Question

Нужно найти дифференциал. Мат. анализ, 1 курс института! Срочно! y=x^2*arctg*\sqrt{x^2-1}-\sqrt{x^2-1}Заранее спасибо!

Answer 1

Ответ:

$y = {x}^{2} \times arctg( \sqrt{ {x}^{2} - 1} ) - \sqrt{ {x}^{2} - 1 }$

$y' = 2xarctg( \sqrt{ {x}^{2} - 1} ) + {x}^{2} \times \frac{1}{1 + {x}^{2} - 1 } \times \frac{1}{2 \sqrt{ {x}^{2} - 1 } } \times 2x - \frac{2x}{2 \sqrt{ {x}^{2} - 1} } = \\ = 2xarctg( \sqrt{ {x}^{2} - 1} ) + \frac{ {x}^{2} \times x }{ {x}^{2} \times \sqrt{ {x}^{2} - 1} } - \frac{x}{ \sqrt{ {x}^{2} - 1} } = \\ = 2xarctg \sqrt{ {x}^{2} - 1 } + \frac{x}{ \sqrt{ {x}^{2} - 1} } - \frac{x}{ \sqrt{ {x}^{2} - 1} } = \\ = 2xarctg( \sqrt{ {x}^{2} - 1} )$

Answer 2

Finding the Differential

To find the differential of the given function, we'll start by differentiating each term separately and then combining the differentials.

The given function is: y = x^2 * arctg(sqrt(x^2-1)) - sqrt(x^2-1)

Let's differentiate each term step by step.

1. Differentiating x^2 - The differential of x^2 is given by 2x*dx.

2. Differentiating arctg(sqrt(x^2-1)) - The differential of arctg(u) is given by du / (1 + u^2). - Here, u = sqrt(x^2-1), so the differential is (sqrt(x^2-1)) / (1 + (x^2-1)).

3. Differentiating sqrt(x^2-1) - The differential of sqrt(u) is given by du / (2*sqrt(u)). - Here, u = x^2-1, so the differential is (x*dx) / (2*sqrt(x^2-1)).

Now, let's combine these differentials to find the overall differential of the given function.

Overall Differential

The overall differential dy of the given function y is the sum of the differentials of each term: dy = (2x*dx) * arctg(sqrt(x^2-1)) + x^2 * ((sqrt(x^2-1)) / (1 + (x^2-1))) - ((x*dx) / (2*sqrt(x^2-1)))

This is the differential of the given function y = x^2 * arctg(sqrt(x^2-1)) - sqrt(x^2-1).

I hope this helps! If you have further questions or need additional assistance, feel free to ask.