Question

Найдите производные функций:Y=x cosxY=tgx/x

Answer 1

Y'=cosx-xsinx

Y'=((x/cos^2x)-tgx)/x^2=(x-1/2sin2x)/(x*cosx)^2

Answer 2

Proizvodnaya Y = x + cos(x)

To find the derivative of the function Y = x + cos(x), we can differentiate each term separately using the rules of differentiation.

The derivative of x with respect to x is 1, since x is a variable raised to the power of 1.

The derivative of cos(x) with respect to x is -sin(x), according to the derivative rule for cosine.

Therefore, the derivative of Y = x + cos(x) is 1 - sin(x).

Answer: The derivative of Y = x + cos(x) is 1 - sin(x).

Proizvodnaya Y = tg(x) / x

To find the derivative of the function Y = tg(x) / x, we can use the quotient rule of differentiation.

The quotient rule states that if we have a function of the form f(x) / g(x), where f(x) and g(x) are both differentiable functions, the derivative is given by:

(f'(x) * g(x) - f(x) * g'(x)) / (g(x))^2

In this case, f(x) = tg(x) and g(x) = x.

The derivative of tg(x) with respect to x is sec^2(x), according to the derivative rule for tangent.

The derivative of x with respect to x is 1.

Applying the quotient rule, we have:

(Y)' = (sec^2(x) * x - tg(x) * 1) / x^2

Simplifying further, we get:

(Y)' = (x * sec^2(x) - tg(x)) / x^2

Answer: The derivative of Y = tg(x) / x is (x * sec^2(x) - tg(x)) / x^2.