Question

encontre a derivação da função
a) f (x) = sen ( -3x)
b) g (t) = $t^{3} cos t$
c) y = $\frac{x}{2-tgx}$
d) y = $\frac{t sen t}{1+t}$

Answer 1

a) $\mathtt{f(x)=sen(-3 x)}$

Derive usando a Regra da Cadeia:

$\mathtt{\dfrac{df}{dx}=\dfrac{d}{dx}(sen(-3 x))}\\\\\\ \mathtt{\dfrac{df}{dx}=cos(-3x)\cdot \dfrac{d}{dx}(-3x)}\\\\\\ \mathtt{\dfrac{df}{dx}=cos(-3x)\cdot (-3)}\\\\\\ \mathtt{\dfrac{df}{dx}=-3\,cos(-3x)}$


Como a função cosseno é uma função par,

$\boxed{\begin{array}{c}\mathtt{\dfrac{df}{dx}=-3\,cos(3x)} \end{array}}$

_________

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

Deriva usando a Regra do Produto:

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

_________

c) $\mathtt{y=\dfrac{x}{2-tg\,x}}$

Deriva usando a Regra do Quociente:

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


_________

d) $\mathtt{y=\dfrac{t\,sen\,t}{1+t}}$

Deriva usando a Regra do Quociente, combinada com a Regra do Produto:

$\mathtt{\dfrac{dy}{dt}=\dfrac{d}{dt}\!\left(\dfrac{t\,sen\,t}{1+t}\right)}\\\\\\ \mathtt{\dfrac{dy}{dt}=\dfrac{\frac{d}{dt}(t\,sen\,t)\cdot (1+t)-t\,sen\,t\cdot \frac{d}{dt}(1+t)}{(1+t)^2}}\\\\\\ \mathtt{\dfrac{dy}{dt}=\dfrac{\left(\frac{d}{dt}(t)\cdot sen\,t+t\cdot \frac{d}{dt}(sen\,t) \right )\cdot (1+t)-t\,sen\,t\cdot (0+1)}{(1+t)^2}}\\\\\\ \mathtt{\dfrac{dy}{dt}=\dfrac{(1\cdot sen\,t+t\cdot cos\,t)\cdot (1+t)-t\,sen\,t\cdot 1}{(1+t)^2}}\\\\\\ \mathtt{\dfrac{dy}{dt}=\dfrac{(sen\,t+t\,cos\,t)\cdot (1+t)-t\,sen\,t}{(1+t)^2}}$


Fazendo a distributiva,

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


Dúvidas? Comente.


Bons estudos! :-)