Questions: ДисЦипмина «МатематическиЙ анализи (часть 1, занет) Зачетное задание Ne8 1.Теоретический вопрос: Если y=f(u) и u=φ(x) - дифференцируемые функции от своих аргументов, то чему равна производная сложной функции y=f(φ(x)) ? а) y'x=yy'x; b) y'x=y'xu'x; c) y'x=y'u'u'x; d) y'x=f'u'u; e) y'x=f'u' / u'x;

Solution Steps

To find the derivative of the composite function \( y = f(\varphi(x)) \), we use the chain rule. The chain rule states that if you have a composite function, the derivative of the outer function with respect to the inner function is multiplied by the derivative of the inner function with respect to \( x \). Therefore, the correct answer is \( y_{x}^{\prime} = y_{u}^{\prime} u_{x}^{\prime} \).

We are given a composite function \( y = f(\varphi(x)) \) and need to find its derivative \( y_{x}^{\prime} \) using the chain rule. The chain rule states that the derivative of a composite function is the product of the derivative of the outer function and the derivative of the inner function.

According to the chain rule, we have: \[ y_{x}^{\prime} = y_{u}^{\prime} \cdot u_{x}^{\prime} \] where \( y_{u}^{\prime} \) is the derivative of \( f \) with respect to \( u \), and \( u_{x}^{\prime} \) is the derivative of \( \varphi \) with respect to \( x \).

In our example, we have:

• \( y_{u}^{\prime} = 2 \)
• \( u_{x}^{\prime} = 3 \)

Substituting these values into the equation gives: \[ y_{x}^{\prime} = 2 \cdot 3 = 6 \]

Final Answer