Question

Ребят,помогите с математикой,пожалуйста!Вообще ноль в этом(∫(arcsinx)/(sqrt(1-x^2))=иy=e^x/sin^2*3x, найти производную yЗаранее огромное спасибо!

Answer 1

Решение интеграла методом подмены переменной:
$(ArcSinx)'= \frac{1}{\sqrt{1-x^2}}$
Пусть $u=ArcSinx$ тогда $=>du = \frac{dx}{\sqrt{1-x^2}}$, следовательно: $\frac{ArcSinx}{\sqrt{1-x^2}}dx=ArcSinx \frac{dx}{\sqrt{1-x^2}}=u*du$
Получили простой интеграл для $u$: $\int {u} \, du = \frac{u^2}{2} +Const$
Подставляем $u=ArcSinx$ и получаем: $\int \frac{ArcSinx}{\sqrt{1-x^2}}dx=\int udu =\frac{u^2}{2}+Const=\frac{(ArcSinx)^2}{2}+Const$

Второй вопрос: просто правила нахождения производной при делении и знание производных "простых" функций:
$( \frac{e^x}{sin^23x})'=\frac{e^xSin^23x-2*Sin3x*Cos3x*3}{Sin^43x}=\frac{e^xSin3x(Sin3x-6Cos3x)}{Sin^43x}$

Answer 2

Solving the Derivative of a Function

To solve the derivative of the given function, we'll start by finding the integral of the function ∫(arcsinx)/(sqrt(1-x^2)) and then proceed to find its derivative.

The integral of the function is given by:

∫(arcsinx)/(sqrt(1-x^2))

Let's find the integral first and then proceed to find its derivative.

Finding the Integral

The integral of the function ∫(arcsinx)/(sqrt(1-x^2)) can be found using the substitution method. We can let u = arcsinx, then du = 1/(sqrt(1-x^2)) dx. This will simplify the integral.

After the substitution, the integral becomes:

∫(u)/(sqrt(1-x^2)) du

Now, we can solve this integral to find the antiderivative.

Solving the Integral

The antiderivative of the function ∫(u)/(sqrt(1-x^2)) du can be found using the substitution method. After solving the integral, we get:

∫(u)/(sqrt(1-x^2)) du = sqrt(1-x^2) + C

Where C is the constant of integration.

Finding the Derivative

Now that we have found the antiderivative of the given function, we can proceed to find its derivative.

The derivative of the function y = e^x/sin^2*3x can be found using the chain rule and the quotient rule.

Let's proceed to find the derivative of the function.

Derivative of the Function

The derivative of the function y = e^x/sin^2*3x can be found using the chain rule and the quotient rule. After finding the derivative, we get:

y' = (e^x * sin^2*3x - e^x * 2 * sin3x * cos3x) / (sin^4*3x)

Where y' represents the derivative of y with respect to x.

So, the derivative of the given function y = e^x/sin^2*3x is:

y' = (e^x * sin^2*3x - e^x * 2 * sin3x * cos3x) / (sin^4*3x)

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