Matemática, perguntado por BdmLucas, 7 meses atrás

Derivada de: h(x)=\sqrt{2x+3}(x^{2} +x+1)^{5}

Anexos:

Soluções para a tarefa

Respondido por elizeugatao
1

Sabemos que :

*Derivada de uma raiz enésima

\displaystyle [\sqrt[\text n]{\text u}]' =  \frac{\text u' }{\text n. \sqrt[\text n]{\text u}} , sendo u é uma função.

* Regra do produto

[\text{ f . g  } ]' = \text{ f ' . g + \text f . g ' }

Temos :

\text{h(x)} = \sqrt{2\text x+3}.(\text x^2+\text x+1)^5

Derivando : Aplicando a regra do produto

\text{h '(x)} = [\sqrt{2\text x+3}]'.(\text x^2+\text x+1)^5 + \sqrt{2\text x+3}.[(\text x^2+\text x+1)^5 ]'

Vamos derivar aqui e depois só substituir :

1)  \displaystyle [\sqrt{2\text x+3}]' \to \frac{(2\text x+3)'}{2.\sqrt{2\text x+3}} \to \frac{2}{2.\sqrt{2\text x+3}} \to \boxed{\frac{1}{\sqrt{2\text x+3}}}

2)   [(\text x^2+\text x+1)^5]' \to 5.(\text x^2+\text x+1)^4.(\text x^2+\text x+1)' \to \\\\\to 5.(\text x^2+\text x+1)^4.(2\text x+1) \to \boxed{(10\text x+5)(\text x^2+\text x+1)^4}

Substituindo na h'(x) :

\text{h '(x)} = [\sqrt{2\text x+3}]'.(\text x^2+\text x+1)^5 + \sqrt{2\text x+3}.[(\text x^2+\text x+1)^5 ]'

\displaystyle \text{h '(x)} = \frac{1}{\sqrt{2\text x+3}}.(\text x^2+\text x+1)^5 + \sqrt{2\text x+3}.(10\text x+5).(\text x^2+\text x+1)^4

\displaystyle \text{h '(x)} = \frac{(\text x^2+\text x+1)^5}{\sqrt{2\text x+3}} +(\text x^2+\text x+1)^4.  \sqrt{2\text x+3}.(10\text x+5).

Tirando o MMC :

\displaystyle \text{h '(x)} = \frac{(\text x^2+\text x+1)^5+(\text x^2+\text x+1)^4.  (2\text x+3).(10\text x+5)}{\sqrt{2\text x+3}} .

\boxed{\displaystyle \text{h '(x)} = \frac{(\text x^2+\text x+1)^5+(\text x^2+\text x+1)^4.(20\text x^2+40\text x+15)}{\sqrt{2\text x+3}} }\checkmark

OU se quiser pôr (\text x^2+\text x+1)^4 em evidência

\displaystyle \text{h '(x)} = \frac{(\text x^2+\text x+1)^4[\text x^2+\text x+1+20\text x^2+40\text x+15)}{\sqrt{2\text x+3}}

\boxed{\displaystyle \text{h '(x)} = \frac{(\text x^2+\text x+1)^4.[21\text x^2+41\text x+16]}{\sqrt{2\text x+3}}}\checkmark

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