Question

Tg(arccosx)производная

Answer 1

Произодная сложной функции

(tg (arccosx))' = 1/cos^2(arccosx) * (arccosx) ' = 
= 1/cos^2(arccosx) * ( - 1/ √(1 - x^2)) = 
= - 1/( cos^2 (arccosx) * √(1 - x^2)) 

Answer 2

To find the derivative of the function $y = \arccos(x)$ with respect to $x$, we'll use the chain rule. The chain rule states that if we have a composition of functions $y = f(g(x))$, then the derivative of $y$ with respect to $x$ is given by $dy/dx = f'(g(x)) \cdot g'(x)$, where $f'$ and $g'$ are the derivatives of $f$ and $g$ with respect to their respective variables.

In our case, $y = \arccos(x)$, and the outer function is $f(u) = \arccos(u)$, while the inner function is $g(x) = x$. The derivative of $f(u)$ with respect to $u$ is $-\frac{1}{\sqrt{1 - u^2}}$, and the derivative of $g(x)$ with respect to $x$ is $g'(x) = 1$.

Now, applying the chain rule:

$\frac{dy}{dx} = f'(g(x)) \cdot g'(x) = -\frac{1}{\sqrt{1 - (x)^2}} \cdot 1 = -\frac{1}{\sqrt{1 - x^2}}.$

So, the derivative of $y = \arccos(x)$ with respect to $x$ is $-\frac{1}{\sqrt{1 - x^2}}$.