Resposta final para o problema
$10\left(2x+1\right)^{4}\left(x^4-3\right)^6+24\left(2x+1\right)^5\left(x^4-3\right)^{5}x^{3}$
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Solução explicada passo a passo
1
Aplicamos a regra: $\frac{d}{dx}\left(ab\right)$$=\frac{d}{dx}\left(a\right)b+a\frac{d}{dx}\left(b\right)$, onde $d/dx=\frac{d}{dx}$, $ab=\left(2x+1\right)^5\left(x^4-3\right)^6$, $a=\left(2x+1\right)^5$, $b=\left(x^4-3\right)^6$ e $d/dx?ab=\frac{d}{dx}\left(\left(2x+1\right)^5\left(x^4-3\right)^6\right)$
$\frac{d}{dx}\left(\left(2x+1\right)^5\right)\left(x^4-3\right)^6+\left(2x+1\right)^5\frac{d}{dx}\left(\left(x^4-3\right)^6\right)$
Passos
Aplicamos a regra: $\frac{d}{dx}\left(x^a\right)$$=ax^{\left(a-1\right)}\frac{d}{dx}\left(x\right)$, onde $a=5$ e $x=2x+1$
$5\left(2x+1\right)^{5-1}\frac{d}{dx}\left(2x+1\right)\left(x^4-3\right)^6+\left(2x+1\right)^5\frac{d}{dx}\left(\left(x^4-3\right)^6\right)$
Aplicamos a regra: $a+b$$=a+b$, onde $a=5$, $b=-1$ e $a+b=5-1$
$5\left(2x+1\right)^{4}\frac{d}{dx}\left(2x+1\right)\left(x^4-3\right)^6+\left(2x+1\right)^5\frac{d}{dx}\left(\left(x^4-3\right)^6\right)$
Aplicamos a regra: $\frac{d}{dx}\left(x^a\right)$$=ax^{\left(a-1\right)}\frac{d}{dx}\left(x\right)$, onde $a=5$ e $x=2x+1$
$5\left(2x+1\right)^{5-1}\frac{d}{dx}\left(2x+1\right)\left(x^4-3\right)^6$
Aplicamos a regra: $a+b$$=a+b$, onde $a=5$, $b=-1$ e $a+b=5-1$
$5\left(2x+1\right)^{4}\frac{d}{dx}\left(2x+1\right)\left(x^4-3\right)^6$
2
Aplicamos a regra: $\frac{d}{dx}\left(x^a\right)$$=ax^{\left(a-1\right)}\frac{d}{dx}\left(x\right)$, onde $a=5$ e $x=2x+1$
$5\left(2x+1\right)^{4}\frac{d}{dx}\left(2x+1\right)\left(x^4-3\right)^6+\left(2x+1\right)^5\frac{d}{dx}\left(\left(x^4-3\right)^6\right)$
Explique melhor esta etapa
Passos
Aplicamos a regra: $\frac{d}{dx}\left(x^a\right)$$=ax^{\left(a-1\right)}\frac{d}{dx}\left(x\right)$, onde $a=6$ e $x=x^4-3$
$5\left(2x+1\right)^{4}\frac{d}{dx}\left(2x+1\right)\left(x^4-3\right)^6+6\left(2x+1\right)^5\left(x^4-3\right)^{6-1}\frac{d}{dx}\left(x^4-3\right)$
Aplicamos a regra: $\frac{d}{dx}\left(x^a\right)$$=ax^{\left(a-1\right)}\frac{d}{dx}\left(x\right)$, onde $a=5$ e $x=2x+1$
$5\left(2x+1\right)^{5-1}\frac{d}{dx}\left(2x+1\right)\left(x^4-3\right)^6+\left(2x+1\right)^5\frac{d}{dx}\left(\left(x^4-3\right)^6\right)$
Aplicamos a regra: $a+b$$=a+b$, onde $a=5$, $b=-1$ e $a+b=5-1$
$5\left(2x+1\right)^{4}\frac{d}{dx}\left(2x+1\right)\left(x^4-3\right)^6+\left(2x+1\right)^5\frac{d}{dx}\left(\left(x^4-3\right)^6\right)$
Aplicamos a regra: $\frac{d}{dx}\left(x^a\right)$$=ax^{\left(a-1\right)}\frac{d}{dx}\left(x\right)$, onde $a=5$ e $x=2x+1$
$5\left(2x+1\right)^{5-1}\frac{d}{dx}\left(2x+1\right)\left(x^4-3\right)^6$
Aplicamos a regra: $a+b$$=a+b$, onde $a=5$, $b=-1$ e $a+b=5-1$
$5\left(2x+1\right)^{4}\frac{d}{dx}\left(2x+1\right)\left(x^4-3\right)^6$
Aplicamos a regra: $a+b$$=a+b$, onde $a=6$, $b=-1$ e $a+b=6-1$
$5\left(2x+1\right)^{4}\frac{d}{dx}\left(2x+1\right)\left(x^4-3\right)^6+6\left(2x+1\right)^5\left(x^4-3\right)^{5}\frac{d}{dx}\left(x^4-3\right)$
Aplicamos a regra: $\frac{d}{dx}\left(x^a\right)$$=ax^{\left(a-1\right)}\frac{d}{dx}\left(x\right)$, onde $a=6$ e $x=x^4-3$
$6\left(2x+1\right)^5\left(x^4-3\right)^{6-1}\frac{d}{dx}\left(x^4-3\right)$
Aplicamos a regra: $a+b$$=a+b$, onde $a=6$, $b=-1$ e $a+b=6-1$
$6\left(2x+1\right)^5\left(x^4-3\right)^{5}\frac{d}{dx}\left(x^4-3\right)$
3
Aplicamos a regra: $\frac{d}{dx}\left(x^a\right)$$=ax^{\left(a-1\right)}\frac{d}{dx}\left(x\right)$, onde $a=6$ e $x=x^4-3$
$5\left(2x+1\right)^{4}\frac{d}{dx}\left(2x+1\right)\left(x^4-3\right)^6+6\left(2x+1\right)^5\left(x^4-3\right)^{5}\frac{d}{dx}\left(x^4-3\right)$
Explique melhor esta etapa
Passos
Aplicamos a regra: $\frac{d}{dx}\left(c\right)$$=0$, onde $c=1$
$5\left(2x+1\right)^{4}\frac{d}{dx}\left(2x\right)\left(x^4-3\right)^6+6\left(2x+1\right)^5\left(x^4-3\right)^{5}\frac{d}{dx}\left(x^4-3\right)$
4
A derivada da soma de duas ou mais funções é equivalente à soma das derivadas de cada função separadamente
$5\left(2x+1\right)^{4}\frac{d}{dx}\left(2x\right)\left(x^4-3\right)^6+6\left(2x+1\right)^5\left(x^4-3\right)^{5}\frac{d}{dx}\left(x^4-3\right)$
Explique melhor esta etapa
Passos
Aplicamos a regra: $\frac{d}{dx}\left(c\right)$$=0$, onde $c=-3$
$5\left(2x+1\right)^{4}\frac{d}{dx}\left(2x\right)\left(x^4-3\right)^6+6\left(2x+1\right)^5\left(x^4-3\right)^{5}\frac{d}{dx}\left(x^4\right)$
5
A derivada da soma de duas ou mais funções é equivalente à soma das derivadas de cada função separadamente
$5\left(2x+1\right)^{4}\frac{d}{dx}\left(2x\right)\left(x^4-3\right)^6+6\left(2x+1\right)^5\left(x^4-3\right)^{5}\frac{d}{dx}\left(x^4\right)$
Explique melhor esta etapa
Passos
Aplicamos a regra: $\frac{d}{dx}\left(cx\right)$$=c\frac{d}{dx}\left(x\right)$
$10\left(2x+1\right)^{4}\frac{d}{dx}\left(x\right)\left(x^4-3\right)^6$
Aplicamos a regra: $\frac{d}{dx}\left(x\right)$$=1$
$10\left(2x+1\right)^{4}\left(x^4-3\right)^6$
6
Aplicamos a regra: $\frac{d}{dx}\left(nx\right)$$=n\frac{d}{dx}\left(x\right)$, onde $n=2$
$10\left(2x+1\right)^{4}\frac{d}{dx}\left(x\right)\left(x^4-3\right)^6+6\left(2x+1\right)^5\left(x^4-3\right)^{5}\frac{d}{dx}\left(x^4\right)$
Explique melhor esta etapa
7
Aplicamos a regra: $\frac{d}{dx}\left(x\right)$$=1$
$10\left(2x+1\right)^{4}\left(x^4-3\right)^6+6\left(2x+1\right)^5\left(x^4-3\right)^{5}\frac{d}{dx}\left(x^4\right)$
Passos
Aplicamos a regra: $\frac{d}{dx}\left(x^a\right)$$=ax^{\left(a-1\right)}$, onde $a=4$
$24\left(2x+1\right)^5\left(x^4-3\right)^{5}x^{\left(4-1\right)}$
Aplicamos a regra: $a+b$$=a+b$, onde $a=4$, $b=-1$ e $a+b=4-1$
$24\left(2x+1\right)^5\left(x^4-3\right)^{5}x^{3}$
8
Aplicamos a regra: $\frac{d}{dx}\left(x^a\right)$$=ax^{\left(a-1\right)}$, onde $a=4$
$10\left(2x+1\right)^{4}\left(x^4-3\right)^6+6\cdot 4\left(2x+1\right)^5\left(x^4-3\right)^{5}x^{3}$
Explique melhor esta etapa
9
Aplicamos a regra: $ab$$=ab$, onde $ab=6\cdot 4\left(2x+1\right)^5\left(x^4-3\right)^{5}x^{3}$, $a=6$ e $b=4$
$10\left(2x+1\right)^{4}\left(x^4-3\right)^6+24\left(2x+1\right)^5\left(x^4-3\right)^{5}x^{3}$
Resposta final para o problema $10\left(2x+1\right)^{4}\left(x^4-3\right)^6+24\left(2x+1\right)^5\left(x^4-3\right)^{5}x^{3}$
