Question

Найти d²z: z=tg(3x·y)

Answer 1

$\displaystyle \mathrm{d^2z}={\partial^2z\over\partial{x^2}}\m\mathrm{dx^2}+2{\partial^2z\over\partial{x}\partial{y}}\mathrm{dx}\mathrm{dy}+{\partial^2z\over\partial{y^2}}\mathrm{dy^2}={\partial^2z\over\partial{x^2}}\m\mathrm{dx^2}+2{\partial^2z\over\partial{y}\partial{x}}\mathrm{dy}\mathrm{dx}+{\partial^2z\over\partial{y^2}}\mathrm{dy^2}$

Answer 2

To find the second derivative of z with respect to x, denoted as d²z/dx², we'll need to take the derivative of z with respect to x twice. Let's start with the first derivative:

Given: z = tan(3xy)

Step 1: Find dz/dx (the first derivative of z with respect to x) To find dz/dx, we use the chain rule. The chain rule states that for a function u = g(v) and v = h(x), the derivative of u with respect to x is du/dx = g'(v) * h'(x).

Let u = z, g(v) = tan(v), and v = 3xy. Now, we find the derivative of g(v) = tan(v), which is g'(v) = sec^2(v).

The derivative of v with respect to x, h'(x), is 3y.

So, dz/dx = sec^2(3xy) * 3y.

Step 2: Find d²z/dx² (the second derivative of z with respect to x) Now, we need to find the second derivative by differentiating dz/dx with respect to x.

Using the chain rule again, the derivative of dz/dx with respect to x is: d²z/dx² = d/dx (dz/dx) = d/dx (sec^2(3xy) * 3y).

Now, we find the derivative of sec^2(3xy) with respect to x. The derivative of sec^2(u) is 2sec(u) * tan(u).

So, d²z/dx² = d/dx (sec^2(3xy) * 3y) = 2 * sec(3xy) * tan(3xy) * d/dx(3y).

The derivative of 3y with respect to x is zero since y is a constant with respect to x.

Therefore, the second derivative d²z/dx² is simply:

d²z/dx² = 2 * sec(3xy) * tan(3xy) * 0 = 0.

Thus, the second derivative d²z/dx² is equal to 0.