Question

Sendo z = ln $\sqrt{x^{2} + y^{2} }$ calcular o valor de x.$\frac{dz}{dx}$ + y.$\frac{dz}{dy}$. a resposta é 1

Answer 1

Resposta:

z= ln √(x²+y²)

z =(1/2) * ln (x²+y²)

dz/dx=(1/2) * (x²+y²)'/(x²+y²)

dz/dx=(1/2) * (2x)/(x²+y²)

dz/dx=x/(x²+y²)

dz/dy=(1/2) * (x²+y²)'/(x²+y²)

dz/dy=(1/2) * (2y)/(x²+y²)

dz/dy=y/(x²+y²)

x* dz/dx + y * dz/dy

=x²/(x²+y²) + y²/(x²+y²)

=(x²+y²)/(x²+y²) =  1

Answer 2

Antes, convém lembrar a seguinte regra:

$y=ln|u(x)|$

$\frac{dy}{dx} = \frac{\frac{du}{dx}}{u(x)}$

Que também vale para duas variáveis considerando a derivação parcial.

Então, temos que:

$z = ln\sqrt{x^2+y^2}$

$z = ln({x^2+y^2})^{\frac{1}{2}$

Assim, temos:

$\frac{dz}{dx} = \frac{\frac{1}{2}(x^2+y^2)^{\frac{1}{2}-1}(2x)}{(x^2+y^2)^{\frac{1}{2}}}$

$\frac{dz}{dx} = \frac{x(x^2+y^2)^{\frac{-1}{2}}}{(x^2+y^2)^{\frac{1}{2}}}$

$\frac{dz}{dx} = x(x^2+y^2)^{\frac{-1}{2}-\frac{1}{2}}$

$\frac{dz}{dx} = x(x^2+y^2)^{-1}$

$\frac{dz}{dx} = \frac{x}{x^2+y^2}$

Além disso:

$\frac{dz}{dy} = \frac{\frac{1}{2}(x^2+y^2)^{\frac{1}{2}-1}(2y)}{(x^2+y^2)^{\frac{1}{2}}}}$

$\frac{dz}{dy} = \frac{y(x^2+y^2)^{\frac{-1}{2}}}{(x^2+y^2)^\frac{1}{2}}}}$

$\frac{dz}{dy} = y(x^2+y^2)^{\frac{-1}{2}-\frac{1}{2}}$

$\frac{dz}{dy} = y(x^2+y^2)^{-1}$

$\frac{dz}{dy} = \frac{y}{x^2+y^2}$

Devemos calcular:

$=x\frac{dz}{dx}+y\frac{dz}{dy}$

$= x\frac{x}{x^2+y^2}+y\frac{y}{x^2+y^2}$

$= \frac{x^2}{x^2+y^2}+\frac{y^2}{x^2+y^2}$

$= \frac{x^2+y^2}{x^2+y^2}$

$=1$