1
Aqui apresentamos um exemplo resolvido passo a passo de pré-cálculo. Esta solução foi gerada automaticamente pela nossa calculadora inteligente:
d d x ( x x ) \frac{d}{dx}\left(x^x\right) d x d ( x x )
2
Aplicamos a regra: d d x ( a b ) \frac{d}{dx}\left(a^b\right) d x d ( a b ) = y = a b =y=a^b = y = a b , onde d / d x = d d x d/dx=\frac{d}{dx} d / d x = d x d , a = x a=x a = x , b = x b=x b = x , a b = x x a^b=x^x a b = x x e d / d x ? a b = d d x ( x x ) d/dx?a^b=\frac{d}{dx}\left(x^x\right) d / d x ? a b = d x d ( x x )
3
Aplicamos a regra: y = a b y=a^b y = a b → ln ( y ) = ln ( a b ) \to \ln\left(y\right)=\ln\left(a^b\right) → ln ( y ) = ln ( a b ) , onde a = x a=x a = x e b = x b=x b = x
ln ( y ) = ln ( x x ) \ln\left(y\right)=\ln\left(x^x\right) ln ( y ) = ln ( x x )
4
Aplicamos a regra: ln ( x a ) \ln\left(x^a\right) ln ( x a ) = a ln ( x ) =a\ln\left(x\right) = a ln ( x ) , onde a = x a=x a = x
ln ( y ) = x ln ( x ) \ln\left(y\right)=x\ln\left(x\right) ln ( y ) = x ln ( x )
5
Aplicamos a regra: ln ( y ) = x \ln\left(y\right)=x ln ( y ) = x → d d x ( ln ( y ) ) = d d x ( x ) \to \frac{d}{dx}\left(\ln\left(y\right)\right)=\frac{d}{dx}\left(x\right) → d x d ( ln ( y ) ) = d x d ( x ) , onde x = x ln ( x ) x=x\ln\left(x\right) x = x ln ( x )
d d x ( ln ( y ) ) = d d x ( x ln ( x ) ) \frac{d}{dx}\left(\ln\left(y\right)\right)=\frac{d}{dx}\left(x\ln\left(x\right)\right) d x d ( ln ( y ) ) = d x d ( x ln ( x ) )
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Aplicamos a regra: d d x ( a b ) \frac{d}{dx}\left(ab\right) d x d ( ab ) = d d x ( a ) b + a d d x ( b ) =\frac{d}{dx}\left(a\right)b+a\frac{d}{dx}\left(b\right) = d x d ( a ) b + a d x d ( b ) , onde d / d x = d d x d/dx=\frac{d}{dx} d / d x = d x d , a b = x ln ( x ) ab=x\ln\left(x\right) ab = x ln ( x ) , a = x a=x a = x , b = ln ( x ) b=\ln\left(x\right) b = ln ( x ) e d / d x ? a b = d d x ( x ln ( x ) ) d/dx?ab=\frac{d}{dx}\left(x\ln\left(x\right)\right) d / d x ? ab = d x d ( x ln ( x ) )
d d x ( ln ( y ) ) = d d x ( x ) ln ( x ) + x d d x ( ln ( x ) ) \frac{d}{dx}\left(\ln\left(y\right)\right)=\frac{d}{dx}\left(x\right)\ln\left(x\right)+x\frac{d}{dx}\left(\ln\left(x\right)\right) d x d ( ln ( y ) ) = d x d ( x ) ln ( x ) + x d x d ( ln ( x ) )
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Aplicamos a regra: d d x ( x ) \frac{d}{dx}\left(x\right) d x d ( x ) = 1 =1 = 1
d d x ( ln ( y ) ) = ln ( x ) + x d d x ( ln ( x ) ) \frac{d}{dx}\left(\ln\left(y\right)\right)=\ln\left(x\right)+x\frac{d}{dx}\left(\ln\left(x\right)\right) d x d ( ln ( y ) ) = ln ( x ) + x d x d ( ln ( x ) )
Passos
Aplicamos a regra: d d x ( ln ( x ) ) \frac{d}{dx}\left(\ln\left(x\right)\right) d x d ( ln ( x ) ) = 1 x d d x ( x ) =\frac{1}{x}\frac{d}{dx}\left(x\right) = x 1 d x d ( x )
1 y d d x ( y ) = ln ( x ) + x 1 x d d x ( x ) \frac{1}{y}\frac{d}{dx}\left(y\right)=\ln\left(x\right)+x\frac{1}{x}\frac{d}{dx}\left(x\right) y 1 d x d ( y ) = ln ( x ) + x x 1 d x d ( x )
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Aplicamos a regra: d d x ( ln ( x ) ) \frac{d}{dx}\left(\ln\left(x\right)\right) d x d ( ln ( x ) ) = 1 x d d x ( x ) =\frac{1}{x}\frac{d}{dx}\left(x\right) = x 1 d x d ( x )
1 y d d x ( y ) = ln ( x ) + x 1 x d d x ( x ) \frac{1}{y}\frac{d}{dx}\left(y\right)=\ln\left(x\right)+x\frac{1}{x}\frac{d}{dx}\left(x\right) y 1 d x d ( y ) = ln ( x ) + x x 1 d x d ( x )
Explique melhor esta etapa
Passos
Aplicamos a regra: d d x ( x ) \frac{d}{dx}\left(x\right) d x d ( x ) = 1 =1 = 1
y ′ y = ln ( x ) + x 1 x \frac{y^{\prime}}{y}=\ln\left(x\right)+x\frac{1}{x} y y ′ = ln ( x ) + x x 1
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Aplicamos a regra: d d x ( x ) \frac{d}{dx}\left(x\right) d x d ( x ) = 1 =1 = 1
y ′ y = ln ( x ) + x 1 x \frac{y^{\prime}}{y}=\ln\left(x\right)+x\frac{1}{x} y y ′ = ln ( x ) + x x 1
Explique melhor esta etapa
Passos
Aplicamos a regra: a b x a\frac{b}{x} a x b = a b x =\frac{ab}{x} = x ab , onde a = x a=x a = x e b = 1 b=1 b = 1
y ′ y = ln ( x ) + 1 x x \frac{y^{\prime}}{y}=\ln\left(x\right)+\frac{1x}{x} y y ′ = ln ( x ) + x 1 x
Aplicamos a regra: 1 x 1x 1 x = x =x = x
y ′ y = ln ( x ) + x x \frac{y^{\prime}}{y}=\ln\left(x\right)+\frac{x}{x} y y ′ = ln ( x ) + x x
Aplicamos a regra: a a \frac{a}{a} a a = 1 =1 = 1 , onde a = x a=x a = x e a / a = x x a/a=\frac{x}{x} a / a = x x
y ′ y = ln ( x ) + 1 \frac{y^{\prime}}{y}=\ln\left(x\right)+1 y y ′ = ln ( x ) + 1
10
Aplicamos a regra: a b x a\frac{b}{x} a x b = a b x =\frac{ab}{x} = x ab , onde a = x a=x a = x e b = 1 b=1 b = 1
y ′ y = ln ( x ) + 1 \frac{y^{\prime}}{y}=\ln\left(x\right)+1 y y ′ = ln ( x ) + 1
Explique melhor esta etapa
11
Aplicamos a regra: a b = c \frac{a}{b}=c b a = c → a = c b \to a=cb → a = c b , onde a = y ′ a=y^{\prime} a = y ′ , b = y b=y b = y e c = ln ( x ) + 1 c=\ln\left(x\right)+1 c = ln ( x ) + 1
y ′ = ( ln ( x ) + 1 ) y y^{\prime}=\left(\ln\left(x\right)+1\right)y y ′ = ( ln ( x ) + 1 ) y
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Substitua o valor de y y y pelo valor da função original: x x x^x x x
y ′ = ( ln ( x ) + 1 ) x x y^{\prime}=\left(\ln\left(x\right)+1\right)x^x y ′ = ( ln ( x ) + 1 ) x x
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A derivada da função é então
( ln ( x ) + 1 ) x x \left(\ln\left(x\right)+1\right)x^x ( ln ( x ) + 1 ) x x
Resposta final para o problema
( ln ( x ) + 1 ) x x \left(\ln\left(x\right)+1\right)x^x ( ln ( x ) + 1 ) x x