Question

Найти y’ и y’’ y2 = x + ln(y/x)

Answer 1

$y^2=x+ln\frac{y}{x}\\\\2yy'=1+\frac{x}{y}\cdot \frac{1}{x}\cdot y'\\\\y'(2y-\frac{1}{y})=1\\\\y'\cdot \frac{2y^2-1}{y} =1\\\\y'=\frac{y}{2y^2-1}$

Answer 2

To find the first and second derivatives of y with respect to x, we'll start by differentiating the given equation y^2 = x + ln(y/x) with respect to x. Let's proceed step by step.

Given equation: y^2 = x + ln(y/x)

Step 1: Differentiate both sides of the equation with respect to x using the chain rule.

d/dx(y^2) = d/dx(x) + d/dx(ln(y/x))

Step 2: Apply the chain rule and the derivative of the natural logarithm.

2y * dy/dx = 1 + (1/x) * (y * dy/dx - y'/x)

Step 3: Now, solve for dy/dx (the first derivative).

2y * dy/dx = 1 + (1/x) * (y * dy/dx - y'/x) 2y * dy/dx = 1 + (1/x) * (y * dy/dx) - y'/x dy/dx * (2y - (1/x) * y) = 1 - (1/x) * y' dy/dx = (1 - (1/x) * y') / (2y - (1/x) * y)

Step 4: Now, to find the second derivative, differentiate the expression obtained for dy/dx once again with respect to x:

y'' = d/dx(dy/dx)

y'' = d/dx[(1 - (1/x) * y') / (2y - (1/x) * y)]

Step 5: Use the quotient rule and the chain rule to find the second derivative.

y'' = [(2y - (1/x) * y) * d/dx(1 - (1/x) * y') - (1 - (1/x) * y') * d/dx(2y - (1/x) * y)] / (2y - (1/x) * y)^2

Step 6: Simplify the expression for y''.

y'' = [(2y - (1/x) * y) * (y'' * x - y' + (1/x^2) * y') - (1 - (1/x) * y') * (2dy/dx - (1/x) * dy/dx)] / (2y - (1/x) * y)^2

Now we have the expressions for both the first derivative (dy/dx) and the second derivative (y''). The calculations can become quite complex, and it's important to double-check the algebra to ensure accuracy. If you have specific values for x and y or further instructions, I can help you perform the actual calculations.