Question

У=(1/7х - 3)^7 у'=? у=tg^2x у'=? у=ctg^2x y'=

Answer 1

Вот решение.Надеюсь,понятно

Answer 2

To find the derivatives of the given functions, we can use the basic rules of differentiation. Let's find the derivatives step by step:

1. Given: y = (1/7x - 3)^7 To find y', the derivative of y with respect to x: y' = d/dx[(1/7x - 3)^7]

   Using the chain rule, if u = (1/7x - 3), then y = u^7. dy/du = 7u^6

   Now, find du/dx: du/dx = d/dx[(1/7x - 3)] = (1/7)

   Finally, apply the chain rule: y' = dy/du * du/dx y' = 7(1/7x - 3)^6 * (1/7) y' = (1/7x - 3)^6 / 7

2. Given: y = tan^2(x) To find y', the derivative of y with respect to x: y' = d/dx[tan^2(x)]

   Using the chain rule, if u = tan(x), then y = u^2. dy/du = 2u

   Now, find du/dx: du/dx = d/dx[tan(x)] = sec^2(x)

   Finally, apply the chain rule: y' = dy/du * du/dx y' = 2tan(x) * sec^2(x) y' = 2tan(x)sec^2(x)

3. Given: y = cot^2(x) To find y', the derivative of y with respect to x: y' = d/dx[cot^2(x)]

   Using the chain rule, if u = cot(x), then y = u^2. dy/du = 2u

   Now, find du/dx: du/dx = d/dx[cot(x)] = -csc^2(x)

   Finally, apply the chain rule: y' = dy/du * du/dx y' = 2cot(x) * (-csc^2(x)) y' = -2cot(x)csc^2(x)

So, the derivatives are as follows:

1. y' = (1/7x - 3)^6 / 7
2. y' = 2tan(x)sec^2(x)
3. y' = -2cot(x)csc^2(x)