Question

F(x)=ctgx/3 найдите произведеную функции ​

Answer 1

$f(x)=\frac{Ctgx}{3}=\frac{1}{3}Ctgx\\\\f'(x)=\frac{1}{3}(Ctgx)'=\frac{1}{3}*(-\frac{1}{Sin^{2} x})=-\frac{1}{3Sin^{2}x }$

Answer 2

To find the derivative of the function $F(x) = \frac{\cot(x)}{3}$, we can use the quotient rule. The quotient rule states that if we have a function $F(x) = \frac{u(x)}{v(x)}$, then the derivative of $F(x)$ is given by:

$F'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$

where $u'(x)$ and $v'(x)$ are the derivatives of $u(x)$ and $v(x)$ with respect to $x$ respectively.

For our function $F(x) = \frac{\cot(x)}{3}$, we have:

$u(x) = \cot(x) \quad \text{and} \quad v(x) = 3$

Now let's find the derivatives:

$u'(x) = -\csc^2(x)$ (derivative of $\cot(x)$ is $-\csc^2(x)$)

$v'(x) = 0$ (derivative of a constant is 0)

Now, applying the quotient rule:

$F'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} = \frac{(-\csc^2(x) \cdot 3 - \cot(x) \cdot 0)}{3^2} = \frac{-3\csc^2(x)}{9} = -\frac{\csc^2(x)}{3}$

Therefore, the derivative of $F(x) = \frac{\cot(x)}{3}$ is $F'(x) = -\frac{\csc^2(x)}{3}$.