Math Analysis Homework - Week 4
Class 1
T1
{ f ∈ C [ a , b ] ∀ x ∈ [ a , b ] , ∃ y ∈ [ a , b ] , ∣ f ( y ) ∣ ≤ 1 2 ∣ f ( x ) ∣ ⟹ ∃ ξ ∈ [ a , b ] , f ( ξ ) = 0 \begin{gathered}
\begin{cases}
f\in C[a,b] \\
\forall x\in [a,b],\exists y\in [a,b], \vert f(y) \vert \le \dfrac{1}{2} \vert f(x) \vert
\end{cases} \\
\implies \exists \xi\in [a,b],f(\xi)=0
\end{gathered} ⎩ ⎨ ⎧ f ∈ C [ a , b ] ∀ x ∈ [ a , b ] , ∃ y ∈ [ a , b ] , ∣ f ( y ) ∣ ≤ 2 1 ∣ f ( x ) ∣ ⟹ ∃ ξ ∈ [ a , b ] , f ( ξ ) = 0
let x 0 ∈ [ a , b ] , ∃ x i ∈ [ a , b ] s . t . ∣ f ( x i ) ∣ ≤ 1 2 ∣ f ( x i − 1 ) ∣ x i ∈ [ a , b ] ⟹ ∃ { k n } s . t . { x k n } is convergent ⟹ lim n → ∞ x k n = X ∣ f ( x k n ) ∣ ≤ 1 2 ∣ f ( x k n − 1 ) ∣ ⟹ ∣ f ( x k n ) ∣ ≤ ∣ f ( x 0 ) 1 2 n ∣ ⟹ lim n → ∞ ∣ f ( x k n ) ∣ = 0 ∴ lim n → ∞ ∣ f ( x k n ) ∣ = ∣ f ( lim n → ∞ x k n ) ∣ = ∣ f ( X ) ∣ = 0 Q.E.D \begin{gathered}
\text{let } x_0\in [a,b], \\
\exists x_i\in[a,b] \ s.t.\
\vert f(x_i) \vert \le \dfrac{1}{2} \vert f(x_{i-1}) \vert \\
x_i\in [a,b] \implies \exists \{ k_n \} \ s.t.\
\{ x_{k_n} \} \text{ is convergent} \\
\implies \lim_{n \to \infty} x_{k_n}=X \\
\vert f(x_{k_n})\vert \le \dfrac{1}{2} \vert f(x_{k_{n-1}})\vert
\implies \vert f(x_{k_n})\vert\le \vert f(x_{0})\dfrac{1}{2^n}\vert \\
\implies \lim_{n \to \infty} \vert f(x_{k_n})\vert=0 \\
\therefore \lim_{n \to \infty} \vert f(x_{k_n}) \vert =\vert f(\lim_{n \to \infty} x_{k_n}) \vert =\vert f(X) \vert =0
\\
\text{Q.E.D}
\end{gathered} let x 0 ∈ [ a , b ] , ∃ x i ∈ [ a , b ] s . t . ∣ f ( x i ) ∣ ≤ 2 1 ∣ f ( x i − 1 ) ∣ x i ∈ [ a , b ] ⟹ ∃ { k n } s . t . { x k n } is convergent ⟹ n → ∞ lim x k n = X ∣ f ( x k n ) ∣ ≤ 2 1 ∣ f ( x k n − 1 ) ∣ ⟹ ∣ f ( x k n ) ∣ ≤ ∣ f ( x 0 ) 2 n 1 ∣ ⟹ n → ∞ lim ∣ f ( x k n ) ∣ = 0 ∴ n → ∞ lim ∣ f ( x k n ) ∣ = ∣ f ( n → ∞ lim x k n ) ∣ = ∣ f ( X ) ∣ = 0 Q.E.D
T2
{ ψ ∈ C ( R ) lim x → + ∞ ψ ( x ) x n = lim x → = − ∞ ψ ( x ) x n = 0 ⟹ { n m o d 2 = 1 ⟹ ∃ x 0 , x 0 n + ψ ( x 0 ) = 0 n m o d 2 = 0 ⟹ ∃ y , ∀ x ∈ R , y n + ψ ( y ) ≤ x n + ψ ( x ) \begin{gathered}
\begin{cases}
\psi\in C(R) \\
\lim_{x \to +\infty} \dfrac{\psi(x)}{x^n} =\lim_{x \to =-\infty} \dfrac{\psi(x)}{x^n} =0
\end{cases} \\
\implies \begin{cases}
n \bmod 2=1 \implies \exists x_0,x_0^n+\psi(x_0)=0 \\
n \bmod 2=0 \implies \exists y,\forall x\in R,y^n+\psi(y)\le x^n+\psi(x)
\end{cases}
\end{gathered} ⎩ ⎨ ⎧ ψ ∈ C ( R ) lim x → + ∞ x n ψ ( x ) = lim x →= − ∞ x n ψ ( x ) = 0 ⟹ { n mod 2 = 1 ⟹ ∃ x 0 , x 0 n + ψ ( x 0 ) = 0 n mod 2 = 0 ⟹ ∃ y , ∀ x ∈ R , y n + ψ ( y ) ≤ x n + ψ ( x )
f ( x ) : = ψ ( x ) x n \begin{gathered}
f(x):=\dfrac{\psi(x)}{x^n}
\end{gathered} f ( x ) := x n ψ ( x ) (1):
if ψ ( 0 ) = 0 : x 0 = 0 else ψ ( 0 ) ≠ 0 , without lossing generality,let ψ ( 0 ) = a < 0 ⟹ lim x → 0 ψ ( x ) = a < 0 ∴ lim x → 0 + f ( x ) = lim x → 0 + a x n = − ∞ ⟹ let M = − 2 , ∃ x 1 , f ( x 1 ) < M = − 2 lim x → + ∞ f ( x ) = 0 ⟹ let ϵ = 0.5 , ∃ x 2 , ∣ f ( x 2 ) ∣ < ϵ f ( x ) is continuous in ( 0 , + ∞ ) ⟹ Intermediate Theorem ∃ x 0 , f ( x 0 ) = − 1 , x 0 n + ψ ( x 0 ) = 0 Q.E.D \begin{gathered}
\text{if } \psi(0)=0 : x_0=0 \\
\text{else } \psi(0)\ne 0, \\
\text{without lossing generality,let } \psi(0)=a<0 \\
\implies \lim_{x \to 0} \psi(x)=a<0 \\
\therefore \lim_{x \to 0^+} f(x) =\lim_{x \to 0^+} \dfrac{a}{x^n} =-\infty \\
\implies \text{let }M=-2,\exists x_1,f(x_1) <M=-2 \\
\lim_{x \to +\infty} f(x)=0 \\
\implies \text{let } \epsilon=0.5,\exists x_2,\vert f(x_2) \vert <\epsilon \\
f(x) \text{ is continuous in }(0,+\infty) \\
\stackrel{\text{ Intermediate Theorem }}{\Longrightarrow}\exists x_0,f(x_0)=-1,x_0^n+\psi(x_0)=0
\\
\text{Q.E.D}
\end{gathered} if ψ ( 0 ) = 0 : x 0 = 0 else ψ ( 0 ) = 0 , without lossing generality,let ψ ( 0 ) = a < 0 ⟹ x → 0 lim ψ ( x ) = a < 0 ∴ x → 0 + lim f ( x ) = x → 0 + lim x n a = − ∞ ⟹ let M = − 2 , ∃ x 1 , f ( x 1 ) < M = − 2 x → + ∞ lim f ( x ) = 0 ⟹ let ϵ = 0.5 , ∃ x 2 , ∣ f ( x 2 ) ∣ < ϵ f ( x ) is continuous in ( 0 , + ∞ ) ⟹ Intermediate Theorem ∃ x 0 , f ( x 0 ) = − 1 , x 0 n + ψ ( x 0 ) = 0 Q.E.D (2):
f ( x ) = x n + ψ ( x ) lim x → ∞ f ( x ) x n = 1 let ϵ = 0.1 , ∃ X , ∣ x > X ∣ ⟹ f ( x ) ∈ ( 0.9 x n , 1.1 x n ) ∃ y 1 ∈ [ − X , X ] , ∀ x ∈ [ − X , X ] , f ( y 1 ) ≤ f ( x ) ∃ X 2 , 0.9 X 2 n > f ( y 1 ) ∃ y 2 ∈ [ − X 2 , X 2 ] , ∀ x ∈ [ − X 2 , X 2 ] , f ( y 2 ) ≤ f ( x ) ∀ x ∉ [ − X 2 , X 2 ] , f ( x ) > 0.9 x n > 0.9 X 2 n > f ( y 1 ) ≥ f ( y 2 ) ∴ ∀ x , f ( y 2 ) ≤ f ( x ) Q.E.D \begin{gathered}
f(x)=x^n+\psi(x) \\
\lim_{x \to \infty} \dfrac{f(x)}{x^n} =1 \\
\text{let } \epsilon=0.1,\exists X,\vert x>X \vert \implies f(x)\in (0.9x^n,1.1x^n) \\
\exists y_1\in [-X,X],\forall x\in [-X,X],f(y_1)\le f(x) \\
\exists X_2,0.9X_2^n>f(y_1) \\
\exists y_2\in [-X_2,X_2],\forall x\in [-X_2,X_2],f(y_2)\le f(x) \\
\forall x\notin [-X_2,X_2],f(x)>0.9x^n>0.9X_2^n>f(y_1)\ge f(y_2) \\
\therefore \forall x,f(y_2)\le f(x)\\
\text{Q.E.D}
\end{gathered} f ( x ) = x n + ψ ( x ) x → ∞ lim x n f ( x ) = 1 let ϵ = 0.1 , ∃ X , ∣ x > X ∣ ⟹ f ( x ) ∈ ( 0.9 x n , 1.1 x n ) ∃ y 1 ∈ [ − X , X ] , ∀ x ∈ [ − X , X ] , f ( y 1 ) ≤ f ( x ) ∃ X 2 , 0.9 X 2 n > f ( y 1 ) ∃ y 2 ∈ [ − X 2 , X 2 ] , ∀ x ∈ [ − X 2 , X 2 ] , f ( y 2 ) ≤ f ( x ) ∀ x ∈ / [ − X 2 , X 2 ] , f ( x ) > 0.9 x n > 0.9 X 2 n > f ( y 1 ) ≥ f ( y 2 ) ∴ ∀ x , f ( y 2 ) ≤ f ( x ) Q.E.D
T3
{ f ∈ C ( R ) , lim x → ∞ f ( x ) = + ∞ min f ( x ) = f ( a ) < a ⟹ ∃ x 1 , x 2 , f ( f ( x i ) ) = f ( a ) \begin{gathered}
\begin{cases}
f\in C(R),\lim_{x \to \infty} f(x)=+\infty \\
\min f(x)=f(a)<a
\end{cases} \\
\implies \exists x_1,x_2,f(f(x_i))=f(a)
\end{gathered} { f ∈ C ( R ) , lim x → ∞ f ( x ) = + ∞ min f ( x ) = f ( a ) < a ⟹ ∃ x 1 , x 2 , f ( f ( x i )) = f ( a )
∃ X , ∣ x ∣ > X ⟹ f ( x ) > 2 a ⟹ ∃ X 1 < a , X 2 > a , f ( X 1 ) > a , f ( X 2 ) > a { f ( a ) < a f ( X 1 ) > a f ( x ) is continuous ⟹ Intermediate Theorem ∃ x 1 ∈ [ X 1 , a ] , f ( x 1 ) = a , f ( f ( x 1 ) ) = f ( a ) { f ( a ) < a f ( X 2 ) > a f ( x ) is continuous ⟹ Intermediate Theorem ∃ x 2 ∈ [ a , X 2 ] , f ( x 2 ) = a , f ( f ( x 2 ) ) = f ( a ) Q.E.D \begin{gathered}
\exists X,\vert x \vert >X \implies f(x)>2a \\
\implies \exists X_1<a,X_2>a,f(X_1)>a,f(X_2)>a \\
\begin{cases}
f(a)<a \\
f(X_1)>a \\
f(x) \text{ is continuous}
\end{cases} \\
\stackrel{\text{ Intermediate Theorem }}{\Longrightarrow} \\
\exists x_1\in [X_1,a],f(x_1)=a,f(f(x_1))=f(a) \\
\begin{cases}
f(a)<a \\
f(X_2)>a \\
f(x) \text{ is continuous}
\end{cases} \\
\stackrel{\text{ Intermediate Theorem }}{\Longrightarrow} \\
\exists x_2\in [a,X_2],f(x_2)=a,f(f(x_2))=f(a)
\\
\text{Q.E.D}
\end{gathered} ∃ X , ∣ x ∣ > X ⟹ f ( x ) > 2 a ⟹ ∃ X 1 < a , X 2 > a , f ( X 1 ) > a , f ( X 2 ) > a ⎩ ⎨ ⎧ f ( a ) < a f ( X 1 ) > a f ( x ) is continuous ⟹ Intermediate Theorem ∃ x 1 ∈ [ X 1 , a ] , f ( x 1 ) = a , f ( f ( x 1 )) = f ( a ) ⎩ ⎨ ⎧ f ( a ) < a f ( X 2 ) > a f ( x ) is continuous ⟹ Intermediate Theorem ∃ x 2 ∈ [ a , X 2 ] , f ( x 2 ) = a , f ( f ( x 2 )) = f ( a ) Q.E.D
T4
f : C [ a , b ] → R , D ( f ( x ) ) + D ( x ) = 1 , ( D ( x ) = [ x ∈ Q ] ) ⟹ f ∉ C [ a , b ] \begin{gathered}
f:C[a,b]\to R,D(f(x))+D(x)=1,(D(x)=[x\in Q]) \\
\implies f\notin C[a,b]
\end{gathered} f : C [ a , b ] → R , D ( f ( x )) + D ( x ) = 1 , ( D ( x ) = [ x ∈ Q ]) ⟹ f ∈ / C [ a , b ]
By Contradiction ∀ [ x 1 , x 2 ] ⟹ Intermediate Theorem ∀ y ∈ [ f ( x 1 ) , f ( x 2 ) ] , ∃ x 0 , f ( x 0 ) = y { f ( x ) ∣ x ∈ Q ∩ [ x 1 , x 2 ] } ⊃ { y ∣ y ∈ Q C ∩ [ f ( x 1 ) , f ( x 2 ) ] } ⟹ [ f ( x 1 ) , f ( x 2 ) ] is countable Ridiculous! Q.E.D \begin{gathered}
\text{By Contradiction} \\
\forall [x_1,x_2] \\
\stackrel{ \text{Intermediate Theorem} }{\Longrightarrow} \\
\forall y\in [f(x_1),f(x_2)],\exists x_0,f(x_0)=y \\
\{ f(x) \vert x\in Q\cap [x_1,x_2] \} \supset \{ y \vert y\in Q^C \cap [f(x_1),f(x_2)] \} \\
\implies [f(x_1),f(x_2)] \text{ is countable} \\
\text{Ridiculous!}\\
\text{Q.E.D}
\end{gathered} By Contradiction ∀ [ x 1 , x 2 ] ⟹ Intermediate Theorem ∀ y ∈ [ f ( x 1 ) , f ( x 2 )] , ∃ x 0 , f ( x 0 ) = y { f ( x ) ∣ x ∈ Q ∩ [ x 1 , x 2 ]} ⊃ { y ∣ y ∈ Q C ∩ [ f ( x 1 ) , f ( x 2 )]} ⟹ [ f ( x 1 ) , f ( x 2 )] is countable Ridiculous! Q.E.D
T5
f ∈ C [ 0 , 2 a ] , f ( 0 ) = f ( 2 a ) , a > 0 ⟹ ∃ ξ ∈ [ 0 , a ] , f ( ξ ) = f ( ξ + a ) \begin{gathered}
f\in C[0,2a],f(0)=f(2a),a>0 \\
\implies \exists \xi\in [0,a],f(\xi)=f(\xi+a)
\end{gathered} f ∈ C [ 0 , 2 a ] , f ( 0 ) = f ( 2 a ) , a > 0 ⟹ ∃ ξ ∈ [ 0 , a ] , f ( ξ ) = f ( ξ + a )
if f ( a ) = 0 , ξ = a else: f ( a ) ≠ 0 without lossing generality,let f ( a ) > 0 let g ( x ) ∈ C [ a , 2 a ] = f ( x ) − f ( x − a ) g ( a ) > 0 , g ( 2 a ) < 0 ⟹ Intermediate Theorem ∃ ξ , g ( ξ ) = 0 = f ( ξ ) = f ( ξ − a ) Q.E.D \begin{gathered}
\text{if } f(a)=0,\xi=a \\
\text{else: } f(a)\ne 0 \\
\text{without lossing generality,let } f(a)>0 \\
\text{let } g(x)\in C[a,2a]= f(x)-f(x-a) \\
g(a)>0,g(2a)<0 \\
\stackrel{\text{ Intermediate Theorem }}{\Longrightarrow} \exists \xi,g(\xi)=0=f(\xi)=f(\xi-a) \\
\text{Q.E.D}
\end{gathered} if f ( a ) = 0 , ξ = a else: f ( a ) = 0 without lossing generality,let f ( a ) > 0 let g ( x ) ∈ C [ a , 2 a ] = f ( x ) − f ( x − a ) g ( a ) > 0 , g ( 2 a ) < 0 ⟹ Intermediate Theorem ∃ ξ , g ( ξ ) = 0 = f ( ξ ) = f ( ξ − a ) Q.E.D
T6
{ f ∈ C [ a , b ] { x n } , x i ∈ [ a , b ] ⟹ ∃ ξ ∈ [ min i ∈ [ 1 , n ] { x i } , max i ∈ [ 1 , n ] { x i } ] : f ( ξ ) = 1 n ∑ i = 1 n f ( x i ) \begin{gathered}
\begin{cases}
f\in C[a,b] \\
\{ x_n \} ,x_i\in[a,b]
\end{cases} \\
\implies \exists \xi\in [\min_{i\in [1,n]} \{ x_i \} ,\max_{i\in[1,n]} \{ x_i \} ]:f(\xi)=\dfrac{1}{n} \sum _{i = 1} ^{n} f(x_i)
\end{gathered} { f ∈ C [ a , b ] { x n } , x i ∈ [ a , b ] ⟹ ∃ ξ ∈ [ i ∈ [ 1 , n ] min { x i } , i ∈ [ 1 , n ] max { x i }] : f ( ξ ) = n 1 i = 1 ∑ n f ( x i )
L : = min i ∈ [ 1 , n ] { x i } , R : = max i ∈ [ 1 , n ] { x i } f ∈ C [ a , b ] ⟹ Extreme Value Theorem ∃ x 1 , x 2 ∈ [ L , R ] , ∀ x ∈ [ L , R ] , f ( x 1 ) ≤ f ( x ) ≤ f ( x 2 ) 1 n ∑ i = 1 n f ( x i ) ∈ [ f ( x 1 ) , f ( x 2 ) ] ⟹ Intermediate Theorem ∃ ξ , f ( ξ ) = 1 n ∑ i = 1 n f ( x i ) Q.E.D \begin{gathered}
L:=\min_{i\in [1,n]} \{ x_i \} ,R:=\max_{i\in[1,n]} \{ x_i \} \\
f\in C[a,b] \stackrel{\text{ Extreme Value Theorem }}{\Longrightarrow} \exists x_1,x_2\in [L,R], \\
\forall x\in [L,R],f(x_1)\le f(x)\le f(x_2) \\
\dfrac{1}{n} \sum _{i = 1} ^{n} f(x_i)\in [f(x_1),f(x_2)] \\
\stackrel{\text{ Intermediate Theorem }}{\Longrightarrow}\exists \xi,f(\xi)=\dfrac{1}{n} \sum _{i = 1} ^{n} f(x_i)
\\
\text{Q.E.D}
\end{gathered} L := i ∈ [ 1 , n ] min { x i } , R := i ∈ [ 1 , n ] max { x i } f ∈ C [ a , b ] ⟹ Extreme Value Theorem ∃ x 1 , x 2 ∈ [ L , R ] , ∀ x ∈ [ L , R ] , f ( x 1 ) ≤ f ( x ) ≤ f ( x 2 ) n 1 i = 1 ∑ n f ( x i ) ∈ [ f ( x 1 ) , f ( x 2 )] ⟹ Intermediate Theorem ∃ ξ , f ( ξ ) = n 1 i = 1 ∑ n f ( x i ) Q.E.D
T7
f ∈ C ( R ) , ∀ open interval I , f ( I ) is open interval ⟹ f ( x ) is monotonic \begin{gathered}
f\in C(R),\forall \text{open interval } I,f(I) \text{ is open interval} \\
\implies f(x) \text{ is monotonic}
\end{gathered} f ∈ C ( R ) , ∀ open interval I , f ( I ) is open interval ⟹ f ( x ) is monotonic
By Contradiction if ∃ x 1 < x 2 < x 3 , f ( x 2 ) ≤ min ( f ( x 1 ) , f ( x 3 ) ) ⟹ ∃ x 4 ∈ ( x 1 , x 2 ) , ∀ x ∈ [ x 1 , x 2 ] , f ( x 4 ) ≥ f ( x ) let I = ( x 1 , x 2 ) , f ( I ) = { f ( x ) ∣ x ∈ I } . but f ( x 4 ) ∈ f ( I ) , f ( x 4 ) = sup f ( I ) , f ( I ) is not open . Ridiculous! ⟹ ∄ x 1 < x 2 < x 3 , f ( x 2 ) ≤ min ( f ( x 1 ) , f ( x 3 ) ) same for ∄ x 1 < x 2 < x 3 , f ( x 2 ) ≥ max ( f ( x 1 ) , f ( x 3 ) ) ∴ ∀ x 1 < x 2 < x 3 , f ( x 2 ) ∈ ( f ( x 1 ) , f ( x 3 ) ) ∴ f ( x ) is monotonic \begin{gathered}
\text{By Contradiction} \\
\text{if } \exists x_1<x_2<x_3,f(x_2)\le \min(f(x_1),f(x_3)) \\
\implies \exists x_4\in (x_1,x_2),\forall x\in[x_1,x_2],f(x_4)\ge f(x) \\
\text{let } I=(x_1,x_2),f(I)=\{ f(x) \vert x\in I \} . \\
\text{but } f(x_4)\in f(I),f(x_4) = \sup f(I),f(I) \text{ is not open}. \\
\text{Ridiculous!} \\
\implies \not\exists x_1<x_2<x_3,f(x_2)\le \min(f(x_1),f(x_3)) \\
\text{same for} \not\exists x_1<x_2<x_3,f(x_2)\ge \max(f(x_1),f(x_3)) \\
\therefore \forall x_1<x_2<x_3,f(x_2)\in (f(x_1),f(x_3)) \\
\therefore f(x) \text{ is monotonic}
\end{gathered} By Contradiction if ∃ x 1 < x 2 < x 3 , f ( x 2 ) ≤ min ( f ( x 1 ) , f ( x 3 )) ⟹ ∃ x 4 ∈ ( x 1 , x 2 ) , ∀ x ∈ [ x 1 , x 2 ] , f ( x 4 ) ≥ f ( x ) let I = ( x 1 , x 2 ) , f ( I ) = { f ( x ) ∣ x ∈ I } . but f ( x 4 ) ∈ f ( I ) , f ( x 4 ) = sup f ( I ) , f ( I ) is not open . Ridiculous! ⟹ ∃ x 1 < x 2 < x 3 , f ( x 2 ) ≤ min ( f ( x 1 ) , f ( x 3 )) same for ∃ x 1 < x 2 < x 3 , f ( x 2 ) ≥ max ( f ( x 1 ) , f ( x 3 )) ∴ ∀ x 1 < x 2 < x 3 , f ( x 2 ) ∈ ( f ( x 1 ) , f ( x 3 )) ∴ f ( x ) is monotonic
T8
f ∈ C ( R ) , lim x → ∞ f ( f ( x ) ) = ∞ ⟹ lim x → ∞ f ( x ) = ∞ \begin{gathered}
f\in C(R),\lim_{x \to \infty} f(f(x))=\infty \\
\implies \lim_{x \to \infty} f(x)=\infty
\end{gathered} f ∈ C ( R ) , x → ∞ lim f ( f ( x )) = ∞ ⟹ x → ∞ lim f ( x ) = ∞
By Contradiction ∃ M , a n , ∣ a n ∣ > n , ∣ f ( a n ) ∣ < M ⟹ ∃ k i , f ( a k i ) is convergent ⟹ lim n → ∞ f ( a k i ) = A < M lim n → ∞ f ( f ( a n ) ) = ∞ ⟹ lim n → ∞ f ( f ( a k i ) ) = ∞ ⟹ lim n → ∞ f ( f ( a k i ) ) = f ( lim n → ∞ f ( a k i ) ) = M Ridiculous! Q.E.D \begin{gathered}
\text{By Contradiction} \\
\exists M,a_n,\vert a_n \vert >n,\vert f(a_n) \vert <M \\
\implies \exists k_i,f(a_{k_i}) \text{ is convergent} \\
\implies \lim_{n \to \infty} f(a_{k_i})=A<M \\
\lim_{n \to \infty} f(f(a_n))=\infty \\
\implies \lim_{n \to \infty} f(f(a_{k_i}))=\infty \\
\implies \lim_{n \to \infty} f(f(a_{k_i})) \\
=f(\lim_{n \to \infty} f(a_{k_i})) \\
=M \\
\text{Ridiculous!} \\
\text{Q.E.D}
\end{gathered} By Contradiction ∃ M , a n , ∣ a n ∣ > n , ∣ f ( a n ) ∣ < M ⟹ ∃ k i , f ( a k i ) is convergent ⟹ n → ∞ lim f ( a k i ) = A < M n → ∞ lim f ( f ( a n )) = ∞ ⟹ n → ∞ lim f ( f ( a k i )) = ∞ ⟹ n → ∞ lim f ( f ( a k i )) = f ( n → ∞ lim f ( a k i )) = M Ridiculous! Q.E.D
T9
∀ x , y ∈ R , ∣ f ( x ) − f ( y ) ∣ ≤ k ∣ x − y ∣ , 0 < k < 1 ⟹ { x − f ( x ) is increasing ∃ ! ξ ∈ R , f ( ξ ) = ξ \begin{gathered}
\forall x,y\in R,\vert f(x)-f(y) \vert \le k \vert x-y \vert ,0<k<1 \\
\implies \begin{cases}
x-f(x) \text{ is increasing} \\
\exists! \xi\in R,f(\xi)=\xi
\end{cases}
\end{gathered} ∀ x , y ∈ R , ∣ f ( x ) − f ( y ) ∣ ≤ k ∣ x − y ∣ , 0 < k < 1 ⟹ { x − f ( x ) is increasing ∃ ! ξ ∈ R , f ( ξ ) = ξ
(1)
∀ x 1 > x 2 ( x 1 − f ( x 1 ) ) − ( x 2 − f ( x 2 ) ) = ( x 1 − x 2 ) − ( f ( x 1 ) − f ( x 2 ) ) ≥ ( x 1 − x 2 ) − k ∣ x 1 − x 2 ∣ ≥ 0 Q.E.D \begin{gathered}
\forall x_1>x_2 \\
(x_1-f(x_1))-(x_2-f(x_2)) \\
=(x_1-x_2)-(f(x_1)-f(x_2)) \\
\ge (x_1-x_2)-k \vert x_1-x_2 \vert \\
\ge 0
\\
\text{Q.E.D}
\end{gathered} ∀ x 1 > x 2 ( x 1 − f ( x 1 )) − ( x 2 − f ( x 2 )) = ( x 1 − x 2 ) − ( f ( x 1 ) − f ( x 2 )) ≥ ( x 1 − x 2 ) − k ∣ x 1 − x 2 ∣ ≥ 0 Q.E.D (2)
g ( x ) = x − f ( x ) by (1) g ( x ) is increasing ∀ x > 0 , g ( x ) − g ( 0 ) = ( x − 0 ) − ( f ( x ) − f ( 0 ) ) ≥ ( 1 − k ) x ⟹ lim x → + ∞ g ( x ) − g ( 0 ) ≥ lim x → + ∞ ( 1 − k ) x ⟹ lim x → + ∞ g ( x ) = + ∞ same for lim x → − ∞ g ( x ) = − ∞ { lim x → + ∞ g ( x ) = + ∞ lim x → − ∞ g ( x ) = − ∞ g ( x ) is continuous g ( x ) is increasing ⟹ Intermediate Theorem ∃ ! ξ ∈ R , g ( ξ ) = 0 , f ( ξ ) = ξ \begin{gathered}
g(x)=x-f(x) \\
\text{by (1) } g(x) \text{ is increasing} \\
\forall x>0, \\
g(x)-g(0) \\
=(x-0)-(f(x)-f(0)) \\
\ge (1-k)x \\
\implies \lim_{x \to +\infty} g(x)-g(0)\ge \lim_{x \to +\infty} (1-k)x \\
\implies \lim_{x \to +\infty} g(x)=+\infty \\
\text{same for } \lim_{x \to -\infty} g(x)=-\infty \\
\begin{cases}
\lim_{x \to +\infty} g(x)=+\infty \\
\lim_{x \to -\infty} g(x)=-\infty \\
g(x) \text{ is continuous} \\
g(x) \text{ is increasing}
\end{cases} \\
\stackrel{\text{ Intermediate Theorem }}{\Longrightarrow} \\
\exists! \xi \in R,g(\xi)=0,f(\xi)=\xi \\
\end{gathered} g ( x ) = x − f ( x ) by (1) g ( x ) is increasing ∀ x > 0 , g ( x ) − g ( 0 ) = ( x − 0 ) − ( f ( x ) − f ( 0 )) ≥ ( 1 − k ) x ⟹ x → + ∞ lim g ( x ) − g ( 0 ) ≥ x → + ∞ lim ( 1 − k ) x ⟹ x → + ∞ lim g ( x ) = + ∞ same for x → − ∞ lim g ( x ) = − ∞ ⎩ ⎨ ⎧ lim x → + ∞ g ( x ) = + ∞ lim x → − ∞ g ( x ) = − ∞ g ( x ) is continuous g ( x ) is increasing ⟹ Intermediate Theorem ∃ ! ξ ∈ R , g ( ξ ) = 0 , f ( ξ ) = ξ
Class 2
T1
lim h → 0 f 2 ( x 0 + 2 h ) − f 2 ( x 0 − h ) h \begin{gathered}
\lim_{h \to 0} \dfrac{f^2(x_0+2h)-f^2(x_0-h)}{h}
\end{gathered} h → 0 lim h f 2 ( x 0 + 2 h ) − f 2 ( x 0 − h )
lim h → 0 f 2 ( x 0 + 2 h ) − f 2 ( x 0 − h ) h = lim h → 0 ( f ( x 0 + 2 h ) + f ( x 0 − h ) ) ( f ( x 0 + 2 h ) − f ( x 0 − h ) ) h = lim h → 0 2 f ( x 0 ) f ( x 0 + 2 h ) − f ( x 0 ) + f ( x 0 ) − f ( x 0 − h ) h = 4 f ( x 0 ) lim h → 0 f ( x 0 + 2 h ) − f ( x 0 ) 2 h + 2 f ( x 0 ) f ( x 0 ) − f ( x 0 − h ) h = 6 f ( x 0 ) f ′ ( x 0 ) \begin{gathered}
\lim_{h \to 0} \dfrac{f^2(x_0+2h)-f^2(x_0-h)}{h} \\
=\lim_{h \to 0} \dfrac{(f(x_0+2h)+f(x_0-h))(f(x_0+2h)-f(x_0-h))}{h} \\
=\lim_{h \to 0} 2f(x_0)\dfrac{f(x_0+2h)-f(x_0)+f(x_0)-f(x_0-h)}{h} \\
=4f(x_0)\lim_{h \to 0} \dfrac{f(x_0+2h)-f(x_0)}{2h} +2f(x_0)\dfrac{f(x_0)-f(x_0-h)}{h} \\
=6f(x_0)f'(x_0)
\end{gathered} h → 0 lim h f 2 ( x 0 + 2 h ) − f 2 ( x 0 − h ) = h → 0 lim h ( f ( x 0 + 2 h ) + f ( x 0 − h )) ( f ( x 0 + 2 h ) − f ( x 0 − h )) = h → 0 lim 2 f ( x 0 ) h f ( x 0 + 2 h ) − f ( x 0 ) + f ( x 0 ) − f ( x 0 − h ) = 4 f ( x 0 ) h → 0 lim 2 h f ( x 0 + 2 h ) − f ( x 0 ) + 2 f ( x 0 ) h f ( x 0 ) − f ( x 0 − h ) = 6 f ( x 0 ) f ′ ( x 0 )
T2
lim x → x 0 x f ( x 0 ) − x 0 f ( x ) x − x 0 \begin{gathered}
\lim_{x \to x_0} \dfrac{xf(x_0)-x_0f(x)}{x-x_0}
\end{gathered} x → x 0 lim x − x 0 x f ( x 0 ) − x 0 f ( x )
lim x → x 0 x f ( x 0 ) − x 0 f ( x ) x − x 0 = lim δ → 0 ( x 0 + δ ) f ( x 0 ) − x 0 f ( x 0 + δ ) δ = lim δ → 0 ( x 0 + δ ) f ( x 0 ) − x 0 f ( x 0 ) δ + x 0 f ( x 0 ) − x 0 f ( x 0 + δ ) δ = f ( x 0 ) − x 0 f ′ ( x 0 ) \begin{gathered}
\lim_{x \to x_0} \dfrac{xf(x_0)-x_0f(x)}{x-x_0} \\
=\lim_{\delta \to 0} \dfrac{(x_0+\delta)f(x_0)-x_0f(x_0+\delta)}{\delta} \\
=\lim_{\delta \to 0} \dfrac{(x_0+\delta)f(x_0)-x_0f(x_0)}{\delta} +\dfrac{x_0f(x_0)-x_0f(x_0+\delta)}{\delta} \\
=f(x_0)-x_0f'(x_0)
\end{gathered} x → x 0 lim x − x 0 x f ( x 0 ) − x 0 f ( x ) = δ → 0 lim δ ( x 0 + δ ) f ( x 0 ) − x 0 f ( x 0 + δ ) = δ → 0 lim δ ( x 0 + δ ) f ( x 0 ) − x 0 f ( x 0 ) + δ x 0 f ( x 0 ) − x 0 f ( x 0 + δ ) = f ( x 0 ) − x 0 f ′ ( x 0 )
T3
solve f s . t . f ( x + y ) = f ( x ) f ( y ) , f ′ ( 0 ) = 1 \begin{gathered}
\text{solve } f \\ s.t.\\
f(x+y)=f(x)f(y),f'(0)=1
\end{gathered} solve f s . t . f ( x + y ) = f ( x ) f ( y ) , f ′ ( 0 ) = 1
f ( x + 0 ) = f ( x ) f ( 0 ) ⟹ f ( 0 ) = 1 f ( x + y ) = f ( x ) f ( y ) ⟹ d d x f ′ ( x + y ) = f ′ ( x ) f ( y ) ⟹ f ( x + y ) = f ( x ) f ( y ) f ′ ( x + y ) f ( x + y ) = f ′ ( x ) f ( x ) ⟹ f ( 0 ) = f ′ ( 0 ) f ( x ) = f ′ ( x ) ⟹ f ′ ( x ) f ( x ) = 1 ⟹ ln ( f ( x ) ) ′ = 1 ⟹ ln ( f ( x ) ) = x + C ⟹ f ( x ) = e x + C ⟹ f ( 0 ) = 1 C = 0 ⟹ f ( x ) = e x \begin{gathered}
f(x+0)=f(x)f(0) \implies f(0)=1 \\
f(x+y)=f(x)f(y) \\
\stackrel{ \frac{d}{dx} }{\Longrightarrow}f'(x+y)=f'(x)f(y) \\
\stackrel{ f(x+y)=f(x)f(y) }{\Longrightarrow}\dfrac{f'(x+y)}{f(x+y)} =\dfrac{f'(x)}{f(x)} \\
\stackrel{ f(0)=f'(0) }{\Longrightarrow}f(x)=f'(x) \\
\implies \dfrac{f'(x)}{f(x)} =1 \\
\implies \ln(f(x))'=1 \\
\implies \ln(f(x))=x+C \\
\implies f(x)=e^{x+C} \\
\stackrel{ f(0)=1 }{\Longrightarrow}C=0 \\
\implies f(x)=e^x
\end{gathered} f ( x + 0 ) = f ( x ) f ( 0 ) ⟹ f ( 0 ) = 1 f ( x + y ) = f ( x ) f ( y ) ⟹ d x d f ′ ( x + y ) = f ′ ( x ) f ( y ) ⟹ f ( x + y ) = f ( x ) f ( y ) f ( x + y ) f ′ ( x + y ) = f ( x ) f ′ ( x ) ⟹ f ( 0 ) = f ′ ( 0 ) f ( x ) = f ′ ( x ) ⟹ f ( x ) f ′ ( x ) = 1 ⟹ ln ( f ( x ) ) ′ = 1 ⟹ ln ( f ( x )) = x + C ⟹ f ( x ) = e x + C ⟹ f ( 0 ) = 1 C = 0 ⟹ f ( x ) = e x
T4
{ f ( x ) ∈ C [ a , b ] , f ( a ) = f ( b ) = 0 f + ′ ( a ) f − ′ ( b ) > 0 ⟹ ∃ ξ ∈ ( a , b ) : f ( ξ ) = 0 \begin{gathered}
\begin{cases}
f(x)\in C[a,b],f(a)=f(b)=0 \\
f'_+(a)f'_-(b)>0
\end{cases} \\
\implies \exists \xi\in (a,b):f(\xi)=0
\end{gathered} { f ( x ) ∈ C [ a , b ] , f ( a ) = f ( b ) = 0 f + ′ ( a ) f − ′ ( b ) > 0 ⟹ ∃ ξ ∈ ( a , b ) : f ( ξ ) = 0
f + ′ ( a ) > 0 ⟹ lim Δ x → 0 f ( a + Δ x ) − f ( a ) Δ x > 0 ⟹ lim Δ x → 0 f ( a + Δ x ) > 0 ∃ x 1 ∈ N ∗ ( a ) , x 1 > a , f ( x 1 ) > 0 same for ∃ x 2 ∈ N ∗ ( b ) , x 2 < b , f ( x 2 ) < 0 ⟹ Intermediate Theorem ∃ ξ ∈ [ x 1 , x 2 ] , f ( ξ ) = 0 \begin{gathered}
f'_+(a)>0 \implies \lim_{\Delta x \to 0} \dfrac{f(a+\Delta x)-f(a)}{\Delta x}>0 \\
\implies \lim_{\Delta x \to 0} f(a+\Delta x)>0 \\
\exists x_1\in N^*(a),x_1>a,f(x_1)>0 \\
\text{same for } \exists x_2 \in N^*(b),x_2<b,f(x_2)<0 \\
\stackrel{\text{ Intermediate Theorem }}{\Longrightarrow}\exists \xi \in [x_1,x_2],f(\xi)=0
\end{gathered} f + ′ ( a ) > 0 ⟹ Δ x → 0 lim Δ x f ( a + Δ x ) − f ( a ) > 0 ⟹ Δ x → 0 lim f ( a + Δ x ) > 0 ∃ x 1 ∈ N ∗ ( a ) , x 1 > a , f ( x 1 ) > 0 same for ∃ x 2 ∈ N ∗ ( b ) , x 2 < b , f ( x 2 ) < 0 ⟹ Intermediate Theorem ∃ ξ ∈ [ x 1 , x 2 ] , f ( ξ ) = 0
T5
f ( x ) = ∣ x − a ∣ g ( x ) g ( x ) is continuous at x = a solve under what condition f is differentiable at x = a \begin{gathered}
f(x)=\vert x-a \vert g(x) \\
g(x) \text{ is continuous at } x=a \\
\text{solve under what condition } f \text{ is differentiable at } x=a
\end{gathered} f ( x ) = ∣ x − a ∣ g ( x ) g ( x ) is continuous at x = a solve under what condition f is differentiable at x = a
f is differentiable at x = a ⟺ lim Δ x → 0 f ( a + Δ x ) Δ x exists lim Δ x → 0 + f ( a + Δ x ) Δ x = lim Δ x → 0 ∣ Δ x ∣ g ( a + Δ x ) Δ x = g ( a ) same for lim Δ x → 0 − f ( a + Δ x ) Δ x = − g ( a ) thus f is differentiable at x = a ⟺ lim Δ x → 0 f ( a + Δ x ) Δ x exists ⟺ g ( a ) = − g ( a ) ⟺ g ( a ) = 0 \begin{gathered}
f \text{ is differentiable at } x=a \\
\iff \lim_{\Delta x \to 0} \dfrac{f(a+\Delta x)}{\Delta x} \text{ exists} \\
\lim_{\Delta x \to 0^+} \dfrac{f(a+\Delta x)}{\Delta x} \\
=\lim_{\Delta x \to 0} \dfrac{\vert \Delta x \vert g(a+\Delta x)}{\Delta x} \\
=g(a) \\
\text{same for } \lim_{\Delta x \to 0^-} \dfrac{f(a+\Delta x)}{\Delta x} = -g(a) \\
\text{thus } f \text{ is differentiable at } x=a \\
\iff \lim_{\Delta x \to 0} \dfrac{f(a+\Delta x)}{\Delta x} \text{ exists} \\
\iff
g(a)=-g(a) \\
\iff g(a)=0
\end{gathered} f is differentiable at x = a ⟺ Δ x → 0 lim Δ x f ( a + Δ x ) exists Δ x → 0 + lim Δ x f ( a + Δ x ) = Δ x → 0 lim Δ x ∣Δ x ∣ g ( a + Δ x ) = g ( a ) same for Δ x → 0 − lim Δ x f ( a + Δ x ) = − g ( a ) thus f is differentiable at x = a ⟺ Δ x → 0 lim Δ x f ( a + Δ x ) exists ⟺ g ( a ) = − g ( a ) ⟺ g ( a ) = 0
T6
{ f ( x ) is continuous at x = 0 f ( 0 ) = 0 lim x → 0 f ( 2 x ) − f ( x ) x = A ⟹ f ′ ( 0 ) = A \begin{gathered}
\begin{cases}
f(x) \text{ is continuous at } x=0 \\
f(0)=0 \\
\lim_{x \to 0} \dfrac{f(2x)-f(x)}{x} = A
\end{cases}
\implies f'(0)=A
\end{gathered} ⎩ ⎨ ⎧ f ( x ) is continuous at x = 0 f ( 0 ) = 0 lim x → 0 x f ( 2 x ) − f ( x ) = A ⟹ f ′ ( 0 ) = A
without lossing generality, let A > 0 , x > 0 f ′ ( x ) = lim x → 0 f ( x ) − f ( 0 ) x = lim x → 0 f ( x ) x lim x → 0 f ( 2 x ) − f ( x ) A x = 1 ⟹ ∀ ϵ , ∃ δ , f ( 2 x ) − f ( x ) ∈ ( ( 1 − ϵ ) A x , ( 1 + ϵ ) A x ) ⟹ f ( x ) = f ( x 2 n ) + ∑ i = 1 n f ( x 2 i − 1 ) − f ( x 2 i ) ∈ ( f ( x 2 n ) + A x ( 1 − ϵ ) ∑ i = 1 n 1 2 i , f ( x 2 n ) + A x ( 1 + ϵ ) ∑ i = 1 n 1 2 i ) lim n → ∞ f ( x ) ∈ [ A x ( 1 − ϵ ) , A x ( 1 + ϵ ) ] ⟹ lim x → 0 f ( x ) x = A \begin{gathered}
\text{without lossing generality, let } A>0,x>0 \\
f'(x)=\lim_{x \to 0} \dfrac{f(x)-f(0)}{x} =\lim_{x \to 0} \dfrac{f(x)}{x} \\
\lim_{x \to 0} \dfrac{f(2x)-f(x)}{Ax} =1 \\
\implies \forall \epsilon,\exists \delta, f(2x)-f(x) \in ((1-\epsilon)Ax,(1+\epsilon)Ax)\\
\implies f(x) = f(\dfrac{x}{2^n} )+\sum _{i = 1} ^{n} f(\dfrac{x}{2^{i-1}})-f(\dfrac{x}{2^i} ) \\
\in (f(\dfrac{x}{2^n} )+Ax(1-\epsilon) \sum _{i = 1} ^{n} \dfrac{1}{2^i},f(\dfrac{x}{2^n} )+Ax(1+\epsilon) \sum _{i = 1} ^{n} \dfrac{1}{2^i}) \\
\lim_{n \to \infty} f(x)\in [Ax(1-\epsilon),Ax(1+\epsilon)] \\
\implies \lim_{x \to 0} \dfrac{f(x)}{x} =A
\end{gathered} without lossing generality, let A > 0 , x > 0 f ′ ( x ) = x → 0 lim x f ( x ) − f ( 0 ) = x → 0 lim x f ( x ) x → 0 lim A x f ( 2 x ) − f ( x ) = 1 ⟹ ∀ ϵ , ∃ δ , f ( 2 x ) − f ( x ) ∈ (( 1 − ϵ ) A x , ( 1 + ϵ ) A x ) ⟹ f ( x ) = f ( 2 n x ) + i = 1 ∑ n f ( 2 i − 1 x ) − f ( 2 i x ) ∈ ( f ( 2 n x ) + A x ( 1 − ϵ ) i = 1 ∑ n 2 i 1 , f ( 2 n x ) + A x ( 1 + ϵ ) i = 1 ∑ n 2 i 1 ) n → ∞ lim f ( x ) ∈ [ A x ( 1 − ϵ ) , A x ( 1 + ϵ )] ⟹ x → 0 lim x f ( x ) = A
T7
f ( x ) is differentiable at x 0 ⟹ ∀ { a n } , { b n } , lim n → ∞ a n = x 0 − , lim n → ∞ b n = x 0 + f ′ ( x 0 ) = lim n → ∞ f ( b n ) − f ( a n ) b n − a n \begin{gathered}
f(x) \text{ is differentiable at } x_0 \\
\implies \forall \{ a_n \} ,\{ b_n \} , \lim_{n \to \infty} a_n=x_0^-,\lim_{n \to \infty} b_n=x_0^+ \\
f'(x_0)=\lim_{n \to \infty} \dfrac{f(b_n)-f(a_n)}{b_n-a_n}
\end{gathered} f ( x ) is differentiable at x 0 ⟹ ∀ { a n } , { b n } , n → ∞ lim a n = x 0 − , n → ∞ lim b n = x 0 + f ′ ( x 0 ) = n → ∞ lim b n − a n f ( b n ) − f ( a n )
lim n → ∞ f ( b n ) − f ( x 0 ) b n − x 0 = lim x → x 0 f ( x ) − f ( x 0 ) x − x 0 = f ′ ( x 0 ) same for f ( x 0 ) − f ( a n ) x 0 − a n = f ′ ( x 0 ) f ( b n ) − f ( a n ) b n − a n = f ( b n ) − f ( x 0 ) + f ( x 0 ) − f ( a n ) b n − x 0 + x 0 − a n ∈ ( f ( b n ) − f ( x 0 ) b n − x 0 , f ( x 0 ) − f ( a n ) a n − x 0 ) ⟹ Squeeze Theorem lim n → ∞ f ( b n ) − f ( a n ) b n − a n = f ′ ( x 0 ) \begin{gathered}
\lim_{n \to \infty} \dfrac{f(b_n)-f(x_0)}{b_n-x_0} =\lim_{x \to x_0} \dfrac{f(x)-f(x_0)}{x-x_0}=f'(x_0) \\
\text{same for } \dfrac{f(x_0)-f(a_n)}{x_0-a_n} =f'(x_0) \\
\dfrac{f(b_n)-f(a_n)}{b_n-a_n} \\
=\dfrac{f(b_n)-f(x_0)+f(x_0)-f(a_n)}{b_n-x_0+x_0-a_n} \\
\in (\dfrac{f(b_n)-f(x_0)}{b_n-x_0} ,\dfrac{f(x_0)-f(a_n)}{a_n-x_0} )
\stackrel{\text{ Squeeze Theorem }}{\Longrightarrow} \\
\lim_{n \to \infty} \dfrac{f(b_n)-f(a_n)}{b_n-a_n} =f'(x_0)
\end{gathered} n → ∞ lim b n − x 0 f ( b n ) − f ( x 0 ) = x → x 0 lim x − x 0 f ( x ) − f ( x 0 ) = f ′ ( x 0 ) same for x 0 − a n f ( x 0 ) − f ( a n ) = f ′ ( x 0 ) b n − a n f ( b n ) − f ( a n ) = b n − x 0 + x 0 − a n f ( b n ) − f ( x 0 ) + f ( x 0 ) − f ( a n ) ∈ ( b n − x 0 f ( b n ) − f ( x 0 ) , a n − x 0 f ( x 0 ) − f ( a n ) ) ⟹ Squeeze Theorem n → ∞ lim b n − a n f ( b n ) − f ( a n ) = f ′ ( x 0 )
Class 3
Calc the following function's directive
T1
y = x sin x + sin x x \begin{gathered}
y=x\sin x+\dfrac{\sin x}{x}
\end{gathered} y = x sin x + x sin x
y ′ = sin ( x ) + x cos ( x ) + x cos x − sin x x 2 = sin x + x cos x + cos x x − sin x x 2 \begin{gathered}
y'=\sin(x)+x\cos(x)+\dfrac{x\cos x-\sin x}{x^2} \\
=\sin x+x\cos x+\dfrac{\cos x}{x} -\dfrac{\sin x}{x^2}
\end{gathered} y ′ = sin ( x ) + x cos ( x ) + x 2 x cos x − sin x = sin x + x cos x + x cos x − x 2 sin x
T2
y = x e x − 1 sin x \begin{gathered}
y=\dfrac{xe^x-1}{\sin x}
\end{gathered} y = sin x x e x − 1
y ′ = ( x + 1 ) e x sin x − ( x e x − 1 ) cos x sin 2 x = ( x + 1 ) e x sin x − ( x e x − 1 ) cos x sin 2 x \begin{gathered}
y'=\dfrac{(x+1)e^x\sin x-(xe^x-1)\cos x}{\sin^2 x} \\
=\dfrac{(x+1)e^x}{\sin x} -\dfrac{(xe^x-1)\cos x}{\sin^2 x}
\end{gathered} y ′ = sin 2 x ( x + 1 ) e x sin x − ( x e x − 1 ) cos x = sin x ( x + 1 ) e x − sin 2 x ( x e x − 1 ) cos x
T3
y = ( x 3 + 1 ) 4 \begin{gathered}
y=(x^3+1)^4
\end{gathered} y = ( x 3 + 1 ) 4
y ′ = 4 ( x 3 + 1 ) 3 3 x 2 = 12 x 2 ( x 3 + 1 ) 3 \begin{gathered}
y'=4(x^3+1)^3 3x^2 \\
=12x^2(x^3+1)^3
\end{gathered} y ′ = 4 ( x 3 + 1 ) 3 3 x 2 = 12 x 2 ( x 3 + 1 ) 3
T4
y = e x 3 + 1 \begin{gathered}
y=e^{\sqrt{x^3+1}}
\end{gathered} y = e x 3 + 1
y ′ = e x 3 + 1 ⋅ 3 x 2 2 x 3 + 1 \begin{gathered}
y'=e^{\sqrt{x^3+1}} \cdot \dfrac{3x^2}{2\sqrt {x^3+1}}
\end{gathered} y ′ = e x 3 + 1 ⋅ 2 x 3 + 1 3 x 2
T5
y = 2 sin ( x 2 ) + 2 tan 1 x \begin{gathered}
y=2^{\sin(x^2)}+2^{\tan \frac{1}{x} }
\end{gathered} y = 2 s i n ( x 2 ) + 2 t a n x 1
y ′ = 2 sin x 2 + 1 ln 2 cos x 2 x − 2 tan 1 x ln 2 sec 2 1 x 1 x 2 \begin{gathered}
y'=2^{\sin x^2+1}\ln 2 \cos x^2 x-2^{\tan \frac{1}{x}}\ln 2 \sec^2 \dfrac{1}{x} \dfrac{1}{x^2}
\end{gathered} y ′ = 2 s i n x 2 + 1 ln 2 cos x 2 x − 2 t a n x 1 ln 2 sec 2 x 1 x 2 1
T6
y = sin ( sin ( sin ( x 2 + 1 ) ) ) \begin{gathered}
y=\sin(\sin(\sin(\sqrt{x^2+1})))
\end{gathered} y = sin ( sin ( sin ( x 2 + 1 )))
y ′ = cos ( sin ( sin ( x 2 + 1 ) ) ) cos ( sin x 2 + 1 ) cos ( x 2 + 1 ) x x 2 + 1 \begin{gathered}
y'=\cos(\sin(\sin(\sqrt{x^2+1})))\cos(\sin \sqrt{x^2+1})\cos (\sqrt {x^2+1})\dfrac{x}{\sqrt {x^2+1}}
\end{gathered} y ′ = cos ( sin ( sin ( x 2 + 1 ))) cos ( sin x 2 + 1 ) cos ( x 2 + 1 ) x 2 + 1 x
T7
y = arctan 2 x 1 − x 2 \begin{gathered}
y=\arctan \dfrac{2x}{1-x^2}
\end{gathered} y = arctan 1 − x 2 2 x
arctan'=1/1+x^2
y ′ = 1 1 + ( 2 x 1 − x 2 ) 2 2 ( 1 − x 2 ) + 4 x 2 ( 1 − x 2 ) 2 = 2 ( 1 + x 2 ) \begin{gathered}
y'=\dfrac{1}{1+(\dfrac{2x}{1-x^2} )^2}\dfrac{2(1-x^2)+4x^2}{(1-x^2)^2} \\
=\dfrac{2}{(1+x^2)}
\end{gathered} y ′ = 1 + ( 1 − x 2 2 x ) 2 1 ( 1 − x 2 ) 2 2 ( 1 − x 2 ) + 4 x 2 = ( 1 + x 2 ) 2
T8
y = ln 1 + cos x 1 − cos x \begin{gathered}
y=\ln \sqrt{\dfrac{1+\cos x}{1-\cos x} }
\end{gathered} y = ln 1 − cos x 1 + cos x
1 + cos x 1 − cos x = cot x 2 ⟹ y = ln cot x 2 ⟹ y ′ = − tan x 2 csc 2 x 2 ⋅ 1 2 = − 1 2 sin x 2 cos x 2 = − csc x \begin{gathered}
\sqrt {\dfrac{1+\cos x}{1-\cos x} } \\
=\cot \dfrac{x}{2} \\
\implies y=\ln \cot \dfrac{x}{2} \\
\implies y'=-\tan \dfrac{x}{2} \csc^2 \dfrac{x}{2} \cdot \dfrac{1}{2} \\
=-\dfrac{1}{2\sin \frac{x}{2}\cos \frac{x}{2}} \\
=-\csc x
\end{gathered} 1 − cos x 1 + cos x = cot 2 x ⟹ y = ln cot 2 x ⟹ y ′ = − tan 2 x csc 2 2 x ⋅ 2 1 = − 2 sin 2 x cos 2 x 1 = − csc x
T9
y = x a a + a a x + a x a \begin{gathered}
y=x^{a^a}+a^{a^x}+a^{x^a}
\end{gathered} y = x a a + a a x + a x a
y ′ = a a x a a − 1 + a a x + x ln 2 a + a x a + 1 ln a x a − 1 \begin{gathered}
y'=a^ax^{a^a-1}+a^{a^x+x}\ln^2 a+ a^{x^a+1}\ln ax^{a-1}
\end{gathered} y ′ = a a x a a − 1 + a a x + x ln 2 a + a x a + 1 ln a x a − 1
T10
y = sin ( f ( sin ( x ) ) ) \begin{gathered}
y=\sin(f(\sin(x)))
\end{gathered} y = sin ( f ( sin ( x )))
y ′ = cos ( f ( sin ( x ) ) ) f ′ ( sin x ) cos x \begin{gathered}
y'=\cos(f(\sin(x)))f'(\sin x)\cos x
\end{gathered} y ′ = cos ( f ( sin ( x ))) f ′ ( sin x ) cos x
T11
y = ( sin ( f ( x ) ) x ) f ( f ( x ) ) \begin{gathered}
y={\left( \dfrac{\sin(f(x))}{x} \right)} ^{f(f(x))} \\
\end{gathered} y = ( x sin ( f ( x )) ) f ( f ( x ))
y = exp f ( f ( x ) ) ( ln sin ( f ( x ) ) − ln x ) = ( sin ( f ( x ) ) x ) f ( f ( x ) ) ( f ′ ( f ( x ) ) f ′ ( x ) ln sin f ( x ) x + f ( f ( x ) ) ( x sin f ( x ) x f ′ x cos f ( x ) − sin f ( x ) x 2 ) ) = ( sin ( f ( x ) ) x ) f ( f ( x ) ) ( f ′ ( f ( x ) ) f ′ ( x ) ln sin f ( x ) x + f ( f ( x ) ) ( x f ′ x cos f x − sin f x x sin f x ) \begin{gathered}
y=\exp f(f(x)) (\ln \sin(f(x))-\ln x) \\
={\left( \dfrac{\sin(f(x))}{x} \right)} ^{f(f(x))} {\left( f'(f(x))f'(x)\ln\dfrac{\sin f(x)}{x} +f(f(x)) {\left( \dfrac{x}{\sin f(x)} \dfrac{xf'x \cos f(x)-\sin f(x)}{x^2} \right)} \right)} \\
={\left( \dfrac{\sin(f(x))}{x} \right)} ^{f(f(x))} {\left( f'(f(x))f'(x)\ln\dfrac{\sin f(x)}{x} +\dfrac{f(f(x))(xf'x\cos f x-\sin f x}{x\sin f x} \right)}
\end{gathered} y = exp f ( f ( x )) ( ln sin ( f ( x )) − ln x ) = ( x sin ( f ( x )) ) f ( f ( x )) ( f ′ ( f ( x )) f ′ ( x ) ln x sin f ( x ) + f ( f ( x )) ( sin f ( x ) x x 2 x f ′ x cos f ( x ) − sin f ( x ) ) ) = ( x sin ( f ( x )) ) f ( f ( x )) ( f ′ ( f ( x )) f ′ ( x ) ln x sin f ( x ) + x sin f x f ( f ( x )) ( x f ′ x cos f x − sin f x )