Question

derivada de:
a)(x^3)/(1-x^2)

b)t-√t/t^1/3

Answer 1

Regra para a derivada de uma função quociente.

Dadas duas funções $\mathtt{u}$ e $\mathtt{v}\ne 0$ na variável $\mathtt{x}$ deriváveis, a função $\mathtt{w=\dfrac{u}{v}}$ também é derivável, e sua derivada é dada por

$\mathtt{\dfrac{dw}{dx}=\dfrac{d}{dx}\!\left(\dfrac{u}{v}\right)=\dfrac{\frac{du}{dx}\cdot v-u\cdot \frac{dv}{dx}}{v^2}}$

_________

a) $\mathtt{f(x)=\dfrac{x^3}{1-x^2}}$

Derivando, usando a regra do quociente:

$\mathtt{\dfrac{df}{dx}=\dfrac{d}{dx}\!\left(\dfrac{x^3}{1-x^2} \right )}\\\\\\ \mathtt{\dfrac{df}{dx}=\dfrac{\frac{d}{dx}(x^3)\cdot (1-x^2)-x^3\cdot \frac{d}{dx}(1-x^2)}{(1-x^2)^2}}\\\\\\ \mathtt{\dfrac{df}{dx}=\dfrac{3x^{3-1}\cdot (1-x^2)-x^3\cdot (0-2x^{2-1})}{(1-x^2)^2}}\\\\\\ \mathtt{\dfrac{df}{dx}=\dfrac{3x^2\cdot (1-x^2)-x^3\cdot (-2x)}{(1-x^2)^2}}\\\\\\ \mathtt{\dfrac{df}{dx}=\dfrac{3x^2-3x^4+2x^4}{(1-x^2)^2}}\\\\\\ \boxed{\begin{array}{c} \mathtt{\dfrac{df}{dx}=\dfrac{3x^2-x^4}{(1-x^2)^2}} \end{array}}$


b) $\mathtt{g(t)=\dfrac{t-\sqrt{t}}{t^{\frac{1}{3}}}}$

$\mathtt{g(t)=\dfrac{t-t^{\frac{1}{2}}}{t^{\frac{1}{3}}}}$


Derivando, usando a regra do quociente:

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

$\boxed{\begin{array}{c}\mathtt{\dfrac{dg}{dt}=\dfrac{\left(1-\frac{1}{2}t^{-\frac{1}{2}}\right)\cdot t^{\frac{1}{3}}-(t-t^{\frac{1}{2}})\cdot \frac{1}{3}t^{-\frac{2}{3}}}{t^{\frac{2}{3}}}} \end{array}}$


Poderia parar por aqui, já que não há mais nada a derivar. Contudo, caso queira continuar fazendo as simplificações,

$\mathtt{\dfrac{dg}{dt}=\dfrac{t^{\frac{1}{3}}-\frac{1}{2}t^{-\frac{1}{2}+\frac{1}{3}}-\left(\frac{1}{3}t^{1-\frac{2}{3}}-\frac{1}{3}t^{\frac{1}{2}-\frac{2}{3}}\right)}{t^{\frac{2}{3}}}}\\\\\\ \mathtt{\dfrac{dg}{dt}=\dfrac{t^{\frac{1}{3}}-\frac{1}{2}t^{-\frac{1}{6}}-\frac{1}{3}t^{\frac{1}{3}}+\frac{1}{3}t^{-\frac{1}{6}}}{t^{\frac{2}{3}}}}\\\\\\ \mathtt{\dfrac{dg}{dt}=\dfrac{t^{\frac{1}{3}}-\frac{1}{2}t^{-\frac{1}{6}}-\frac{1}{3}t^{\frac{1}{3}}+\frac{1}{3}t^{-\frac{1}{6}}}{t^{\frac{2}{3}}}}\\\\\\ \mathtt{\dfrac{dg}{dt}=\dfrac{\frac{2}{3}t^{\frac{1}{3}}-\frac{1}{6}t^{-\frac{1}{6}}}{t^{\frac{2}{3}}}}$


Como $\mathtt{t>0},$ podemos multiplicar e dividir por $\mathtt{6t^{\frac{1}{6}}}:$

$\mathtt{\dfrac{dg}{dt}=\dfrac{\frac{2}{3}t^{\frac{1}{3}}-\frac{1}{6}t^{-\frac{1}{6}}}{t^{\frac{2}{3}}}\cdot \dfrac{6t^{\frac{1}{6}}}{6t^{\frac{1}{6}}}}\\\\\\ \mathtt{\dfrac{dg}{dt}=\dfrac{4t^{\frac{1}{3}+\frac{1}{6}}-1}{6t^{\frac{2}{3}+\frac{1}{6}}}}\\\\\\ \mathtt{\dfrac{dg}{dt}=\dfrac{4t^{\frac{1}{2}}-1}{6t^{\frac{5}{6}}}}\\\\\\ \boxed{\begin{array}{c}\mathtt{\dfrac{dg}{dt}=\dfrac{4\sqrt{t}-1}{6\,^5\!\!\!\sqrt{t^6}}} \end{array}}$


Dúvidas? Comente.


Bons estudos! :-)