Question

U=lntg(y/x) найти d2u/dx2

Answer 1

$\frac{du}{dx} =-\frac{y}{x^2cos^2(\frac{y}{x})tg(\frac{y}{x})}=-\frac{2y}{x^2sin(\frac{2y}{x})} \\\\ \frac{d^2y}{dx^2}=-2y(\frac{x^{-2}}{sin(\frac{2y}{x})})'=-2y\frac{-2x^{-3}sin(\frac{2y}{x})+cos(\frac{2y}{x})}{sin^2(\frac{2y}{x})}$

Answer 2

To find the second derivative of the function U with respect to x, we'll first need to find the first derivative of U with respect to x, and then take its derivative again.

Given: U = ln(tg(y/x))

Step 1: Find the first derivative of U with respect to x (du/dx). Step 2: Find the second derivative of U with respect to x (d^2u/dx^2).

Let's proceed with Step 1:

Step 1: Find du/dx

To find du/dx, we'll use the chain rule. The chain rule states that if we have a function U = f(g(x)), then the derivative of U with respect to x is given by dU/dx = f'(g(x)) * g'(x).

In our case, f(u) = ln(u) and g(x) = tg(y/x). Let's find f'(u) and g'(x):

f'(u) = d/du [ln(u)] = 1/u g'(x) = d/dx [tg(y/x)]

Now, apply the chain rule:

du/dx = f'(g(x)) * g'(x) du/dx = (1/tg(y/x)) * d/dx [tg(y/x)]

To find d/dx [tg(y/x)], we'll use trigonometric identities. The derivative of tangent is:

d/dx [tg(u)] = sec^2(u) * du/dx

In our case, u = y/x:

d/dx [tg(y/x)] = sec^2(y/x) * d/dx [y/x]

d/dx [y/x] can be found using the quotient rule:

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

However, since y is a constant with respect to x (y does not depend on x), its derivative is zero:

d/dx [y] = 0

Therefore, d/dx [y/x] = 0, and the equation simplifies to:

d/dx [tg(y/x)] = 0

Now, we can go back to the expression for du/dx:

du/dx = (1/tg(y/x)) * 0 = 0

Step 1 is complete. We found that du/dx = 0.

Step 2: Find d^2u/dx^2

Now, to find the second derivative d^2u/dx^2, we need to take the derivative of du/dx (which we found to be 0) with respect to x:

d^2u/dx^2 = d/dx [du/dx]

Since du/dx is constant (0), the derivative of a constant with respect to x is 0:

d/dx [du/dx] = d/dx [0] = 0

Thus, the second derivative d^2u/dx^2 is 0.

In summary:

U = ln(tg(y/x)) du/dx = 0 d^2u/dx^2 = 0