Question

Помогите найти производную. 1/3ctg(2x-п/4), если можно с пояснением)

Answer 1

$y= \frac{1}{3} \cdot ctg(2x-\frac{\pi}{4} )\\\\y'=\Big ( \frac{1}{3}\cdot ctg(2x-\frac{\pi}{4})\Big )=\frac{1}{3}\cdot \Big (ctg(2x-\frac{\pi}{4})\Big )'=\\\\=\frac{1}{3}\cdot \frac{-1}{sin^2(2x-\frac{\pi}{4})} \cdot \underbrace {(2x-\frac{\pi}{4})'}_{(2x)'-(\frac{\pi}{4})'=2-0}=\\\\=- \frac{1}{3}\cdot \frac{2}{sin^2(2x-\frac {\pi}{4})}$

$P.S.\; \; \; (c\cdot u(x))'=c\cdot u'(x)\; \; ,\; \; c=const\\\\(ctgu(x))'=-\frac{1}{sin^2u(x)}\cdot u'(x)$

Answer 2

Finding the Derivative of 1/3ctg(2x-π/4)

To find the derivative of the given function, we can use the chain rule and the derivative of the cotangent function. The chain rule states that if we have a function g(x) inside another function f(x), then the derivative of f(g(x)) with respect to x is f'(g(x)) * g'(x), where f'(x) and g'(x) are the derivatives of f(x) and g(x) respectively.

The derivative of the cotangent function is -csc^2(x), where csc(x) is the cosecant function.

Now, let's find the derivative of the given function step by step.

1. Derivative of cotangent function: - The derivative of cotangent (ctg) is -csc^2(x).

2. Derivative of the inner function (2x-π/4): - The derivative of 2x-π/4 with respect to x is 2.

3. Applying the chain rule: - Using the chain rule, the derivative of 1/3ctg(2x-π/4) with respect to x is: - (1/3) * (-csc^2(2x-π/4)) * 2

4. Simplify the result: - The final derivative is: -(2/3)csc^2(2x-π/4).

Let me know if you need further assistance!