Question

Calcule a derivada:

y=$x \sqrt{a+bx}$

Answer 1

Caso esteja pelo app, experimente abrir esta resposta pelo navegador:  https://brainly.com.br/tarefa/8290537

—————

$\large\begin{array}{l} \texttt{Calcular a derivada da fun\c{c}\~ao}\\\\ \mathtt{y=x\sqrt{a+bx}}\\\\ \mathtt{y=x\cdot (a+bx)^{1/2}\qquad\qquad com~a,\,b~constantes.} \end{array}$


$\large\begin{array}{l} \texttt{Aqui, usamos a Regra do Produto:}\\\\ \mathtt{\dfrac{dy}{dx}=\dfrac{d}{dx}\big[x\cdot (a+bx)^{1/2}\big]}\\\\ \mathtt{\dfrac{dy}{dx}=\dfrac{d}{dx}(x)\cdot (a+bx)^{1/2}+x\cdot \dfrac{d}{dx}\big[(a+bx)^{1/2}\big]}\\\\ \mathtt{\dfrac{dy}{dx}=1\cdot (a+bx)^{1/2}+x\cdot \dfrac{d}{dx}\big[(a+bx)^{1/2}\big]} \end{array}$


$\large\begin{array}{l} \texttt{Agora, para a \'ultima derivada no lado direito, usamos}\\\texttt{a Regra da Cadeia, pois temos uma fun\c{c}\~ao composta:}\\\\ \mathtt{\dfrac{dy}{dx}=1\cdot (a+bx)^{1/2}+x\cdot \left[\dfrac{1}{2}\cdot (a+bx)^{(1/2)-1}\cdot \dfrac{d}{dx}\,(a+bx)\right]}\\\\ \mathtt{\dfrac{dy}{dx}=1\cdot (a+bx)^{1/2}+x\cdot \left[\dfrac{1}{2}\cdot (a+bx)^{-1/2}\cdot (0+b)\right]}\\\\ \mathtt{\dfrac{dy}{dx}=(a+bx)^{1/2}+\dfrac{b}{2}\,x\cdot (a+bx)^{-1/2}} \end{array}$

$\large\begin{array}{l} \mathtt{\dfrac{dy}{dx}=(a+bx)^{1/2}+\dfrac{b}{2}\,x\cdot \dfrac{1}{(a+bx)^{1/2}}} \end{array}$


$\large\begin{array}{l} \therefore~~\boxed{\begin{array}{c}\mathtt{\dfrac{dy}{dx}=\sqrt{a+bx}+\dfrac{bx}{2\sqrt{a+bx}}} \end{array}}\qquad\quad\checkmark \end{array}$


$\large\texttt{Bons estudos! :-)}$


Tags:  derivada função produto polinômio raiz quadrada sqrt irracional função composta regra do produto regra da cadeia cálculo diferencial integral