Question

Considere a função f(x,y)=2x²-cos(xy). Determine resolvendo passo a passo o valor da derivada ∂f∂x no ponto (2,1)

Answer 1

Resposta:

$\sf f(x,y)=2x^2-cos(xy)$

Derivando em relação a x, considere x a função e y a constante:

$\sf f_x(x,y)=\frac{\partial }{\partial x}[2x^2-cos(xy)]$

$\sf f_x(x,y)=\frac{\partial }{\partial x}2x^2-\frac{\partial }{\partial x}cos(xy)$

$\sf f_x(x,y)=4x-\frac{\partial }{\partial x}cos(xy)$

Aplique a regra da cadeia fazendo u = xy:

$\sf f_x(x,y)=4x-\big[\frac{\partial }{\partial u}cos(u)\cdot\frac{\partial }{\partial x}u\big]$

$\sf f_x(x,y)=4x-\big[\!\!-\,sen(u)\cdot\frac{\partial }{\partial x}u\big]$

$\sf f_x(x,y)=4x-\big[\!\!-\,sen(xy)\cdot\frac{\partial }{\partial x}xy\big]$

$\sf f_x(x,y)=4x-\big[\!\!-\,sen(xy)\cdot y\cdot\frac{\partial }{\partial x}x\big]$

$\sf f_x(x,y)=4x-\big[\!\!-\,sen(xy)\cdot y\cdot1\big]$

$\sf f_x(x,y)=4x+sen(xy)\cdot y$

Assim, a derivada de f em relação a x no ponto (2, 1) é:

$\sf f_x(2,1)=4\cdot2+sen(2\cdot1)\cdot 1$

$\red{\boxed{\sf f_x(2,1)=8+sen(2)}}$