Question

derive e simplifique 2x+1/4(x^2+x+1)+3/8(2x+1+2*raiz de x^2+x+1)

Answer 1

Olá

$\displaystyle \mathsf{y= \frac{2x+1}{4}\cdot \sqrt{x^2+x+1}~+~ \frac{3}{8}\ell n(2x+1+2 \sqrt{x^2+x+1} ) }$

Vamos derivar por partes, para deixar claro o que foi feito.
Então vamos derivar o primeiro termo.

$\displaystyle \mathsf{f= \frac{2x+1}{4}\cdot \sqrt{x^2+x+1} }$

Temos que aplicar a regra do produto ... (f.g)' = f'g + fg'

Derivando

$\displaystyle \mathsf{f'= \frac{2}{4}\cdot \sqrt{x^2+x+1} ~+~ \frac{2x+1}{4}\cdot \left( \frac{2x+1}{2 \sqrt{x^2+x+1} } \right) }\\\\\\\mathsf{f'= \frac{2 \sqrt{x^2+x+1} }{4}~+~ \frac{(2x+1)\cdot (2x+1)}{4\cdot 2 \sqrt{x^2+x+1} } }\\\\\\\text{Expande os termos}\\\\\mathsf{f'=\frac{2 \sqrt{x^2+x+1} }{4}~+~ \frac{4x^2+4x+1}{8 \sqrt{x^2+x+1} } }\\\\\\\text{Faz o MMC}\\\\\\\mathsf{f'= \frac{16\cdot \left( \sqrt{x^2+x+1} \right)^2+4\codt (4x^2+4x+1)}{4\cdot 8\sqrt{x^2+x+1} } }\\\\\\\text{Cancela a raiz }$

$\displaystyle \mathsf{f'= \frac{16\cdot (x^2+x+1)~+~4(4x^2+4x+1)}{32 \sqrt{x^2+x+1} } }\\\\\\\text{Aplica a distributiva}\\\\\\\mathsf{f'= \frac{16x^2+16x+16+16x^2+16x+4}{32 \sqrt{x^2+x+1} } }\\\\\\\text{Agrupa os termos em comum}\\\\\\\mathsf{f'= \frac{32x^2+32x+20}{32 \sqrt{x^2+x+1} } }\\\\\\\text{Poe o 4 em evidencia, para simplificarmos}\\\\\\\mathsf{f'= \frac{\diagup\!\!\!\!4\cdot (8x^2+8x+5)}{\diagup\!\!\!\!\!\!32 \sqrt{x^2+x+1} } }\\\\\\\boxed{\mathsf{f'= \frac{8x^2+8x+5}{8 \sqrt{x^2+x+1} } }}$



Derivamos o primeiro termo. Então vamos derivar o segundo.


$\displaystyle\mathsf{g= \frac{3}{8}\ell n(2x+1+2 \sqrt{x^2+x+1} ) }\\\\\\\\\text{Lembrando que a derivada de ln e: } \frac{u'}{u} \\\\\\\mathsf{g'= \frac{3}{8}\cdot \left( \frac{2+ \diagup\!\!\!\!2\cdot \frac{2x+1}{\diagup\!\!\!\!2 \sqrt{x^2+x+1} } }{2x+1+2 \sqrt{x^2+x+1} } \right) }\\\\\\\mathsf{g'= \frac{3}{8}\left( \frac{2+ \frac{2x+1}{ \sqrt{x^2+x+1} } }{2x+1+2 \sqrt{x^2+x+1} } \right) }\\\\\\\text{Faz o MMC}$

$\displaystyle \mathsf{g'= \frac{3}{8}\left( \frac{ \frac{2 \sqrt{x^2+x+1}+2x+1 }{ \sqrt{x^2+x+1} } }{2x+1+2 \sqrt{x^2+x+1} } \right) }\\\\\\\text{Divisao de fracoes, multiplica a primeira pelo inverso da segunda}\\\\\\\mathsf{g'= \frac{3}{8}\left( \frac{2 \sqrt{x^2+x+1}+2x+1 }{ \sqrt{x^2+x+1} }~\cdot ~ \frac{1}{2 \sqrt{x^2+x+1}+2x+1} \right) }\\\\\\\mathsf{Simplifica}\\\\\\\mathsf{g'= \frac{3}{8}\left( \frac{1}{ \sqrt{x^2+x+1} \right)} }\\\\\\\boxed{\mathsf{g'= \frac{3}{8 \sqrt{x^2+x+1} } }}$



Agora que derivamos o segundo termo, juntamos tudo.



$\displaystyle \mathsf{y'= \frac{8x^2+8x+5}{8 \sqrt{x^2+x+1} } ~+~ \frac{3}{8 \sqrt{x^2+x+1} } }\\\\\\\text{faz o MMC, os denominadores sao iguais :)}\\\\\\\mathsf{y'= \frac{8x^2+8x+5+3}{8 \sqrt{x^2+x+1} } }\\\\\\\mathsf{y'= \frac{8x^2+8x+8}{8 \sqrt{x^2+x+1} } }\\\\\\\text{Poe o 8 em evidencia}\\\\\\\mathsf{y'= \frac{\diagup\!\!\!\!8(x^2+x+1)}{\diagup\!\!\!\!8 \sqrt{x^2+x+1} } }\\\\\\\mathsf{y'= \frac{x^2+x+1}{ \sqrt{x^2+x+1} } }$


Racionaliza

$\displaystyle \mathsf{y'= \frac{x^2+x+1}{ \sqrt{x^2+x+1} } ~\cdot ~ \frac{\sqrt{x^2+x+1} }{\sqrt{x^2+x+1} } }\\\\\\\mathsf{y'= \frac{(x^2+x+1)\cdot \sqrt{x^2+x+1} }{ (\sqrt{x^2+x+1})^2 } }\\\\\\\mathsf{y'= \frac{(x^2+x+1)\cdot \sqrt{x^2+x+1} }{ x^2+x+1 }}\\\\\\\text{Simplifica}\\\\\\\\\boxed{\boxed{\mathsf{y'= \sqrt{x^2+x+1} }}}$




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