Question

Проиводные ylnx-xlny=0

Answer 1

$y \ln x = x \ln y\\ (y \ln x)' = (x \ln y)' \\ y' \ln x + \frac{y}{x} = \ln y + \frac{x}{y} \cdot y' \\ y' \ln x- \frac{x}{y} \cdot y' = \ln y -\frac{y}{x}\\ ( \ln x- \frac{x}{y}) \cdot y' = \ln y -\frac{y}{x}\\ y' = \dfrac {\ln y -\frac{y}{x}}{\ln x- \frac{x}{y}} = \dfrac{y(x \ln y - y)}{x(y \ln x - x)}$

Ответ: $y' = \dfrac{y(x \ln y - y)}{x(y \ln x - x)}$

Answer 2

To find the derivatives of the given expression yln(x) - xln(y) = 0 with respect to x and y, we can use implicit differentiation. Let's differentiate both sides of the equation with respect to x and y separately.

Given: yln(x) - xln(y) = 0

1. Differentiating with respect to x: To differentiate yln(x) with respect to x, we use the product rule: (d/dx) [yln(x)] = y * (d/dx)[ln(x)] + ln(x) * (d/dx)[y] = y * (1/x) + ln(x) * dy/dx

To differentiate -xln(y) with respect to x, we use the product rule again: (d/dx)[-xln(y)] = -x * (d/dx)[ln(y)] + ln(y) * (d/dx)[(-x)] = -x * (1/y) + ln(y) * (-1)

Now, the derivative of 0 with respect to x is simply 0.

So, the complete differentiation with respect to x is: y * (1/x) + ln(x) * dy/dx - x/y - ln(y) = 0

2. Differentiating with respect to y: To differentiate yln(x) with respect to y, we use the product rule: (d/dy)[yln(x)] = ln(x) * (d/dy)[y] + y * (d/dy)[ln(x)] = ln(x) + y * (1/x) * dx/dy

To differentiate -xln(y) with respect to y, we use the product rule again: (d/dy)[-xln(y)] = -ln(y) * (d/dy)[x] - x * (d/dy)[ln(y)] = -ln(y) * dx/dy - x * (1/y)

Now, the derivative of 0 with respect to y is also 0.

So, the complete differentiation with respect to y is: ln(x) + y * (1/x) * dx/dy - ln(y) * dx/dy - x/y = 0

Now, we have a system of two equations:

1. y * (1/x) + ln(x) * dy/dx - x/y - ln(y) = 0
2. ln(x) + y * (1/x) * dx/dy - ln(y) * dx/dy - x/y = 0

You can solve this system of equations to find the derivatives dy/dx and dx/dy in terms of x, y, ln(x), and ln(y). The exact solutions will depend on the values of x and y.