Question

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

Answer 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

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

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