Question

Derive a função:
a)f(t)=1/2t^6-3t^4+T

b)f(t)=raiz de t - 1/raiz de t

Answer 1

Regras de derivação utilizadas:

Sendo $\mathtt{u}$ e $\mathtt{v}$ funções deriváveis de $\mathtt{t},$ e $\mathtt{a}$ e $\mathtt{b}$ constantes reais, vale que

•   $\mathtt{\dfrac{d}{dt}(au+bv)=a\dfrac{du}{dt}+b\dfrac{dv}{dt}}$

(linearidade)

____________

Regra da derivada da função potência:

•   $\mathtt{\dfrac{d}{dt}(t^n)=nt^{n-1}}$

____________

a)  $\mathtt{f(t) =\dfrac{1}{2}t^6-3t^4+t}$

Derivando,

$\mathtt{\dfrac{df}{dt}=\dfrac{d}{dt}\!\left(\dfrac{1}{2}t^6-3t^4+t \right )}\\\\\\ \mathtt{\dfrac{df}{dt}=\dfrac{d}{dt}\!\left(\dfrac{1}{2}t^6\right)-\dfrac{d}{dt}(3t^4)+\dfrac{d}{dt}(t)}\\\\\\ \mathtt{\dfrac{df}{dt}=\dfrac{1}{2}\dfrac{d}{dt}(t^6)-3\dfrac{d}{dt}(t^4)+\dfrac{d}{dt}(t^1)}\\\\\\ \mathtt{\dfrac{df}{dt}=\dfrac{1}{2}\cdot 6t^{6-1}-3\cdot 4t^{4-1}+1t^{1-1}}\\\\\\ \mathtt{\dfrac{df}{dt}=\dfrac{1}{2}\cdot 6t^5-3\cdot 4t^3+1t^0}\\\\\\ \boxed{\begin{array}{c}\mathtt{\dfrac{df}{dt}=3t^5-12t^3+1} \end{array}}$

_________

b) $\mathtt{f(t)=\sqrt{t}-\dfrac{1}{\sqrt{t}}}$

$\mathtt{f(t)=t^{\frac{1}{2}}-\dfrac{1}{t^{\frac{1}{2}}}}\\\\\\ \mathtt{f(t)=t^{\frac{1}{2}}-t^{-\frac{1}{2}}}$


Derivando,

$\mathtt{\dfrac{df}{dt}=\dfrac{d}{dt}(t^{\frac{1}{2}}-t^{-\frac{1}{2}})}\\\\\\ \mathtt{\dfrac{df}{dt}=\dfrac{d}{dt}(t^{\frac{1}{2}})-\dfrac{d}{dt}(t^{-\frac{1}{2}})}\\\\\\ \mathtt{\dfrac{df}{dt}=\dfrac{1}{2}t^{\frac{1}{2}-1}-\left(-\dfrac{1}{2}\right)\!t^{-\frac{1}{2}-1}}\\\\\\ \mathtt{\dfrac{df}{dt}=\dfrac{1}{2}t^{-\frac{1}{2}}+\dfrac{1}{2}t^{-\frac{3}{2}}}\\\\\\ \mathtt{\dfrac{df}{dt}=\dfrac{1}{2}\cdot\dfrac{1}{t^{\frac{1}{2}}}+\dfrac{1}{2}\cdot \dfrac{1}{t^{\frac{3}{2}}}}$

$\mathtt{\dfrac{df}{dt}=\dfrac{1}{2}\cdot\dfrac{1}{\sqrt{t}}+\dfrac{1}{2}\cdot \dfrac{1}{\sqrt{t^3}}}\\\\\\ \boxed{\begin{array}{c}\mathtt{\dfrac{df}{dt}=\dfrac{1}{2\sqrt{t}}+\dfrac{1}{2\sqrt{t^3}}} \end{array}}$


Dúvidas? Comente.


Bons estudos! :-)