3. Differential calculus

Definition 0

The number

f(a)=limExaf(x)f(a)xaf^{\prime}(a)=\lim _{E \ni x \rightarrow a} \frac{f(x)-f(a)}{x-a}

is called the derivative of the function ff at aa.

Definition 1

A function f:ERf : E \to R defined on a set ERE\subset R is differentiable at a point x ∈ E that is a limit point of E if f(x+h)f(x)=A(x)h+α(x;h)f (x + h) − f (x) = A(x)h + \alpha(x;h), where hA(x)hh \mapsto A(x) h is a linear function in hh and α(x;h)=o(h)\alpha(x;h) = o(h) as h0h \to 0, x+hEx +h \in E.

Definition 2

The function hA(x)hh \mapsto A(x) h of Definition 1, which is linear in hh, is called the differential of the functionf:ER f : E \to \mathcal R at the point xEx\in E and is denoted df(x)\mathrm d f (x) or Df(x)\mathrm D f (x). Thus, df(x)(h)=A(x)h\mathrm d f (x)(h) = A(x)h.

We obtain

df(x)(h)dx(h)=f(x)\frac{\mathrm{d} f(x)(h)}{\mathrm{d} x(h)}=f^{\prime}(x)

We denote the set of all such vectors by TR(x0)T\mathbb R(x_0) or TRx0T\mathbb R_{x_0}. Similarly, we denote by TR(x0)T\mathbb R(x_0) or TRy0T\mathbb R_{y_0} the set of all displacement vectors from the point y0y_0 along the y-axis. It can then be seen from the definition of the differential that the mapping

df(x0):TR(x0)TR(f(x0))\mathrm{d} f\left(x_{0}\right): T \mathbb{R}\left(x_{0}\right) \rightarrow T \mathbb{R}\left(f\left(x_{0}\right)\right)

The derivative of an inverse function

If a function ff is differentiable at a point x0 and its differential df(x0):TR(x0)TR(y0)a\mathrm{d} f\left(x_{0}\right): T \mathbb{R}\left(x_{0}\right) \rightarrow T \mathbb{R}\left(y_0\right)a is invertible at that point, then the differential of the function f1f^{ −1} inverse to ff exists at the point y0=f(x0)y_0 = f (x_0) and is the mapping

df1(y0)=[df(x0)]1:TR(y0)TR(x0)\mathrm{d} f^{-1}\left(y_{0}\right)=\left[\mathrm{d} f\left(x_{0}\right)\right]^{-1}: T \mathbb{R}\left(y_{0}\right) \rightarrow T \mathbb{R}\left(x_{0}\right)

inverse to df(x0):TR(x0)TR(y0)a\mathrm{d} f\left(x_{0}\right): T \mathbb{R}\left(x_{0}\right) \rightarrow T \mathbb{R}\left(y_0\right)a .

The derivative of some common function formula

  1. (C)=0(C)^{\prime}=0

  2. (xμ)=μxμ1\left(x^{\mu}\right)^{\prime}=\mu x^{\mu-1}

  3. (sinx)=cosx(\sin x)^{\prime}=\cos x

  4. (cosx)=sinx(\cos x)^{\prime}=-\sin x

  5. (tanx)=sec2x(\tan x)^{\prime}=\sec ^{2} x

  6. (cotx)=csc2x(\cot x)^{\prime}=-\csc ^{2} x

  7. (secx)=secxtanx(\sec x)^{\prime}=\sec x \tan x

  8. (cscx)=cscxcotx(\csc x)^{\prime}=-\csc x \cot x

  9. (ax)=axlna(a>0,a1)\left(a^{x}\right)^{\prime}=a^{x} \ln a \quad(a>0, a \neq 1)

  10. (ex)=ex\left(\mathrm{e}^{x}\right)^{\prime}=\mathrm{e}^{x}

  11. (logax)=1xlna(a>0,a1)\left(\log _{a} x\right)^{\prime}=\frac{1}{x \ln a}(a>0, a \neq 1)

  12. (lnx)=1x(\ln x)^{\prime}=\frac{1}{x}

  13. (arcsinx)=11x2(\arcsin x)^{\prime}=\frac{1}{\sqrt{1-x^{2}}}

  14. (arccosx)=11x2(\arccos x)^{\prime}=-\frac{1}{\sqrt{1-x^{2}}}

  15. (arctanx)=11+x2(\arctan x)^{\prime}=\frac{1}{1+x^{2}}

  16. (arccotx)=11+x2(\operatorname{arccot} x)^{\prime}=-\frac{1}{1+x^{2}}

  17. (sinhx)=coshx(\sinh x)'= \cosh x

  18. (coshx)=sinhx(\cosh x)'=\sinh x

  19. (tanhx)=1cosh2x(\tanh x)' =\frac{1}{\cosh ^{2} x}

  20. (cothx)=1sinh2x(\operatorname{coth} x)' =-\frac{1}{\sinh ^{2} x}

  21. (arsinhx)=(ln(x+1+x2))=11+x2(\operatorname{arsinh} x)'=\left(\ln \left(x+\sqrt{1+x^{2}}\right)\right)' = \frac{1}{\sqrt{1+x^{2}}}

  22. (arcoshx)=(ln(x±x21))=±1x21(\operatorname{arcosh} x)'=(\ln \left(x \pm \sqrt{x^{2}-1}\right))'= \pm \frac{1}{\sqrt{x^{2}-1}}

  23. (artanhx)=(12ln1+x1x)=11x2(\operatorname{artanh} x)'=(\frac{1}{2} \ln \frac{1+x}{1-x})' = \frac{1}{1-x^{2}}

  24. (arcothx)=(12lnx+1x1)=1x21(\operatorname{arcoth} x)'=(\frac{1}{2} \ln \frac{x+1}{x-1})' = \frac{1}{x^{2}-1}

theorem Leibniz

L'Hôpital's rule

The theorem states that for functions ff and g g which are differentiable on an open interval II except possibly at a point cc contained in II, if limxcf(x)=limxcg(x)=0 or ±,{\displaystyle \lim {x\to c}f(x)=\lim {x\to c}g(x)=0{\text{ or }}\pm \infty ,}limxcf(x)=limxcg(x)=0{\displaystyle \lim {x\to c}f(x)=\lim {x\to c}g(x)=0}or ±\pm \infty ,and g(x)0 g'(x)\neq 0g(x)0g'(x)\neq 0 for all xx in II with xcx \ne c, and limxcf(x)g(x) \lim_{x\to c}{\frac {f'(x)}{g'(x)}} exists, then

Suppose f,gf,gare the function that defined in the set XX, and also have the nthn_{th} derivative, s.t.

(fg)(n)(x)=k=0n(nk)f(k)(x)g(nk)(x)(f g)^{(n)}(x)=\sum_{k=0}^{n}\left(\begin{array}{l} n \\ k \end{array}\right) f^{(k)}(x) g^{(n-k)}(x)
limxcf(x)g(x)=limxcf(x)g(x)\lim_{x\to c}{\frac {f(x)}{g(x)}}=\lim_{x\to c}{\frac {f'(x)}{g'(x)}}

mean value theorem

Taylor's theorem

Rolle's theorem

Let k1k \geq 1 be an integer and let the function f:RRf :\mathbb R\to\mathbb R be kk times differentiable at the point aRa \in\mathbb R. Then there exists a function Rk:RRR_k : \mathbb R \to\mathbb R such that ,

If a real-valued function ff is continuous on a closed interval [a,b][a, b], differentiable on the open interval (a,b)(a, b), and f(a)=f(b)f (a) = f (b), then there exists at least one cc in the open interval (a,b)(a, b) such that f(c)=0f'(c)=0.

f(a+x)=f(a)+f(a)(x)+f(a)2!x2++f(k)(a)k!xk+Rk(x;a)f(a+x)=f(a)+f'(a)(x)+{\frac {f''(a)}{2!}}x^{2}+\cdots +{\frac {f^{(k)}(a)}{k!}}x^{k}+R_k(x;a)

Lagrange’s Mean Value Theorem

and,

If a real-valued function ff is continuous on a closed interval [a,b][a, b], differentiable on the open interval (a,b)(a, b), then there is exists at least one cc in the open interval (a,b)(a, b) such that

f(b)f(a)=f(c)(ba)f(b)−f(a)=f′(c)(b−a)

Cauchy’s Mean Value Theorem

If a real-valued function ff is continuous on a closed interval [a,b][a, b], differentiable on the open interval (a,b)(a, b), and g(x)0g ' ( x ) \ne 0 for all x(a,b)x \in ( a , b ) , then there is exists at least one cc in the open interval (a,b)(a, b) such that

f(b)f(a)g(b)g(a)=f(c)g(c)\frac{f(b)-f(a)}{g(b)-g(a)}=\frac{f^{\prime}(c)}{g^{\prime}(c)}
Rk(x;a)=aa+xf(k+1)(t)k!(a+xt)kdtR_{k}(x;a)=\int_{a}^{a+x} \frac{f^{(k+1)}(t)}{k !}(a+x-t)^{k} \mathrm d t

prove:

Rk(x;a)=aa+xf(k+1)(t)k!(a+xt)kdt=1k!aa+xf(k+1)(t)(a+xt)kd(a+xt)=1(k+1)!aa+xf(k+1)(t)d(a+xt)k+1=1(k+1)!f(k+1)(t)(a+xt)k+1aa+x+1(k+1)!aa+x(a+xt)k+1df(k+1)(t)=f(k+1)(a)(k+1)!xk+1+Rk+1(x;a)\begin{aligned} R_{k}(x;a)=&\int_{a}^{a+x} \frac{f^{(k+1)}(t)}{k !}(a+x-t)^{k} \mathrm d t\\ =& \frac{-1}{k !}\int_a^{a+x}f^{(k+1)}(t)(a+x-t)^k\mathrm d( a+x-t)\\ =&\frac{-1}{(k+1) !}\int_a^{a+x}f^{(k+1)}(t)\mathrm d( a+x-t)^{k+1}\\ =&\frac{-1}{(k+1) !}f^{(k+1)}(t)( a+x-t)^{k+1}|_{a}^{a+x}+\frac{-1}{(k+1) !}\int_a^{a+x}( a+x-t)^{k+1}\mathrm df^{(k+1)}(t)\\ =&\frac{f^{(k+1)}(a)}{(k+1) !}x^{k+1}+R_{k+1}(x;a) \end{aligned}

q.e.d

remainder term

using little oo notation, Rk(x;a)=o(xk),x0R_{k}(x;a)=o\left(|x|^{k}\right), \quad x \rightarrow 0(The Peano remainder term)

The Lagrange form remainder term( Mean-value forms)

Rn(x;a)=f(n+1)(θ)(n+1)!xn+1(θ(a,a+x))R_{n}(x;a)=\frac{f^{(n+1)}(\theta)}{(n+1) !}x^{n+1}\quad (\theta\in(a,a+x))

\blacksquare

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