Question

Qual é o valor aproximado da derivada parcial fx no ponto (3,2,1) da função f(x,y,z)= e^(xy^2 ) ln⁡(x y z)?a)1290b)1290000c)0d)3e)1

Answer 1

$f(x,\,y,\,z)=e^{xy^2}\,\mathrm{\ell n}(xyz)$

Calcular a derivada parcial de $f$ em relação a $x$ no ponto $(3,\,2,\,1).$

_______________

A derivada parcial de $f$ em relação a $x:$

(Considera $y$ e $z$ como constantes e deriva em relação a $x$ usando as regras já conhecidas)

$f_x(x,\,y,\,z)=\dfrac{\partial}{\partial x}\!\left[e^{xy^2}\,\mathrm{\ell n}(xyz) \right ]\\\\\\ f_x(x,\,y,\,z)=\dfrac{\partial}{\partial x}\!\left(e^{xy^2}\right)\cdot \mathrm{\ell n}(xyz)+e^{xy^2}\cdot \dfrac{\partial}{\partial x}\left[\mathrm{\ell n}(xyz) \right ]\\\\\\ f_x(x,\,y,\,z)=\left(e^{xy^2}\cdot \dfrac{\partial}{\partial x}(xy^2)\right)\cdot \mathrm{\ell n}(xyz)+e^{xy^2}\cdot \left[\dfrac{1}{xyz}\cdot \dfrac{\partial}{\partial x}(xyz) \right ]$

$f_x(x,\,y,\,z)=\left(e^{xy^2}\cdot y^2\right)\cdot \mathrm{\ell n}(xyz)+e^{xy^2}\cdot \left[\dfrac{1}{xyz}\cdot yz \right ]\\\\\\ f_x(x,\,y,\,z)=e^{xy^2}\cdot y^2\cdot \mathrm{\ell n}(xyz)+e^{xy^2}\cdot \dfrac{1}{x}\\\\\\ \therefore~~\boxed{\begin{array}{c} f_x(x,\,y,\,z)=e^{xy^2}\cdot \left[y^2\, \mathrm{\ell n}(xyz)+ \dfrac{1}{x} \right ]\end{array}}$

_______________________

Avaliando a derivada no ponto $(3,\,2,\,1),$ temos

$f_x(x,\,y,\,z)=e^{xy^2}\cdot \left[y^2\, \mathrm{\ell n}(xyz)+ \dfrac{1}{x}\right]\\\\\\ f_x(3,\,2,\,1)=e^{3\,\cdot\, 2^2}\cdot \left[2^2\, \mathrm{\ell n}(3\cdot 2\cdot 1)+ \dfrac{1}{3}\right]\\\\\\ f_x(3,\,2,\,1)=e^{12}\cdot \left[4\, \mathrm{\ell n}(6)+ \dfrac{1}{3}\right]\approx 1\,220\,721$


A resposta que mais se aproxima é a alternativa $b).$