Question

Encontre o gradiente:
grad f(1, −1) = (∂f/∂x(1, −1), ∂f/∂y (1, −1)),

onde

f(x, y) = ln(x + sen(x − y^2)).

Answer 1

Resposta:

Explicação passo-a-passo:

$f(x,y)=ln(x+sen(x-y^{2})\\\\\\\dfrac{\partial f}{\partial x}=\dfrac{1}{x+sen(x-y^{2})}.(1+cos(x-y^{2}).1)=\dfrac{1+cos(x-y^{2})}{x+sen(x-y^{2})}\\\\\\\dfrac{\partial f}{\partial x}(1,-1)=\dfrac{1+cos(1-(-1)^{2})}{1+sen(1-(-1)^{2})}=\dfrac{1+cos(1-1)}{1+sen(1-1)}=\dfrac{1+cos(0)}{1+sen(0)}=\dfrac{1+1}{1+0}\\\\\\\boxed{\dfrac{\partial f}{\partial x}(1,-1)=2}$

$\dfrac{\partial f}{\partial y}=\dfrac{1}{x+sen(x-y^{2})}.(cos(x-y^{2}).(-2y))=\dfrac{-2.y.cos(x-y^{2})}{x+sen(x-y^{2})}$

$\dfrac{\partial f}{\partial y}(1,-1)=\dfrac{-2.(-1).cos(1-(-1)^{2})}{1+sen(1-(-1)^{2})}=\dfrac{2.cos(1-1)}{1+sen(1-1)}=\dfrac{2.cos(0)}{1+sen(0)}\\\\\\\dfrac{\partial f}{\partial y}(1,-1)=\dfrac{2.1}{1+0}\\\\\\\boxed{\dfrac{\partial f}{\partial y}(1,-1)=2}$

$\nabla f(1,-1)=(2,2)$