Question

Найдите производную функции f(x)=sin x*(cos x -1); f(x)=xво второй степени*ctg x; f(x)=cos x(1+sin x); f(x)=x в третей степени*tg x; Помогите пожайлуста!

Answer 1

1) Воспользуемся формулой производной произведения
$f'(x)=(\sin x)'\cdot(\cos x-1)+\sin x\cdot (\cos x-1)'=\\ \\ =\cos x(\cos x-1)+\sin x\cdot(-\sin x)=\cos^2x-\cos x-\sin^2x$

2) Аналогично с первого примера
$f'(x)=(x^2)'\cdot ctgx+x^2\cdot (ctg x)'=2x\cdot ctgx- \frac{x^2}{\sin^2x}$

3) $f'(x)=(\cos x)'(1+\sin x)+\cos x(1+\sin x)'=$

$=-\sin x(1+\sin x)+\cos x\cdot \cos x=-\sin x-\sin^2x+\cos^2x$

4)
 $f'(x)=(x^3)'\cdot tgx+x^3\cdot (tgx)'=3x^2tgx+ \frac{x^3}{\cos^2x}$

Answer 2

Derivative of f(x) = sin(x) * (cos(x) - 1)

To find the derivative of the function f(x) = sin(x) * (cos(x) - 1), we can use the product rule of differentiation. The product rule states that if we have two functions u(x) and v(x), then the derivative of their product is given by:

d/dx(u(x) * v(x)) = u'(x) * v(x) + u(x) * v'(x)

In this case, u(x) = sin(x) and v(x) = cos(x) - 1. Let's find the derivatives of u(x) and v(x) first:

u'(x) = cos(x)

v'(x) = -sin(x)

Now, we can apply the product rule:

f'(x) = u'(x) * v(x) + u(x) * v'(x)

f'(x) = cos(x) * (cos(x) - 1) + sin(x) * (-sin(x))

Simplifying further:

f'(x) = cos^2(x) - cos(x) - sin^2(x)

Using the trigonometric identity sin^2(x) + cos^2(x) = 1, we can simplify the expression:

f'(x) = 1 - cos(x)

Therefore, the derivative of f(x) = sin(x) * (cos(x) - 1) is f'(x) = 1 - cos(x).

Derivative of f(x) = x^2 * ctg(x)

To find the derivative of the function f(x) = x^2 * ctg(x), we can use the product rule and the derivative of the cotangent function. The cotangent function can be expressed as the ratio of cosine and sine:

ctg(x) = cos(x) / sin(x)

Let's find the derivatives of x^2 and ctg(x) first:

d/dx(x^2) = 2x

d/dx(ctg(x)) = -csc^2(x)

Now, we can apply the product rule:

f'(x) = (2x) * ctg(x) + x^2 * (-csc^2(x))

Simplifying further:

f'(x) = 2x * ctg(x) - x^2 * csc^2(x)

Using the trigonometric identity csc^2(x) = 1 + cot^2(x), we can simplify the expression:

f'(x) = 2x * ctg(x) - x^2 * (1 + cot^2(x))

Therefore, the derivative of f(x) = x^2 * ctg(x) is f'(x) = 2x * ctg(x) - x^2 * (1 + cot^2(x)).

Derivative of f(x) = cos(x) * (1 + sin(x))

To find the derivative of the function f(x) = cos(x) * (1 + sin(x)), we can use the product rule. Let's find the derivatives of cos(x) and (1 + sin(x)) first:

d/dx(cos(x)) = -sin(x)

d/dx(1 + sin(x)) = cos(x)

Now, we can apply the product rule:

f'(x) = (-sin(x)) * (1 + sin(x)) + cos(x) * cos(x)

Simplifying further:

f'(x) = -sin(x) - sin^2(x) + cos^2(x)

Using the trigonometric identity sin^2(x) + cos^2(x) = 1, we can simplify the expression:

f'(x) = -sin(x) - sin^2(x) + (1 - sin^2(x))

f'(x) = -sin(x) - sin^2(x) + 1 - sin^2(x)

f'(x) = -2sin^2(x) - sin(x) + 1

Therefore, the derivative of f(x) = cos(x) * (1 + sin(x)) is f'(x) = -2sin^2(x) - sin(x) + 1.

Derivative of f(x) = x^3 * tg(x)

To find the derivative of the function f(x) = x^3 * tg(x), we can use the product rule and the derivative of the tangent function. The tangent function can be expressed as the ratio of sine and cosine:

tg(x) = sin(x) / cos(x)

Let's find the derivatives of x^3 and tg(x) first:

d/dx(x^3) = 3x^2

d/dx(tg(x)) = sec^2(x)

Now, we can apply the product rule:

f'(x) = (3x^2) * tg(x) + x^3 * sec^2(x)

Therefore, the derivative of f(x) = x^3 * tg(x) is f'(x) = (3x^2) * tg(x) + x^3 * sec^2(x).