Question

Найти производную y=sin(lnx)

Answer 1

$y'=(sin(lnx))'=cos(lnx)*(lnx)'=cos(lnx)* \frac{1}{x} = \frac{cos(lnx)}{x}$

Answer 2

Finding the Derivative of y = sin(ln(x))

To find the derivative of the function y = sin(ln(x)), we can use the chain rule. The chain rule states that if we have a composite function, f(g(x)), then the derivative of f(g(x)) with respect to x is given by f'(g(x)) * g'(x).

In this case, we have y = sin(ln(x)), where g(x) = ln(x) and f(u) = sin(u). Applying the chain rule, we can find the derivative of y with respect to x.

The derivative of sin(u) with respect to u is cos(u), and the derivative of ln(x) with respect to x is 1/x. Therefore, the derivative of y = sin(ln(x)) is:

y' = cos(ln(x)) * (1/x)

Let's break it down step by step:

1. The derivative of sin(u) with respect to u is cos(u) [[1]](https://uchi.ru/otvety/questions/nayti-proizvodnuyu-podrobno-y-sin-lnx-cos-lnx). 2. The derivative of ln(x) with respect to x is 1/x.

Therefore, the derivative of y = sin(ln(x)) is y' = cos(ln(x)) * (1/x).

I hope this helps! Let me know if you have any further questions.