Articles in this blog's slice are, for the time being, ony in italian.
Teorema D β
β k = 0 D\;k = 0 D k = 0
Dimostrazione f ( x ) = k f(x) = k f ( x ) = k f ( x + h ) = k f(x+h) = k f ( x + h ) = k
f β² ( x ) = lim β‘ h β 0 f ( x + h ) β f ( x ) h = = lim β‘ h β 0 k β k h = 0 \eq{
f'(x) = &\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h} =\\
=& \lim\limits_{h\to0}\frac{k-k}{h} = 0
} f β² ( x ) = = β h β 0 lim β h f ( x + h ) β f ( x ) β = h β 0 lim β h k β k β = 0 β β Rappresentazione grafica y = k y=k y = k
f ( x ) = 2 β
β βΉ β
β f β² ( x ) = 0 f ( x ) = 15 β
β βΉ β
β f β² ( x ) = 0 f ( x ) = 230 β
β βΉ β
β f β² ( x ) = 0 \eq{
&f(x)= 2 \implies f'(x) = 0 \\
&f(x)= 15 \implies f'(x) = 0 \\
&f(x)= 230 \implies f'(x) = 0 \\
} β f ( x ) = 2 βΉ f β² ( x ) = 0 f ( x ) = 15 βΉ f β² ( x ) = 0 f ( x ) = 230 βΉ f β² ( x ) = 0 β β
Teorema 1 1 1 D β
β x = 1 D\;x = 1 D x = 1
f β² ( x ) = lim β‘ h β 0 f ( x + h ) β f ( x ) h = = lim β‘ h β 0 x + h β x h = = lim β‘ h β 0 h h = 1 \eq{
f'(x) =& \lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h} =\\
=& \lim\limits_{h\to0}\frac{x+h-x}{h} =\\
=&\lim\limits_{h\to0}\frac{h}{h} = 1
} f β² ( x ) = = = β h β 0 lim β h f ( x + h ) β f ( x ) β = h β 0 lim β h x + h β x β = h β 0 lim β h h β = 1 β β Rappresentazione grafica
La funzione identitΓ Γ¨ la bisettrice del primo e terzo quadrante e coincide con la tangente al grafico: il coefficiente angolare Γ¨ uguale a 1 1 1
Esempio f ( x ) = x β
β βΉ β
β f β² ( x ) = 1 f(x)= x \implies f'(x) = 1 f ( x ) = x βΉ f β² ( x ) = 1
Teorema Ξ± β R \alpha \in\R Ξ± β R x > 0 x > 0 x > 0 D x Ξ± = Ξ± x Ξ± β 1 D x^\alpha = \alpha x^{\alpha-1} D x Ξ± = Ξ± x Ξ± β 1 Ξ± β Z \alpha \in \Z Ξ± β Z Ξ± = m n \alpha = \frac{m}{n} Ξ± = n m β n n n x < 0 x<0 x < 0
Inoltre, per n β N β { 0 } n \in \N-\{0\} n β N β { 0 } β x β R \forall x \in\R β x β R D x n = n x n β 1 D x^n = nx^{n-1} D x n = n x n β 1
Dimostrazione
f β² ( x ) = lim β‘ h β 0 f ( x + h ) β f ( x ) h = = lim β‘ h β 0 ( x + h ) Ξ± β x Ξ± h = = lim β‘ h β 0 x Ξ± ( 1 + h x ) Ξ± β x Ξ± h = = lim β‘ h β 0 x Ξ± ( 1 + h x ) Ξ± β 1 h = = lim β‘ h β 0 x Ξ± β 1 ( 1 + h x ) Ξ± β 1 h x = = β
β Ξ± x Ξ± β 1 \eq{
f'(x) =& \lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h} =\\
=& \lim\limits_{h\to0}\frac{(x+h)^\alpha-x^\alpha}{h} =\\
=& \lim\limits_{h\to0}\frac{x^\alpha(1+\frac{h}{x})^\alpha-x^\alpha}{h} =\\
=& \lim\limits_{h\to0}x^\alpha\frac{(1+\frac{h}{x})^\alpha-1}{h} =\\
=& \lim\limits_{h\to0}x^{\alpha-1}\frac{(1+\frac{h}{x})^\alpha-1}{\frac{h}{x}}=\\
=& \;\alpha x^{\alpha-1}
} f β² ( x ) = = = = = = β h β 0 lim β h f ( x + h ) β f ( x ) β = h β 0 lim β h ( x + h ) Ξ± β x Ξ± β = h β 0 lim β h x Ξ± ( 1 + x h β ) Ξ± β x Ξ± β = h β 0 lim β x Ξ± h ( 1 + x h β ) Ξ± β 1 β = h β 0 lim β x Ξ± β 1 x h β ( 1 + x h β ) Ξ± β 1 β = Ξ± x Ξ± β 1 β β Rappresentazione grafica
2Β° Teorema
Siano n β R n \in\R n β R x > 0 x > 0 x > 0
D [ 1 x n ] = n x n + 1 D \left[ \frac{1}{x^n} \right] = \frac{n}{x^{n+1}} D [ x n 1 β ] = x n + 1 n β Teorema Ξ± = 1 2 \alpha = \frac{1}{2} Ξ± = 2 1 β x > 0 x > 0 x > 0 D x = D β
β x Ξ± = 1 2 x D \sqrt{x} = D\;x^\alpha = \frac{1}{2\sqrt{x}} D x β = D x Ξ± = 2 x β 1 β
Dimostrazione ( a + b ) ( a β b ) = a 2 β b 2 (a+b)(a-b)=a^2-b^2 ( a + b ) ( a β b ) = a 2 β b 2
f β² ( x ) = lim β‘ h β 0 f ( x + h ) β f ( x ) h = = lim β‘ h β 0 x + h β x h = = lim β‘ h β 0 ( x + h β x ) ( x + h + x ) h ( x + h + x ) = = lim β‘ h β 0 ( x + h β x ) h ( x + h + x ) = = lim β‘ h β 0 h h ( x + h + x ) = = lim β‘ h β 0 1 h ( x + x ) = 1 2 x \eq{
f'(x) =& \lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h} =\\
=& \lim\limits_{h\to0}\frac{\sqrt{x+h}-\sqrt{x}}{h} =\\
=& \small
\lim\limits_{h\to0}\frac{(\sqrt{x+h}-\sqrt{x})(\sqrt{x+h}+\sqrt{x})}{h(\sqrt{x+h}+\sqrt{x})}=\\
=& \normalsize\lim\limits_{h\to0}\frac{(x+h-x)}{h(\sqrt{x+h}+\sqrt{x})} =\\
=& \lim\limits_{h\to0}\frac{h}{h(\sqrt{x+h}+\sqrt{x})} =\\
=& \lim\limits_{h\to0}\frac{1}{h(\sqrt{x}+\sqrt{x})} = \frac{1}{2\sqrt{x}}
} f β² ( x ) = = = = = = β h β 0 lim β h f ( x + h ) β f ( x ) β = h β 0 lim β h x + h β β x β β = h β 0 lim β h ( x + h β + x β ) ( x + h β β x β ) ( x + h β + x β ) β = h β 0 lim β h ( x + h β + x β ) ( x + h β x ) β = h β 0 lim β h ( x + h β + x β ) h β = h β 0 lim β h ( x β + x β ) 1 β = 2 x β 1 β β β Si ricava che D x = D x 1 2 = 1 2 x 1 2 β 1 = 1 2 x β 1 2 = 1 2 x β x > 0 D \sqrt{x} = D x^{\frac{1}{2}} = \frac{1}{2} x^{\frac{1}{2}-1} = \frac{1}{2} x^{-\frac{1}{2}} = \frac{1}{2 \sqrt{x}} \quad \forall x > 0 D x β = D x 2 1 β = 2 1 β x 2 1 β β 1 = 2 1 β x β 2 1 β = 2 x β 1 β β x > 0
Rappresentazione grafica
Teorema x x x D sin β‘ ( x ) = cos β‘ ( x ) D \sin(x) = \cos(x) D sin ( x ) = cos ( x )
Dimostrazione sin β‘ ( Ξ± + Ξ² ) = sin β‘ ( Ξ± ) β
cos β‘ ( Ξ² ) + cos β‘ ( Ξ± ) β
sin β‘ ( Ξ² ) \sin(\alpha+\beta)=\sin(\alpha)\cdot \cos(\beta)+\cos(\alpha)\cdot \sin(\beta) sin ( Ξ± + Ξ² ) = sin ( Ξ± ) β
cos ( Ξ² ) + cos ( Ξ± ) β
sin ( Ξ² )
f β² ( x ) = lim β‘ h β 0 f ( x + h ) β f ( x ) h = = lim β‘ h β 0 sin β‘ ( x + h ) β sin β‘ ( x ) h = = lim β‘ h β 0 sin β‘ ( x ) β
cos β‘ ( h ) + cos β‘ ( x ) β
sin β‘ ( h ) β sin β‘ ( x ) h = = lim β‘ h β 0 sin β‘ ( x ) [ cos β‘ ( h ) β 1 ] + cos β‘ ( x ) β
sin β‘ ( h ) h = = lim β‘ h β 0 sin β‘ ( x ) cos β‘ ( h ) β 1 h + cos β‘ ( x ) β
sin β‘ ( h ) h = = β
β sin β‘ ( x ) β
0 + cos β‘ ( x ) β
1 = cos β‘ ( x ) \eq{
f'(x) =& \lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h} =\\
=& \lim\limits_{h\to0}\frac{\sin (x+h)- \sin(x)}{h} = \\
=& \lim\limits_{h\to0}\frac{\sin(x)\cdot \cos (h) + \cos (x)\cdot \sin(h)-\sin(x)}{h} = \\
=& \lim\limits_{h\to0}\frac{\sin(x)[\cos (h)-1]+\cos(x)\cdot \sin(h)}{h} = \\
=& \lim\limits_{h\to0} \sin(x)\frac{\cos(h)-1}{h}+ \cos(x)\cdot\frac{\sin(h)}{h} = \\
=& \;\sin(x)\cdot 0 + \cos(x)\cdot 1 = \cos (x)
} f β² ( x ) = = = = = = β h β 0 lim β h f ( x + h ) β f ( x ) β = h β 0 lim β h sin ( x + h ) β sin ( x ) β = h β 0 lim β h sin ( x ) β
cos ( h ) + cos ( x ) β
sin ( h ) β sin ( x ) β = h β 0 lim β h sin ( x ) [ cos ( h ) β 1 ] + cos ( x ) β
sin ( h ) β = h β 0 lim β sin ( x ) h cos ( h ) β 1 β + cos ( x ) β
h sin ( h ) β = sin ( x ) β
0 + cos ( x ) β
1 = cos ( x ) β β Rappresentazione grafica
La funzione seno Γ¨ periodica con periodo 2 Ο 2\pi 2 Ο
Teorema x x x D [ cos β‘ ( x ) ] = β sin β‘ ( x ) D[\cos(x)] = -\sin(x) D [ cos ( x )] = β sin ( x )
Dimostrazione
Rappresentazione grafica
La funzione coseno Γ¨ periodica di periodo 2 Ο 2\pi 2 Ο
Teorema
D β
β tan β‘ ( x ) = 1 cos β‘ 2 ( x ) = 1 + tan β‘ 2 ( x ) D\;\tan(x) = \frac{1}{\cos^2(x)} = 1+\tan^2(x) D tan ( x ) = cos 2 ( x ) 1 β = 1 + tan 2 ( x ) Rappresentazione grafica
Teorema
D β
β arctan ( x ) = tan β‘ β 1 ( x ) = 1 x 2 + 1 D\;\text{arctan}(x) = \tan^{-1}(x) = \frac{1}{x^2+1} D arctan ( x ) = tan β 1 ( x ) = x 2 + 1 1 β Teorema
D β
β cot β‘ ( x ) = β 1 sin β‘ 2 ( x ) = β [ 1 + c o t 2 ( x ) ] D\;\cot(x) = -\frac{1}{\sin^2(x)} = -[1+cot^2(x)] D cot ( x ) = β sin 2 ( x ) 1 β = β [ 1 + co t 2 ( x )] Rappresentazione grafica
Teorema D β
β Ξ± x = Ξ± x β
log β‘ e ( Ξ± ) = Ξ± x β
ln β‘ ( Ξ± ) D\;\alpha^x = \alpha^x \cdot \log_e(\alpha)=\alpha^x \cdot \ln(\alpha) D Ξ± x = Ξ± x β
log e β ( Ξ± ) = Ξ± x β
ln ( Ξ± )
Se Ξ± = e \alpha = e Ξ± = e D β
β Ξ± x = Ξ± x D\;\alpha^x = \alpha^x D Ξ± x = Ξ± x ln β‘ e = 1 \ln e = 1 ln e = 1
Dimostrazione
f β² ( x ) = lim β‘ h β 0 f ( x + h ) β f ( x ) h = = lim β‘ h β 0 Ξ± x + h β Ξ± x h = = lim β‘ h β 0 Ξ± x ( Ξ± h β 1 ) h = = lim β‘ h β 0 ( Ξ± x Ξ± h β 1 h ) = = β
β Ξ± x β
ln β‘ β
β Ξ± \eq{
f'(x) =& \lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h} =\\
=& \lim\limits_{h\to0}\frac{\alpha^{x+h}-\alpha^x}{h} =\\
=& \lim\limits_{h\to0}\frac{\alpha^x(\alpha^h-1)}{h} =\\
=& \lim\limits_{h\to0}(\alpha^x\frac{\alpha^h-1}{h}) =\\
=& \;\alpha^x \cdot \ln\;\alpha
} f β² ( x ) = = = = = β h β 0 lim β h f ( x + h ) β f ( x ) β = h β 0 lim β h Ξ± x + h β Ξ± x β = h β 0 lim β h Ξ± x ( Ξ± h β 1 ) β = h β 0 lim β ( Ξ± x h Ξ± h β 1 β ) = Ξ± x β
ln Ξ± β β Rappresentazione grafica
Esempio f ( x ) = 2 x β
β βΉ β
β f β² ( x ) = 2 x β
ln β‘ ( 2 ) f(x)=2^x \implies f'(x)=2^x \cdot \ln(2) f ( x ) = 2 x βΉ f β² ( x ) = 2 x β
ln ( 2 )
Teorema
D β
β log β‘ Ξ± x = 1 x β
log β‘ Ξ± e D\;\log_\alpha x = \frac{1}{x} \cdot \log_\alpha e D log Ξ± β x = x 1 β β
log Ξ± β e Se Ξ± = e \alpha = e Ξ± = e D β
β ln β‘ x = 1 x D\;\ln x = \frac{1}{x} D ln x = x 1 β D β
β log β‘ x = 1 x D\;\log x = \frac{1}{x} D log x = x 1 β D [ e x ] = e x D[e^x] = e^x D [ e x ] = e x
Dimostrazione log β‘ Ξ± x β log β‘ Ξ± y = log β‘ Ξ± x y \log_\alpha x - \log_\alpha y = \log_\alpha \frac{x}{y} log Ξ± β x β log Ξ± β y = log Ξ± β y x β
2Β° Teorema D log β‘ β£ x β£ = 1 x β x β 0 D \log |x| = \frac{1}{x} \quad \forall x \neq 0 D log β£ x β£ = x 1 β β x ξ = 0
Se x > 0 x>0 x > 0 D [ log β‘ x ] = 1 x D[\log x] = \frac{1}{x} D [ log x ] = x 1 β
Se invece x < 0 x<0 x < 0
d d x log β‘ β£ x β£ = d d x log β‘ ( β x ) = d d x log β‘ y β£ y = β x β
d d x ( β x ) = 1 β x ( β 1 ) = 1 x \frac{d}{dx} \log |x| = \frac{d}{dx} \log(-x) = \frac{d}{dx} \log y|_{y=-x} \cdot \frac{d}{dx}(-x) = \frac{1}{-x}(-1) = \frac{1}{x} d x d β log β£ x β£ = d x d β log ( β x ) = d x d β log y β£ y = β x β β
d x d β ( β x ) = β x 1 β ( β 1 ) = x 1 β
3Β° Teorema
D log β‘ β£ f ( x ) β£ = f β² ( x ) f ( x ) β f ( x ) β 0 D \log |f(x)| = \frac{f'(x)}{f(x)} \quad \forall f(x)\neq 0 D log β£ f ( x ) β£ = f ( x ) f β² ( x ) β β f ( x ) ξ = 0 Dimostrazione
D log β‘ β£ f ( x ) β£ = 1 f ( x ) β
f β² ( x ) = f β² ( x ) f ( x ) D \log |f(x)| = \frac{1}{f(x)} \cdot f'(x) = \frac{f'(x)}{f(x)} D log β£ f ( x ) β£ = f ( x ) 1 β β
f β² ( x ) = f ( x ) f β² ( x ) β Teorema I = ( a ; β
β b ) , β
β f : I βΆ R , β
β x 0 β I I= (a; \; b), \; f: I \lto \R, \; x_0 \in I I = ( a ; b ) , f : I βΆ R , x 0 β β I f f f I I I x 0 x_0 x 0 β f β² ( x 0 ) β 0 f'(x_0) \ne 0 f β² ( x 0 β ) ξ = 0 f β 1 f^{-1} f β 1 y 0 = f ( x 0 ) y_0=f(x_0) y 0 β = f ( x 0 β )
( f β 1 ) β² ( y 0 ) = 1 f β² ( x 0 ) D [ f β 1 ( y ) ] = 1 f β² ( x ) \left( f^{-1} \right)' (y_0) =\frac{1}{f'(x_0)} \\
D[f^{-1}(y)]=\frac{1}{f'(x)} ( f β 1 ) β² ( y 0 β ) = f β² ( x 0 β ) 1 β D [ f β 1 ( y )] = f β² ( x ) 1 β Dimostrazione y = f ( x ) β
β βΊ β
β x = f β 1 ( y ) : y = f(x)\iff x = f^{-1} (y): y = f ( x ) βΊ x = f β 1 ( y ) :
D [ f β 1 ( y ) ] = = lim β‘ y β y 0 f β 1 ( y ) β f β 1 ( y 0 ) y β y 0 = = lim β‘ y β y 0 f β 1 ( f ( x ) ) f β 1 ( f ( x 0 ) ) f ( x ) β f ( x 0 ) = = lim β‘ y β y 0 x β x 0 f ( x ) β f ( x 0 ) = = lim β‘ y β y 0 1 f ( x ) β f ( x 0 ) x β x 0 = 1 f β² ( x 0 ) \eq{
& D[f^{-1}(y)]=\\
=&\lim\limits_{y \to y_0} \frac{f^{-1}(y)-f^{-1}(y_0)}{y - y_0} =\\
=& \lim\limits_{y \to y_0} \frac{f^{-1} \left( f(x)\right) f^{-1}\left( f(x_0) \right)}{f(x)- f(x_0)} =\\
=& \lim\limits_{y \to y_0} \frac{x - x_0}{f(x)- f(x_0)} =\\
=&\lim\limits_{y \to y_0} \frac{1}{\frac{f(x)- f(x_0)}{x - x_0}} = \frac{1}{f'(x_0)} \\
} = = = = β D [ f β 1 ( y )] = y β y 0 β lim β y β y 0 β f β 1 ( y ) β f β 1 ( y 0 β ) β = y β y 0 β lim β f ( x ) β f ( x 0 β ) f β 1 ( f ( x ) ) f β 1 ( f ( x 0 β ) ) β = y β y 0 β lim β f ( x ) β f ( x 0 β ) x β x 0 β β = y β y 0 β lim β x β x 0 β f ( x ) β f ( x 0 β ) β 1 β = f β² ( x 0 β ) 1 β β β f β C ( I ) f \in C(I) f β C ( I ) β
β βΉ β
β f β 1 β C ( I ) \implies f^{-1} \in C(I) βΉ f β 1 β C ( I )
Se x β x 0 β
β βΉ β
β f β 1 ( x ) β f β 1 ( x 0 ) β
β βΊ β
β y β y 0 x \to x_0 \implies f^{-1}(x) \to f^{-1}(x_0) \iff y \to y_0 x β x 0 β βΉ f β 1 ( x ) β f β 1 ( x 0 β ) βΊ y β y 0 β x β x 0 β
β βΊ β
β y β y 0 x \to x_0 \iff y \to y_0 x β x 0 β βΊ y β y 0 β
Esempio 1 f ( x ) = x 3 + x f(x)=x^3+x f ( x ) = x 3 + x R \R R y = 2 y=2 y = 2 x x x y = 2 y=2 y = 2
x 3 + x = 2 β
β βΉ β
β ( x β 1 ) ( x 2 + x + 2 ) = 0 β
β βΉ β
β x = 1 β
β βΉ β
β f β² ( x ) = 3 x 2 + 1 β
β βΉ β
β f β² ( 1 ) = 3 + 1 = 4 \eq{
& x^3+x=2 \\
\implies & (x-1)(x^2+x+2)=0\\
\implies & x=1 \\
\implies & f'(x)=3x^2+1 \\
\implies & f'(1)=3+1=4
} βΉ βΉ βΉ βΉ β x 3 + x = 2 ( x β 1 ) ( x 2 + x + 2 ) = 0 x = 1 f β² ( x ) = 3 x 2 + 1 f β² ( 1 ) = 3 + 1 = 4 β β Si applichi il teorema: D [ f β 1 ( 2 ) ] = 1 f β² ( 1 ) = 1 4 D[f^{-1}(2)]=\frac{1}{f'(1)}=\frac{1}{4} D [ f β 1 ( 2 )] = f β² ( 1 ) 1 β = 4 1 β
Esempio 2
D [ arcsin ( x ) ] = 1 1 β x 2 β x β ( β 1 , β
β 1 ) D [ arcsin ( x ) ] = 1 D sin β‘ y β£ y = arcsin x = = 1 cos β‘ y β£ y = arcsin y = = 1 1 β sin β‘ 2 y β£ y = arcsin x = = 1 1 β x 2 \eq{
D \Big[\text{arcsin}(x)\Big] =& \frac{1}{\sqrt{1-x^2}} \quad \forall x \in (-1, \; 1) \\
D \Big[\text{arcsin}(x)\Big] =& \frac{1}{D\sin y} \LARGE| \normalsize _{ y = \text{arcsin } x } =\\
=& \frac{1}{\cos y |_{y = \text{arcsin } y}} =\\
=& \frac{1}{\sqrt{1-\sin^2 y}} \LARGE| \normalsize _{ y = \text{arcsin } x } =\\
=& \frac{1}{\sqrt{1-x^2}}
} D [ arcsin ( x ) ] = D [ arcsin ( x ) ] = = = = β 1 β x 2 β 1 β β x β ( β 1 , 1 ) D sin y 1 β β£ y = arcsin x β = cos y β£ y = arcsin y β 1 β = 1 β sin 2 y β 1 β β£ y = arcsin x β = 1 β x 2 β 1 β β β Non dimenticarti che:
cos β‘ 2 ( y ) + sin β‘ 2 ( y ) = 1 \cos^2 (y) + \sin^2 (y)=1 cos 2 ( y ) + sin 2 ( y ) = 1
y = log β‘ a ( x ) β
β βΊ β
β a y = x y =\log_a(x) \iff a^y = x y = log a β ( x ) βΊ a y = x
a log β‘ a x = x a^{\log_a x} = x a l o g a β x = x
D [ arccos ( x ) ] = β 1 1 β x 2 β x β ( β 1 , β
β 1 ) D [ arctg ( x ) ] = 1 1 + x 2 β x β R D [ log β‘ a ( x ) ] = 1 x log β‘ a β x > 0 , β
β a > 0 , β
β a β 1 D [ log β‘ a ( x ) ] = 1 D a y β£ y = log β‘ a x = = 1 a y log β‘ a β£ y = log β‘ a x = = 1 x log β‘ a \eq{
D \Big[\text{arccos}(x)\Big] =& - \frac{1}{\sqrt{1-x^2}} \quad \forall x \in (-1, \; 1) \\
D \Big[\text{arctg}(x)\Big] =& \frac{1}{1+x^2} \quad \forall x \in \R \\
D \Big[\log_a (x)\Big] =& \frac{1}{x \log a} \quad \forall x > 0, \; a > 0, \; a \neq 1 \\
D \Big[\log_a (x)\Big] =& \frac{1}{D a^y} \LARGE| \normalsize _{y = \log_a x} = \\
=& \frac{1}{a^y \log a} \LARGE| \normalsize _{y = \log_a x} =\\
=&\frac{1}{x \log a}
} D [ arccos ( x ) ] = D [ arctg ( x ) ] = D [ log a β ( x ) ] = D [ log a β ( x ) ] = = = β β 1 β x 2 β 1 β β x β ( β 1 , 1 ) 1 + x 2 1 β β x β R x log a 1 β β x > 0 , a > 0 , a ξ = 1 D a y 1 β β£ y = l o g a β x β = a y log a 1 β β£ y = l o g a β x β = x log a 1 β β β Teorema I , β
β J β R , β
β g : I βΆ R , β
β f : J βΆ R , β
β g ( I ) β J , β
β x 0 β I I, \; J \subseteq
\R, \; g: I \lto \R, \; f: J \lto\R, \; g(I) \subseteq
J, \; x_0 \in I I , J β R , g : I βΆ R , f : J βΆ R , g ( I ) β J , x 0 β β I g g g x 0 x_0 x 0 β f f f g ( x 0 ) g(x_0) g ( x 0 β ) f β g = f ( g ( x ) ) f \circ g = f(g(x)) f β g = f ( g ( x )) x 0 x_0 x 0 β
D [ f ( g ( x ) ) ] = f β² ( g ( x ) ) β
g β² ( x ) D[f(g(x))]=f'(g(x)) \cdot g'(x) D [ f ( g ( x ))] = f β² ( g ( x )) β
g β² ( x ) Esempio 1 D [ x Ξ± ] = Ξ± x Ξ± β 1 β x > 0 , β
β Ξ± β R D[x^{\alpha}] = \alpha x ^{\alpha -1} \quad \forall x > 0, \; \alpha \in \R D [ x Ξ± ] = Ξ± x Ξ± β 1 β x > 0 , Ξ± β R
D [ x Ξ± ] = D [ e ln β‘ ( x Ξ± ) ] = = D [ e Ξ± ln β‘ ( x ) ] = = β
β f β g t.c. g ( x ) = Ξ± ln β‘ ( x ) , β
β f ( y ) = e y β
β βΉ β
β β
β D [ e y β£ y = Ξ± ln β‘ ( x ) ] β
D [ Ξ± ln β‘ ( x ) ] = = β
β e Ξ± ln β‘ ( x ) Ξ± 1 x = = β
β Ξ± x Ξ± β 1 \eq{
D[x^{\alpha}] =& D[e^{\ln (x^{\alpha})}] =\\
=& D[e^{\alpha \ln (x)}] =\\
=& \; f \circ g \text{ t.c. } g(x) = \alpha \ln (x) , \; f(y) = e^y \\
\implies & \; D[e^y |_{y= \alpha \ln (x)}] \cdot D[\alpha \ln (x)] =\\
=& \; e^{\alpha \ln (x)} \alpha \frac{1}{x} =\\
=& \;\alpha x^{\alpha -1}
} D [ x Ξ± ] = = = βΉ = = β D [ e l n ( x Ξ± ) ] = D [ e Ξ± l n ( x ) ] = f β g t.c. g ( x ) = Ξ± ln ( x ) , f ( y ) = e y D [ e y β£ y = Ξ± l n ( x ) β ] β
D [ Ξ± ln ( x )] = e Ξ± l n ( x ) Ξ± x 1 β = Ξ± x Ξ± β 1 β β Esempio 2 D β
β a x = a x β
log β‘ a β x β R , a > 0 D \; a^x = a^x \cdot \log a \quad \quad \quad \forall x \in \R, a > 0 D a x = a x β
log a β x β R , a > 0
D [ a x ] = D [ e log β‘ a x ] = = D [ e x log β‘ a ] = = β
β f β g ( x ) t.c. g ( x ) = x log β‘ a , β
β f ( y ) = e y β
β βΉ β
β D e y β£ y = x log β‘ a β
D ( x log β‘ a ) = β
β e x log β‘ a β
log β‘ a β
1 = = a x β
log β‘ a \eq{
D[a^x] =& D[e^{\log a^x}] =\\
=& D [e^{x \log a}] =\\
=& \; f \circ g(x) \text{ t.c. }g(x) = x \log a , \; f(y) = e^y \\
\implies & D e^y |_{y= x \log a} \cdot D(x \log a) \\
=& \; e^{x \log a} \cdot \log a \cdot 1 =\\
=& a^x \cdot \log a
} D [ a x ] = = = βΉ = = β D [ e l o g a x ] = D [ e x l o g a ] = f β g ( x ) t.c. g ( x ) = x log a , f ( y ) = e y D e y β£ y = x l o g a β β
D ( x log a ) e x l o g a β
log a β
1 = a x β
log a β β Teorema
D β
β [ k β
f ( x ) ] = k β
f β² ( x ) D\;[k \cdot f(x)]= k \cdot f'(x) D [ k β
f ( x )] = k β
f β² ( x ) Dimostrazione
f β² ( x ) = lim β‘ h β 0 k β
f ( x + h ) β k β
f ( x ) h = = lim β‘ h β 0 k β
[ f ( x + h ) β f ( x ) ] h = = lim β‘ h β 0 k β
β f ( x + h ) β f ( x ) h = k β
β β
β
β f β² ( x ) \eq{
f'(x) =& \lim\limits_{h\to0}\frac{k \cdot f(x+h)- k \cdot f(x)}{h} = \\
=& \lim\limits_{h\to0}\frac{k \cdot [f(x+h)- f(x)]}{h} =\\
=& \lim\limits_{h\to0}k\;\frac{f(x+h)- f(x)}{h} = k\;\cdot\;f'(x)
} f β² ( x ) = = = β h β 0 lim β h k β
f ( x + h ) β k β
f ( x ) β = h β 0 lim β h k β
[ f ( x + h ) β f ( x )] β = h β 0 lim β k h f ( x + h ) β f ( x ) β = k β
f β² ( x ) β β Esempio y = β 3 β
β β
ln β‘ x β
β βΉ β
β y β² = β 3 β
1 x = β 3 x y=-3 \; \cdot \ln x \implies y' = -3 \cdot \frac{1}{x}= -\frac{3}{x} y = β 3 β
ln x βΉ y β² = β 3 β
x 1 β = β x 3 β
Teorema D β
β [ f ( x ) Β± g ( x ) ] = f β² ( x ) Β± g β² ( x ) D\;[f(x) \pm g(x)]= f'(x) \pm g'(x) D [ f ( x ) Β± g ( x )] = f β² ( x ) Β± g β² ( x )
Dimostrazione I ) Sia h ( x ) = f ( x ) + g ( x ) h(x) = f(x)+ g(x) h ( x ) = f ( x ) + g ( x ) D [ h ( x ) ] = f β² ( x ) + g β² ( x ) D[h(x)]=f'(x) + g'(x) D [ h ( x )] = f β² ( x ) + g β² ( x )
h β² ( x ) = lim β‘ h β 0 [ f ( x + h ) + g ( x + h ) ] β [ f ( x ) + g ( x ) ] h = = lim β‘ h β 0 [ f ( x + h ) β f ( x ) ] + [ g ( x + h ) β g ( x ) ] h = = lim β‘ h β 0 f ( x + h ) β f ( x ) h + lim β‘ h β 0 g ( x + h ) β g ( x ) h = = f β² ( x ) + g β² ( x ) \eq{
h'(x) =& \lim\limits_{h\to0}\frac{[f(x+h)+g(x+h)]-[f(x)+g(x)]}{h} =\\
=& \lim\limits_{h\to0}\frac{[f(x+h)-f(x)]+[g(x+h)-g(x)]}{h} =\\
=& \lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h} + \lim\limits_{h\to0}\frac{g(x+h)-g(x)}{h} =\\
=&f'(x) + g'(x)
} h β² ( x ) = = = = β h β 0 lim β h [ f ( x + h ) + g ( x + h )] β [ f ( x ) + g ( x )] β = h β 0 lim β h [ f ( x + h ) β f ( x )] + [ g ( x + h ) β g ( x )] β = h β 0 lim β h f ( x + h ) β f ( x ) β + h β 0 lim β h g ( x + h ) β g ( x ) β = f β² ( x ) + g β² ( x ) β β II ) Sia v ( x ) = f ( x ) β g ( x ) v(x) = f(x)- g(x) v ( x ) = f ( x ) β g ( x ) D [ v ( x ) ] = f β² ( x ) β g β² ( x ) D[v(x)]=f'(x) - g'(x) D [ v ( x )] = f β² ( x ) β g β² ( x )
v β² ( x ) = lim β‘ h β 0 [ f ( x + h ) β g ( x + h ) ] β [ f ( x ) β g ( x ) ] h = = lim β‘ h β 0 [ f ( x + h ) β f ( x ) ] β [ g ( x + h ) β g ( x ) ] h = = lim β‘ h β 0 f ( x + h ) β f ( x ) h β lim β‘ h β 0 g ( x + h ) β g ( x ) h = = f β² ( x ) β g β² ( x ) \eq{
v'(x) =& \lim\limits_{h\to0}\frac{[f(x+h)-g(x+h)]-[f(x)-g(x)]}{h} =\\
& = \lim\limits_{h\to0}\frac{[f(x+h)-f(x)]-[g(x+h)-g(x)]}{h} =\\
& = \lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h} - \lim\limits_{h\to0}\frac{g(x+h)-g(x)}{h} =\\
=& f'(x) - g'(x)
} v β² ( x ) = = β h β 0 lim β h [ f ( x + h ) β g ( x + h )] β [ f ( x ) β g ( x )] β = = h β 0 lim β h [ f ( x + h ) β f ( x )] β [ g ( x + h ) β g ( x )] β = = h β 0 lim β h f ( x + h ) β f ( x ) β β h β 0 lim β h g ( x + h ) β g ( x ) β = f β² ( x ) β g β² ( x ) β β Esempio f ( x ) = x f(x) = x f ( x ) = x g ( x ) = 2 β
sin β‘ ( x ) g(x)=2 \cdot \sin(x) g ( x ) = 2 β
sin ( x ) y = x + 2 β
sin β‘ ( x ) β
β βΉ β
β y β² = 1 + 2 β
cos β‘ ( x ) y = x+2 \cdot \sin(x) \implies y'=1+2 \cdot \cos(x) y = x + 2 β
sin ( x ) βΉ y β² = 1 + 2 β
cos ( x )
Teorema
d d x [ f ( x ) β
g ( x ) ] = f β² ( x ) β
g ( x ) + f ( x ) β
g β² ( x ) \frac{d}{dx} [f(x) \cdot g(x)]= f'(x) \cdot g(x) + f(x) \cdot g'(x) d x d β [ f ( x ) β
g ( x )] = f β² ( x ) β
g ( x ) + f ( x ) β
g β² ( x ) Esempio f ( x ) = x f(x) = x f ( x ) = x g ( x ) = sin β‘ ( x ) g(x)=\sin(x) g ( x ) = sin ( x ) y = x β
sin β‘ ( x ) β
β βΉ β
β y β² = 1 β
sin β‘ ( x ) + x β
cos β‘ ( x ) y = x \cdot \sin(x) \implies y'=1 \cdot \sin(x) + x \cdot \cos(x) y = x β
sin ( x ) βΉ y β² = 1 β
sin ( x ) + x β
cos ( x )
Teorema g ( x ) β 0 g(x) \ne 0 g ( x ) ξ = 0
d d x [ f ( x ) g ( x ) ] = f β² ( x ) β
g ( x ) β f ( x ) β
g β² ( x ) [ g ( x ) ] 2 \frac{d}{dx} \left[ \frac{f(x)}{g(x)} \right]= \frac{f'(x) \cdot g(x) - f(x) \cdot g'(x)}{[g(x)]^2} d x d β [ g ( x ) f ( x ) β ] = [ g ( x ) ] 2 f β² ( x ) β
g ( x ) β f ( x ) β
g β² ( x ) β Teorema f ( x ) β 0 f(x) \ne 0 f ( x ) ξ = 0
d d x [ 1 f ( x ) ] = β f β² ( x ) [ f ( x ) ] 2 \frac{d}{dx} \left[ \frac{1}{f(x)} \right] = -\frac{f'(x)}{ [f(x)]^2} d x d β [ f ( x ) 1 β ] = β [ f ( x ) ] 2 f β² ( x ) β Esempio f ( x ) = sin β‘ ( x ) f(x)=\sin(x) f ( x ) = sin ( x )
y = 1 sin β‘ ( x ) β
β βΉ β
β y β² = β cos β‘ ( x ) sin β‘ 2 ( x ) y=\frac{1}{\sin(x)} \implies y'=-\frac{\cos(x)}{\sin^2(x)} y = sin ( x ) 1 β βΉ y β² = β sin 2 ( x ) cos ( x ) β Teorema n β N , n > 1 n \in \N, n > 1 n β N , n > 1
d d x [ f ( x ) ] n = n β
[ f ( x ) ] n β 1 β
f β² ( x ) \frac{d}{dx}[f(x)]^n=n \cdot [f(x)]^{n-1} \cdot f'(x) d x d β [ f ( x ) ] n = n β
[ f ( x ) ] n β 1 β
f β² ( x ) 
