Вопрос задан 23.02.2019 в 06:27. Предмет Алгебра. Спрашивает Хисамова Алина.

Помогите))) найти производные первого порядка длинных функций, используя правила вычисления

производных:
0 0
Перейти к ответам

Ответы на вопрос

Внимание! Ответы на вопросы дают живые люди. Они могут содержать ошибочную информацию, заблуждения, а также ответы могут быть сгенерированы нейросетями. Будьте внимательны. Если вы уверены, что ответ неверный, нажмите кнопку "Пожаловаться" под ответом.
Отвечает Вавилов Захар.
1)
y =  \frac{8tg(3x)}{1+e^ \frac{x}{4} } \\  \\ 
y' = 8*\frac{\frac{3(1+e^ \frac{x}{4})}{cos^2(3x)}- \frac{tg(3x)*e^ \frac{x}{4}}{4} }{(1+e^ \frac{x}{4})^2}=8*\frac{ \frac{12(1+e^ \frac{x}{4})-sin(3x)cos(3x)*e^ \frac{x}{4}}{4cos^2(3x)}}{(1+e^ \frac{x}{4})^2}=\frac{24(1+e^ \frac{x}{4})-2sin(3x)cos(3x)*e^ \frac{x}{4}}{cos^2(3x)(1+e^ \frac{x}{4})^2}=\frac{24(1+e^ \frac{x}{4})-sin(6x)*e^ \frac{x}{4}}{cos^2(3x)(1+e^ \frac{x}{4})^2}

2)
 \left \{ {{x=ln(5-2t)} \atop {y=arctg(5-2t)}} \right. \\ x'_t=- \frac{2}{5-2t}  \\y'_t=- \frac{2}{1+(5-2t)^2}  \\ y'=\frac{5-2t}{1+(5-2t)^2}

3)
y=x^{sin(x^3)} \\ y' =x^{sin(x^3)}*(sin(x^3)*ln|x|)'=x^{sin(x^3)}*(cos(x^3)*3x^2*ln|x|+sin(x^3)* \frac{1}{x})=x^{sin(x^3)-1}*(cos(x^3)*3x^3*ln|x|+sin(x^3))

0 0
Отвечает Гафуров Юсуф.
Решение
1)  y = 8tg3x/(1 + e^(x/4))
y` = [(8*3/cos²3x)*(1 + e^x) - (1/4)*8*tg3x]/ (1 + e^x)² = 
= [24 + 24*(e^x)  - 2*cos²3x*tg3x] / [cos²3x*(1 + e^x)] =
[24 + 24*(e^x)  - 2*cos²3x*(sin3x/cos3x)] / [cos²3x*(1 + e^x)] =
 [24 + 24*(e^x)  - 2*sin3x*cos3x)] / [cos²3x*(1 + e^x)] =
 [24 + 24*(e^x)  - sin6x] / [cos²3x*(1 + e^x)] 
2)  y = x^(sinx
³)
y` = sinx³ * x^(sinx³ - 1) * cosx³ * 3x² = 
= sinx³ * cosx³ * 3x² *x^(sinx³ - 1) =
= 2*sinx³ * cosx³ * 1,5x² *x^(sinx³ - 1) = 
= sin(2x³) * 1,5x² *x^(sinx³ - 1)
0 0
Отвечает нейросеть ChatGpt. Будьте внимательны, ответы нейросети могут содержать ошибочные или выдуманные данные.

Finding First-Order Derivatives of Long Functions

To find the first-order derivatives of long functions using the rules of differentiation, we can apply various differentiation rules such as the power rule, product rule, quotient rule, chain rule, and sum/difference rule. These rules allow us to find the derivative of a function by differentiating each term separately and then combining the results.

Here are the steps to find the first-order derivatives of long functions:

1. Identify the function you want to differentiate. Let's call it f(x).

2. Break down the function into simpler terms and apply the appropriate differentiation rules to each term.

3. Apply the power rule to differentiate terms with variables raised to a power. The power rule states that if f(x) = x^n, then f'(x) = nx^(n-1).

4. Use the product rule to differentiate terms that are products of two or more functions. The product rule states that if f(x) = u(x)v(x), then f'(x) = u'(x)v(x) + u(x)v'(x).

5. Apply the quotient rule to differentiate terms that are ratios of two functions. The quotient rule states that if f(x) = u(x)/v(x), then f'(x) = (u'(x)v(x) - u(x)v'(x))/[v(x)]^2.

6. Use the chain rule to differentiate composite functions. The chain rule states that if f(x) = g(h(x)), then f'(x) = g'(h(x))h'(x).

7. Apply the sum/difference rule to differentiate terms that are sums or differences of two or more functions. The sum/difference rule states that if f(x) = u(x) ± v(x), then f'(x) = u'(x) ± v'(x).

By applying these rules, you can find the first-order derivatives of long functions.

Please provide the specific long function you would like to find the first-order derivative for, and I will be happy to assist you further.

0 0

Топ вопросов за вчера в категории Алгебра

Последние заданные вопросы в категории Алгебра

Задать вопрос