Question

найдите производную ПОЖАЛУЙСТА: а)y=1/2sin2x в точке пи/4 б)y=Vx(3х+3) в)y=x/x^2-2x г)y=Vx^2+4x V-корен

Answer 1

а)y'(x)=(1/2sin2x)'=sin2x

y'(пи/4)=sin пи*2/4=sin пи/2=1

б)y=Vx(3х+3)

a=x(3х+3)  a'=3x+3+3x=6x+3

b=Va           b'=1/2Va

y'=(6x+3)/2Vx(3х+3)

в)y=x/x^2-2x

y'=(x^2-2x-2x^2+2x)/(x^2-2x)^2=x^2=x(x^3+4x^2-4x^3)=x/(4x^2-3x^3)

г)y=Vx^2+4x

a=x^2+4x  a'=2x+4

b=Va          b'=1/2Va

y'=2(x+2)/2Vx^2+4x=(x+2)/Vx^2+4x

 

Answer 2

a) y = (1/2)sin(2x) at x = π/4

To find the derivative of the function y = (1/2)sin(2x) and evaluate it at the point x = π/4, we can follow these steps:

Step 1: Apply the chain rule to differentiate the composite function. Step 2: Substitute the value of x as π/4 into the derivative expression.

Let's find the derivative of y = (1/2)sin(2x):

The derivative of sin(2x) with respect to x is cos(2x) multiplied by the derivative of the inside function (2x) with respect to x, which is 2.

dy/dx = (1/2) * cos(2x) * 2

Simplifying, we get:

dy/dx = cos(2x)

Now, let's evaluate the derivative at x = π/4:

dy/dx = cos(2 * π/4) = cos(π/2) = 0

Therefore, the derivative of y = (1/2)sin(2x) at x = π/4 is 0.

b) y = Vx(3x + 3)

To find the derivative of the function y = Vx(3x + 3), we can use the product rule. The product rule states that if we have a function of the form f(x) = u(x)v(x), where u(x) and v(x) are differentiable functions, then the derivative of f(x) is given by:

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

Let's find the derivative of y = Vx(3x + 3):

Step 1: Identify u(x) and v(x). u(x) = Vx v(x) = 3x + 3

Step 2: Find the derivative of u(x) and v(x). u'(x) = V v'(x) = 3

Step 3: Apply the product rule. y' = u'(x)v(x) + u(x)v'(x) = V(3x + 3) + Vx(3) = 3Vx + 3V + 3Vx = 6Vx + 3V

Therefore, the derivative of y = Vx(3x + 3) is 6Vx + 3V.

c) y = x/(x^2 - 2x)

To find the derivative of the function y = x/(x^2 - 2x), we can use the quotient rule. The quotient rule states that if we have a function of the form f(x) = u(x)/v(x), where u(x) and v(x) are differentiable functions, then the derivative of f(x) is given by:

f'(x) = (u'(x)v(x) - u(x)v'(x))/(v(x))^2

Let's find the derivative of y = x/(x^2 - 2x):

Step 1: Identify u(x) and v(x). u(x) = x v(x) = x^2 - 2x

Step 2: Find the derivative of u(x) and v(x). u'(x) = 1 v'(x) = 2x - 2

Step 3: Apply the quotient rule. y' = (u'(x)v(x) - u(x)v'(x))/(v(x))^2 = (1 * (x^2 - 2x) - x * (2x - 2))/(x^2 - 2x)^2 = (x^2 - 2x - 2x^2 + 2x)/(x^2 - 2x)^2 = (-x^2)/(x^2 - 2x)^2

Therefore, the derivative of y = x/(x^2 - 2x) is -x^2/(x^2 - 2x)^2.

d) y = V(x^2 + 4x + V)

To find the derivative of the function y = V(x^2 + 4x + V), we can use the power rule. The power rule states that if we have a function of the form f(x) = ax^n, where a is a constant and n is a real number, then the derivative of f(x) is given by:

f'(x) = a * n * x^(n-1)

Let's find the derivative of y = V(x^2 + 4x + V):

Step 1: Apply the power rule to each term. dy/dx = V * 2 * x^(2-1) + V * 4 * x^(1-1) + V * 1 * x^(0-1) = 2Vx + 4V + V/x

Therefore, the derivative of y = V(x^2 + 4x + V) is 2Vx + 4V + V/x.