1
Aqui apresentamos um exemplo resolvido passo a passo de regra da cadeia. Esta solução foi gerada automaticamente pela nossa calculadora inteligente:
d d x ( ( 3 x − 2 x 2 ) 3 ) \frac{d}{dx}\left(\left(3x-2x^2\right)^3\right) d x d ( ( 3 x − 2 x 2 ) 3 )
Passos
Aplicamos a regra: d d x ( x a ) \frac{d}{dx}\left(x^a\right) d x d ( x a ) = a x ( a − 1 ) d d x ( x ) =ax^{\left(a-1\right)}\frac{d}{dx}\left(x\right) = a x ( a − 1 ) d x d ( x ) a = 3 a=3 a = 3 x = 3 x − 2 x 2 x=3x-2x^2 x = 3 x − 2 x 2
3 ( 3 x − 2 x 2 ) 3 − 1 d d x ( 3 x − 2 x 2 ) 3\left(3x-2x^2\right)^{3-1}\frac{d}{dx}\left(3x-2x^2\right) 3 ( 3 x − 2 x 2 ) 3 − 1 d x d ( 3 x − 2 x 2 )
Aplicamos a regra: a + b a+b a + b = a + b =a+b = a + b a = 3 a=3 a = 3 b = − 1 b=-1 b = − 1 a + b = 3 − 1 a+b=3-1 a + b = 3 − 1
3 ( 3 x − 2 x 2 ) 2 d d x ( 3 x − 2 x 2 ) 3\left(3x-2x^2\right)^{2}\frac{d}{dx}\left(3x-2x^2\right) 3 ( 3 x − 2 x 2 ) 2 d x d ( 3 x − 2 x 2 )
Aplicamos a regra: d d x ( x a ) \frac{d}{dx}\left(x^a\right) d x d ( x a ) = a x ( a − 1 ) d d x ( x ) =ax^{\left(a-1\right)}\frac{d}{dx}\left(x\right) = a x ( a − 1 ) d x d ( x ) a = 3 a=3 a = 3 x = 3 x − 2 x 2 x=3x-2x^2 x = 3 x − 2 x 2
3 ( 3 x − 2 x 2 ) 3 − 1 d d x ( 3 x − 2 x 2 ) 3\left(3x-2x^2\right)^{3-1}\frac{d}{dx}\left(3x-2x^2\right) 3 ( 3 x − 2 x 2 ) 3 − 1 d x d ( 3 x − 2 x 2 )
Aplicamos a regra: a + b a+b a + b = a + b =a+b = a + b a = 3 a=3 a = 3 b = − 1 b=-1 b = − 1 a + b = 3 − 1 a+b=3-1 a + b = 3 − 1
3 ( 3 x − 2 x 2 ) 2 d d x ( 3 x − 2 x 2 ) 3\left(3x-2x^2\right)^{2}\frac{d}{dx}\left(3x-2x^2\right) 3 ( 3 x − 2 x 2 ) 2 d x d ( 3 x − 2 x 2 )
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Aplicamos a regra: d d x ( x a ) \frac{d}{dx}\left(x^a\right) d x d ( x a ) = a x ( a − 1 ) d d x ( x ) =ax^{\left(a-1\right)}\frac{d}{dx}\left(x\right) = a x ( a − 1 ) d x d ( x ) a = 3 a=3 a = 3 x = 3 x − 2 x 2 x=3x-2x^2 x = 3 x − 2 x 2
3 ( 3 x − 2 x 2 ) 2 d d x ( 3 x − 2 x 2 ) 3\left(3x-2x^2\right)^{2}\frac{d}{dx}\left(3x-2x^2\right) 3 ( 3 x − 2 x 2 ) 2 d x d ( 3 x − 2 x 2 )
Explique melhor esta etapa
3
A derivada da soma de duas ou mais funções é equivalente à soma das derivadas de cada função separadamente
3 ( 3 x − 2 x 2 ) 2 ( d d x ( 3 x ) + d d x ( − 2 x 2 ) ) 3\left(3x-2x^2\right)^{2}\left(\frac{d}{dx}\left(3x\right)+\frac{d}{dx}\left(-2x^2\right)\right) 3 ( 3 x − 2 x 2 ) 2 ( d x d ( 3 x ) + d x d ( − 2 x 2 ) )
Passos
Aplicamos a regra: d d x ( c x ) \frac{d}{dx}\left(cx\right) d x d ( c x ) = c d d x ( x ) =c\frac{d}{dx}\left(x\right) = c d x d ( x )
3 ( 3 x − 2 x 2 ) 2 ( 3 d d x ( x ) + d d x ( − 2 x 2 ) ) 3\left(3x-2x^2\right)^{2}\left(3\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(-2x^2\right)\right) 3 ( 3 x − 2 x 2 ) 2 ( 3 d x d ( x ) + d x d ( − 2 x 2 ) )
Aplicamos a regra: d d x ( x ) \frac{d}{dx}\left(x\right) d x d ( x ) = 1 =1 = 1
3 ( 3 x − 2 x 2 ) 2 ( 3 + d d x ( − 2 x 2 ) ) 3\left(3x-2x^2\right)^{2}\left(3+\frac{d}{dx}\left(-2x^2\right)\right) 3 ( 3 x − 2 x 2 ) 2 ( 3 + d x d ( − 2 x 2 ) )
4
Aplicamos a regra: d d x ( n x ) \frac{d}{dx}\left(nx\right) d x d ( n x ) = n d d x ( x ) =n\frac{d}{dx}\left(x\right) = n d x d ( x ) n = 3 n=3 n = 3
3 ( 3 x − 2 x 2 ) 2 ( 3 d d x ( x ) + d d x ( − 2 x 2 ) ) 3\left(3x-2x^2\right)^{2}\left(3\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(-2x^2\right)\right) 3 ( 3 x − 2 x 2 ) 2 ( 3 d x d ( x ) + d x d ( − 2 x 2 ) )
Explique melhor esta etapa
5
Aplicamos a regra: d d x ( x ) \frac{d}{dx}\left(x\right) d x d ( x ) = 1 =1 = 1
3 ( 3 x − 2 x 2 ) 2 ( 3 + d d x ( − 2 x 2 ) ) 3\left(3x-2x^2\right)^{2}\left(3+\frac{d}{dx}\left(-2x^2\right)\right) 3 ( 3 x − 2 x 2 ) 2 ( 3 + d x d ( − 2 x 2 ) )
6
Aplicamos a regra: d d x ( c x ) \frac{d}{dx}\left(cx\right) d x d ( c x ) = c d d x ( x ) =c\frac{d}{dx}\left(x\right) = c d x d ( x )
3 ( 3 x − 2 x 2 ) 2 ( 3 − 2 d d x ( x 2 ) ) 3\left(3x-2x^2\right)^{2}\left(3-2\frac{d}{dx}\left(x^2\right)\right) 3 ( 3 x − 2 x 2 ) 2 ( 3 − 2 d x d ( x 2 ) )
Passos
Aplicamos a regra: d d x ( x a ) \frac{d}{dx}\left(x^a\right) d x d ( x a ) = a x ( a − 1 ) =ax^{\left(a-1\right)} = a x ( a − 1 ) a = 2 a=2 a = 2
− 4 x ( 2 − 1 ) -4x^{\left(2-1\right)} − 4 x ( 2 − 1 )
Aplicamos a regra: a + b a+b a + b = a + b =a+b = a + b a = 2 a=2 a = 2 b = − 1 b=-1 b = − 1 a + b = 2 − 1 a+b=2-1 a + b = 2 − 1
7
Aplicamos a regra: d d x ( x a ) \frac{d}{dx}\left(x^a\right) d x d ( x a ) = a x ( a − 1 ) =ax^{\left(a-1\right)} = a x ( a − 1 ) a = 2 a=2 a = 2
3 ( 3 x − 2 x 2 ) 2 ( 3 − 2 ⋅ 2 x ) 3\left(3x-2x^2\right)^{2}\left(3-2\cdot 2x\right) 3 ( 3 x − 2 x 2 ) 2 ( 3 − 2 ⋅ 2 x )
Explique melhor esta etapa
8
Aplicamos a regra: a b ab ab = a b =ab = ab a b = − 2 ⋅ 2 x ab=-2\cdot 2x ab = − 2 ⋅ 2 x a = − 2 a=-2 a = − 2 b = 2 b=2 b = 2
3 ( 3 x − 2 x 2 ) 2 ( 3 − 4 x ) 3\left(3x-2x^2\right)^{2}\left(3-4x\right) 3 ( 3 x − 2 x 2 ) 2 ( 3 − 4 x )
Resposta final para o problema
3 ( 3 x − 2 x 2 ) 2 ( 3 − 4 x ) 3\left(3x-2x^2\right)^{2}\left(3-4x\right) 3 ( 3 x − 2 x 2 ) 2 ( 3 − 4 x )