Question

(ln(cos(sin(x))))'= (cos(sin(ln(x))))'=

Answer 1

Ответ:

Пошаговое объяснение:d/dx((log(sin(x)))/(log(cos(x)))) = (tan(x) log(sin(x))+cot(x) log(cos(x)))/(log^2(cos(x))), надеюсь это то, просто плохо помню, заколебался писать.

Answer 2

To find the derivatives of the given expressions, we'll use the chain rule and the derivative rules for trigonometric functions. Let's calculate them one by one:

1. (ln(cos(sin(x))))':

Let u = cos(sin(x)), and v = ln(u). By the chain rule, we have:

v' = (du/dx) * (1/u).

Now, let's find du/dx:

du/dx = -sin(x) * cos(sin(x)).

Substituting back, we have:

v' = (-sin(x) * cos(sin(x))) * (1/u).

Since u = cos(sin(x)), we can substitute it in:

v' = (-sin(x) * cos(sin(x))) * (1/cos(sin(x))).

Simplifying, we get:

v' = -sin(x).

Therefore, the derivative of ln(cos(sin(x))) is -sin(x).

2. (cos(sin(ln(x))))':

Let u = sin(ln(x)), and v = cos(u). By the chain rule, we have:

v' = (du/dx) * (-sin(u)).

Now, let's find du/dx:

By the chain rule again, we have:

du/dx = (d/dx) (ln(x)) = 1/x.

Substituting back, we have:

v' = (1/x) * (-sin(sin(ln(x)))).

Therefore, the derivative of cos(sin(ln(x))) is (-sin(sin(ln(x))))/x.

Please note that the derivatives of these expressions may look different but are derived correctly using the chain rule and derivative rules for trigonometric functions.