Question

Z=ln(x^2+y^2), x=uv, y=3u−v. Найти ∂z/∂u .

Answer 1

$z=ln(x^2+y^2)\; ,\; \; x=uv\; ,\; \; y=3u-v\\\\\frac{\partial z}{\partial u}=\frac{\partial z}{\partial x}\cdot \frac{\partial x}{\partial u}+\frac{\partial z}{\partial y}\cdot \frac{\partial y}{\partial u}=\frac{2x}{x^2+y^2}\cdot v+\frac{2y}{x^2+y^2}\cdot 3=\\\\=\frac{2xv}{x^2+y^2}+\frac{6y}{x^2+y^2}$

Answer 2

To find ∂z/∂u, we'll first express z in terms of u and v and then differentiate it with respect to u.

Given: Z = ln(x^2 + y^2) (Equation 1) x = uv y = 3u - v

We'll substitute the expressions for x and y into Equation 1:

Z = ln((uv)^2 + (3u - v)^2)

Now, let's simplify Z:

Z = ln(u^2v^2 + (3u - v)^2) Z = ln(u^2v^2 + 9u^2 - 6uv + v^2)

Now, we'll find the partial derivative of Z with respect to u (∂z/∂u):

∂Z/∂u = ∂/∂u ln(u^2v^2 + 9u^2 - 6uv + v^2)

To differentiate the natural logarithm, we'll use the chain rule:

d/dx ln(f(x)) = (1 / f(x)) * f'(x)

where f(x) = u^2v^2 + 9u^2 - 6uv + v^2.

Now, let's find the derivative of f(x) with respect to u:

df/du = d/dx (u^2v^2 + 9u^2 - 6uv + v^2) = 2uv^2 + 18u - 6v

Now, apply the chain rule to find ∂Z/∂u:

∂Z/∂u = (1 / (u^2v^2 + 9u^2 - 6uv + v^2)) * (2uv^2 + 18u - 6v)

So the final expression for ∂Z/∂u is:

∂Z/∂u = (2uv^2 + 18u - 6v) / (u^2v^2 + 9u^2 - 6uv + v^2)