Sabemos que :
*Derivada de uma raiz enésima
[tex]\displaystyle [\sqrt[\text n]{\text u}]' = \frac{\text u' }{\text n. \sqrt[\text n]{\text u}}[/tex] , sendo u é uma função.
* Regra do produto
[tex][\text{ f . g } ]' = \text{ f ' . g + \text f . g ' }[/tex]
Temos :
[tex]\text{h(x)} = \sqrt{2\text x+3}.(\text x^2+\text x+1)^5[/tex]
Derivando : Aplicando a regra do produto
[tex]\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 ]'[/tex]
Vamos derivar aqui e depois só substituir :
1) [tex]\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}}}[/tex]
2) [tex][(\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}[/tex]
Substituindo na h'(x) :
[tex]\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 ]'[/tex]
[tex]\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[/tex]
[tex]\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).[/tex]
Tirando o MMC :
[tex]\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}} .[/tex]
[tex]\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[/tex]
OU se quiser pôr [tex](\text x^2+\text x+1)^4[/tex] em evidência
[tex]\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}}[/tex]
[tex]\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[/tex]
