Question

derivada de:

a)g(t)= t^3 cos t?

b)1+sen(x)/x+cos(x)

Answer 1

Apresento as derivadas de funções trigonométricas simples:

$\mathtt{\dfrac{d}{du}(sen\,u)=cos\,u}\\\\\\ \mathtt{\dfrac{d}{du}(cos\,u)=-sen\,u}$

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a) $\mathtt{g(t)=t^3\,cos\,t}$

Deriva, usando a Regra do Produto:

$\mathtt{\dfrac{dg}{dt}=\dfrac{d}{dt}(t^3\,cos\,t)}\\\\\\ \mathtt{\dfrac{dg}{dt}=\dfrac{d}{dt}(t^3)\cdot cos\,t+t^3\cdot \dfrac{d}{dt}(cos\,t)}\\\\\\ \mathtt{\dfrac{dg}{dt}=3t^{3-1}\cdot cos\,t+t^3\cdot (-sen\,t)}\\\\\\ \boxed{\begin{array}{c}\mathtt{\dfrac{dg}{dt}=3t^2\,cos\,t-t^3\,sen\,t} \end{array}}$

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b) $\mathtt{f(x)=\dfrac{1+sen\,x}{x+cos\, x}}$

Deriva, usando a Regra do Quociente:

$\mathtt{\dfrac{df}{dx}=\dfrac{d}{dx}\!\left(\dfrac{1+sen\,x}{x+cos\, x} \right )}\\\\\\ \mathtt{\dfrac{df}{dx}=\dfrac{\frac{d}{dx}(1+sen\,x)\cdot (x+cos\,x)-(1+sen\,x)\cdot \frac{d}{dx}(x+cos\,x)}{(x+cos\,x)^2}}\\\\\\ \mathtt{\dfrac{df}{dx}=\dfrac{(0+cos\,x)\cdot (x+cos\,x)-(1+sen\,x)\cdot (1-sen\,x)}{(x+cos\,x)^2}}\\\\\\ \mathtt{\dfrac{df}{dx}=\dfrac{x\,cos\,x+cos^2\,x-(1^2-sen^2\,x)}{(x+cos\,x)^2}}\\\\\\ \mathtt{\dfrac{df}{dx}=\dfrac{x\,cos\,x+cos^2\,x-(1-sen^2\,x)}{(x+cos\,x)^2}}\\\\\\ \mathtt{\dfrac{df}{dx}=\dfrac{x\,cos\,x+cos^2\,x-cos^2\,x}{(x+cos\,x)^2}}\\\\\\ \boxed{\begin{array}{c}\mathtt{\dfrac{df}{dx}=\dfrac{x\,cos\,x}{(x+cos\,x)^2}} \end{array}}$


Dúvidas? Comente.


Bons estudos! :-)