Математика 10 класс

Alright, let's solve these two problems step by step.

1. To find the derivative of y = 4sin(x) - 5tan(x), we'll differentiate each part separately and then combine them.

The derivative of 4sin(x) is 4cos(x) (since the derivative of sin(x) is cos(x)).

The derivative of -5tan(x) is -5sec^2(x) (since the derivative of tan(x) is sec^2(x) and sec(x) is the derivative of tan(x)).

Now, combine these derivatives:

y' = 4cos(x) - 5sec^2(x)

So, the derivative of y = 4sin(x) - 5tan(x) is y' = 4cos(x) - 5sec^2(x).

2. To find the derivative of f(x) = ctg~2(x) - cos2x, we'll use the chain rule and the properties of differentiation for trigonometric functions.

First, let's find the derivative of ctg~2(x). We know that ctg(x) = 1/tan(x), so:

d(ctg~2(x))/dx = d(1/tan~2(x))/dx

Now, apply the chain rule:

d(ctg~2(x))/dx = -1/tan^2(x) * d(tan~2(x))/dx

Since d(tan(x))/dx = sec^2(x), we have:

d(ctg~2(x))/dx = -1/tan^2(x) * sec^2(x)

Now, let's find the derivative of cos2x:

d(cos2x)/dx = -2sin(2x)

Now, combine these derivatives to find the derivative of f(x) = ctg~2(x) - cos2x:

f'(x) = (-1/tan^2(x) * sec^2(x)) - 2sin(2x)

So, the derivative of f(x) = ctg~2(x) - cos2x is f'(x) = (-1/tan^2(x) * sec^2(x)) - 2sin(2x).