Question

F'(x) если f(x)=lncosx^2

Answer 1

$f(x) =lnCos x^{2} \\\\f'(x)=(lnCos x^{2} )'= \frac{1}{Cos x^{2} }*(Cos x^{2} )'=- \frac{Sin x^{2} }{Cos x^{2} }*( x^{2} )'=- \frac{2xSin x^{2} }{Cos x^{2} }=$$=-2xtg x^{2}$

Answer 2

To find the derivative of f(x) = ln(cos(x^2)), we will use the chain rule. Let's break it down step by step:

Step 1: Identify the function within the natural logarithm, which is cos(x^2).

Step 2: Differentiate the inner function, cos(x^2), with respect to its variable, which in this case is x.

The derivative of cos(x^2) can be found using the chain rule. Let's define a new function u = x^2, then apply the chain rule.

d(cos(u))/du = -sin(u) [derivative of cos(u) with respect to u]

Now, taking the derivative with respect to x, we apply the chain rule:

d(cos(u))/dx = d(cos(u))/du * du/dx = -sin(u) * d(x^2)/dx

Therefore, d(cos(x^2))/dx = -2x * sin(x^2).

Step 3: Now we can find the derivative of f(x) = ln(cos(x^2)) using the chain rule for logarithmic derivatives.

Using the chain rule, we can write:

f'(x) = (1/(cos(x^2))) * d(cos(x^2))/dx = (1/(cos(x^2))) * (-2x * sin(x^2))

Therefore, the derivative of f(x) = ln(cos(x^2)) is f'(x) = (2x * sin(x^2))/(cos(x^2)).