x) v(x) \Delta x}\\&=\lim _{\Delta x \rightarrow 0} \frac{u(x+\Delta x)-u(x)}{\Delta x} v(x)-u(x) \frac{v(x+\Delta x)-v(x)}{\Delta x}\\&=\frac{u^{\prime}(x) v(x)-u(x) v^{\prime}(x)}{v^{2}(x)}\end{aligned}\end{equation*}

Promotion of formula (1)(2)

    \begin{equation*} \begin{array}{l}{(u+v-w)^{\prime}=u^{\prime}+v^{\prime}-w^{\prime}} \\ {(u v w)^{\prime}=[(u v) w]^{\prime}=(u v)^{\prime} w+(u v) w^{\prime}=\left(u^{\prime} v+u v^{\prime}\right) w+u v w^{\prime}}=u^{\prime} v w+u v^{\prime} w+u v w^{\prime}\end{array}\end{equation*}

Trig derivative

Derivative of y=\tan x

    \begin{equation*} \begin{aligned} y^{\prime} &=(\tan x)^{\prime}=\left(\frac{\sin x}{\cos x}\right)^{\prime}=\frac{(\sin x)^{\prime} \cos x-\sin x(\cos x)^{\prime}}{\cos ^{2} x} \\ &=\frac{\cos ^{2} x+\sin ^{2} x}{\cos ^{2} x}=\frac{1}{\cos ^{2} x}=\sec ^{2} x \end{aligned}\end{equation*}

in a similar way:

    \begin{equation*} \begin{array}{l}(\sec x)^{\prime}=\sec x \tan x\\{(\cot x)^{\prime}=-\csc ^{2} x} \\ {(\csc x)^{\prime}=-\csc x \cot x}\end{array}\end{equation*}

Derivative method of inverse function

If the function x=f(y) is monotone and differentiable in the interval I_{y},and f^{\prime}(y) \neq 0,then y=f^{-1}(x) differentiable in interval I_{x}=\{x | x=f(y), y \in I_{y}\},and:

    \begin{equation*} \left[f^{-1}(x)\right]^{\prime}=\frac{1}{f^{\prime}(y)} \quad \text{or}\quad \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{1}{\frac{\mathrm{d} x}{\mathrm{d} y}}\end{equation*}

Derivative rule of complex functions

If u=g(x) can be derivable at point x,and y=f({u}) can be derivable at point u=g(x),then y=f[g(x)] can be derivable at point x.The derivative is:

    \begin{equation*} \frac{\mathrm{d} y}{\mathrm{d} x}=f^{\prime}(u) \cdot g^{\prime}(x)\quad \text{or}\quad \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\mathrm{d} y}{\mathrm{d} u} \cdot \frac{\mathrm{d} u}{\mathrm{d} x}\end{equation*}

Case 1:If y=e^{x^{3}},for \dfrac{\mathrm{d} y}{\mathrm{d} x}:

    \begin{equation*} \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\mathrm{d} y}{\mathrm{d} u} \cdot \frac{\mathrm{d} u}{\mathrm{d} x}=\mathrm{e}^{u} \cdot 3 x^{2}=3 x^{2} \mathrm{e}^{x^{3}}\end{equation*}

Case 2:If y=\ln \sin x,for \dfrac{\mathrm{d} y}{\mathrm{d} x}.

    \begin{equation*} \frac{\mathrm{d} y}{\mathrm{d} x}=(\ln \sin x)^{\prime}=\frac{1}{\sin x}(\sin x)^{\prime}=\frac{\cos x}{\sin x}=\cot x\end{equation*}

Summary of common derivatives

    \begin{equation*} \begin{array}{ll}{(1)(C)^{\prime}=0,} & {\text { (2) }\left(x^{\mu}\right)^{\prime}=\mu x^{\mu-1}} \\ {(3)(\sin x)^{\prime}=\cos x,} & {\text { (4) }(\cos x)^{\prime}=-\sin x} \\ {(5)(\tan x)^{\prime}=\sec ^{2} x,} & {\text { (6) }(\cot x)^{\prime}=-\csc ^{2} x}\\{\text { (7) }(\sec x)^{\prime}=\sec x \tan x,} & {(8)(\csc x)^{\prime}=-\csc x \cot x} \\ {(9)\left(a^{x}\right)^{\prime}=a^{x} \ln a(a>0, a \neq 1),} & {(10)\left(\mathrm{e}^{x}\right)^{\prime}=\mathrm{e}^{x}} \\ {(11)\left(\log _{a} x\right)^{\prime}=\frac{1}{x \ln a}(a>0, a \neq 1),} & {(12)(\ln x)^{\prime}=\frac{1}{x}}\\{(13)(\arcsin x)^{\prime}=\frac{1}{\sqrt{1-x^{2}}}} & {(14)(\arccos x)^{\prime}=-\frac{1}{\sqrt{1-x^{2}}}} \\ {(15)(\arctan x)^{\prime}=\frac{1}{1+x^{2}},} & {(16)(\operatorname{arccot} x)^{\prime}=-\frac{1}{1+x^{2}}}\end{array}\end{equation*}

higher derivative

accelerated speed 

    \begin{equation*} a=\frac{\mathrm{d} v}{\mathrm{d} t}=\frac{\mathrm{d}}{\mathrm{d} t}\left(\frac{\mathrm{d} s}{\mathrm{d} t}\right)\quad \text{or}\quad a=\left(s^{\prime}\right)^{\prime}\end{equation*}

second derivative 

    \begin{equation*} y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}\quad \text{or}\quad \frac{\mathrm{d}^{2} y}{\mathrm{d} x^{2}}=\frac{\mathrm{d}}{\mathrm{d} x}\left(\frac{\mathrm{d} y}{\mathrm{d} x}\right)\end{equation*}

higher derivative

(1)   \begin{equation*} y^{\prime \prime \prime}, y^{(4)}, \cdots, y^{(n)}\quad \text{or}\quad \frac{\mathrm{d}^{3} y}{\mathrm{d} x^{3}}, \frac{\mathrm{d}^{4} y}{\mathrm{d} x^{4}}, \cdots, \frac{\mathrm{d}^{n} y}{\mathrm{d} x^{n}}\end{equation*}

The higher derivative of the exponential function

for y=e^{x}:

(2)   \begin{equation*} y^{\prime}=\mathrm{e}^{x}, \quad y^{\prime \prime}=\mathrm{e}^{x}, \quad y^{\prime \prime \prime}=\mathrm{e}^{x}, \quad y^{(4)}=\mathrm{e}^{x}\end{equation*}

Higher derivatives of sine and cosine

for \sin x:

    \begin{equation*} \begin{array}{l}{y^{\prime}=\cos x=\sin \left(x+\frac{\pi}{2}\right)} \\ {y^{\prime \prime}=\cos \left(x+\frac{\pi}{2}\right)=\sin \left(x+\frac{\pi}{2}+\frac{\pi}{2}\right)=\sin \left(x+2 \cdot \frac{\pi}{2}\right)} \\ {y^{\prime \prime \prime}=\cos \left(x+2 \cdot \frac{\pi}{2}\right)=\sin \left(x+3 \cdot \frac{\pi}{2}\right)} \\ {y^{(4)}=\cos \left(x+3 \cdot \frac{\pi}{2}\right)=\sin \left(x+4 \cdot \frac{\pi}{2}\right)}\\y^{(n)}=\sin \left(x+n \cdot \frac{\pi}{2}\right)\\\text{so}\quad (\sin x)^{(n)}=\sin \left(x+n \cdot \frac{\pi}{2}\right)\end{array}\end{equation*}

in a similar way,for \cos x:(\cos x)^{(n)}=\cos \left(x+n \cdot \frac{\pi}{2}\right)

Higher derivative of a power function

    \begin{equation*} \begin{array}{l}{y^{\prime}=\mu x^{\mu-1}} \\ {y^{\prime \prime}=\mu(\mu-1) x^{\mu-2}} \\ {y^{\prime \prime \prime}=\mu(\mu-1)(\mu-2) x^{\mu-3}} \\ {y^{(4)}=\mu(\mu-1)(\mu-2)(\mu-3) x^{\mu-4}}\\y^{(n)}=\mu(\mu-1)(\mu-2) \cdots(\mu-n+1) x^{\mu-n}\\so:\left(x^{\mu}\right)^{(n)}=\mu(\mu-1)(\mu-2) \cdots(\mu-n+1) x^{\mu-n}\end{array}\end{equation*}

Higher derivative of the sum difference function

    \begin{equation*} (u \pm v)^{(n)}=u^{(n)} \pm v^{(n)}\end{equation*}

*Higher derivative of the product function

we know:(u v)^{\prime}=u^{\prime} v+u v^{\prime};

then:(u v)^{\prime \prime}=u^{\prime \prime} v+2 u^{\prime} v^{\prime}+u v^{\prime \prime};

then:(u v)^{\prime\prime\prime}=u^{\prime\prime\prime} v+3 u^{\prime \prime} v^{\prime}+3 u^{\prime} v^{\prime \prime}+u v^{\prime \prime \prime};

so:

    \begin{equation*}\begin{aligned}(u v)^{(n)}=& u^{(n)} v+n u^{(n-1)} v^{\prime}+\frac{n(n-1)}{2 !} u^{(n-2)} v^{\prime \prime}+\cdots+\frac{n(n-1) \cdots(n-k+1)}{k !} u^{(n-k)} v^{(k)}+\cdots+u v^{(n)} \\=&\sum_{k=0}^{n} C_{n}^{k} u^{(n-k)} v^{(k)}\end{aligned}\end{equation*}

This is Leibniz formula.

Derivative of an implicit function

General method for differentiating implicit functions

case 1:for function \mathbf{e}^{y}+x y-\mathbf{e}=0,for \frac{d y}{d x}:

(1)derivative for the left-hand side with respect to x:

    \[ \frac{\mathrm{d}}{\mathrm{d} x}\left(\mathrm{e}^{y}+x y-\mathrm{e}\right)=\mathrm{e}^{y} \frac{\mathrm{d} y}{\mathrm{d} x}+y+x \frac{\mathrm{d} y}{\mathrm{d} x} \]

(2)derivative for the right-hand side with respect to x:

    \[ (0)^{\prime}=0 \]

(3)When we differentiate the left and the right sides of this equation for x, we get the same thing:

    \[ \mathrm{e}^{y} \frac{\mathrm{d} y}{\mathrm{d} x}+y+x \frac{\mathrm{d} y}{\mathrm{d} x}=0 \]

(4)then:

    \[\frac{\mathrm{d} y}{\mathrm{d} x}=-\frac{y}{x+\mathrm{e}^{y}} \quad\left(x+\mathrm{e}^{y} \neq 0\right)\]

case 2:for \dfrac{x^{2}}{16}+\dfrac{y^{2}}{9}=1,please get the equation of the tangent line at point \left(2, \dfrac{3}{2} \sqrt{3}\right)

(1)according to geometric meaning of the derivative,we know the slope of this tangent line:

    \[ k=\left.y^{\prime}\right|_{x=2} \]

(2)take the derivative of both sides of this equation with respect to x, we get the same thing:

    \[ \frac{x}{8}+\frac{2}{9} y \cdot \frac{d y}{d x}=0 \]

(3)then:

    \[ \frac{\mathrm{d} y}{\mathrm{d} x}=-\frac{9 x}{16 y} \]

(4)when x=2,y=\dfrac{3}{2} \sqrt{3}:

    \[ \left.\frac{d y}{d x}\right|_{x=2}=-\frac{\sqrt{3}}{4} \]

(5)The tangent equation is:

    \[ y-\frac{3}{2} \sqrt{3}=-\frac{\sqrt{3}}{4}(x-2) \]

(6)Reduction to:

    \[ \sqrt{3} x+4 y-8 \sqrt{3}=0 \]

Take the derivative of an implicit function using logarithmic differentiation

case 1:get derivative of y=x^{\sin x}.

(1)Take the logarithm of both sides of this equation:

    \[ \ln y=\sin x \cdot \ln x \]

(2)Take the derivative of both sides of this equation with respect to x

    \[ \frac{1}{y} y^{\prime}=\cos x \cdot \ln x+\sin x \cdot \frac{1}{x} \]

(3)then:

    \[ y^{\prime}=y\left(\cos x \cdot \ln x+\frac{\sin x}{x}\right)=x^{\sin x}\left(\cos x \cdot \ln x+\frac{\sin x}{x}\right) \]

Derivative of parametric equation

The derivative of an equation whose parameters can be eliminated

Derivative of parabolic parametric equation:

    \[\left\{\begin{array}{l}{x=v_{1} t} \\ {y=v_{2} t-\frac{1}{2} g t^{2}}\end{array}\right.\]

(1)Elimination parameters t:

    \[y=\frac{v_{2}}{v_{1}} x-\frac{g}{2 v_{1}^{2}} x^{2}\]

(2)Using the general implicit differentiation method,Is omitted.

The derivative of an equation that doesn’t cancel out the parameters

for:

    \[\left\{\begin{array}{l}{x=\varphi(t)} \\ {y=\psi(t)}\end{array}\right.\]

we can :

    \[\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\psi^{\prime}(t)}{\varphi^{\prime}(t)}\]

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