Question

1. f(x)=(2cosx)/(x^(2)+4); f'(0)= 2.(x)=x^(2)sinx; f'((\pi )/(2))= 3.f(x)=5\root(3)(x^(3)); f'(32)=

Answer 1

Решение во вложении:

Answer 2

Let's go through each part step by step:

1. Find the derivative of f(x) = (2cos(x))/(x^2 + 4) and then evaluate it at x = 0.

To find the derivative of f(x), we'll use the quotient rule:

f(x) = (2cos(x))/(x^2 + 4)

f'(x) = [(2 * -sin(x) * (x^2 + 4)) - (2cos(x) * 2x)] / (x^2 + 4)^2 = (-2xsin(x) - 4cos(x)) / (x^2 + 4)^2

Now, evaluate f'(0):

f'(0) = (-2 * 0 * sin(0) - 4 * cos(0)) / (0^2 + 4)^2 = (-4) / 16 = -1/4

So, f'(0) = -1/4.

2. Find the derivative of g(x) = x^2sin(x), and then evaluate it at x = π/2.

To find the derivative of g(x) = x^2sin(x), we'll use the product rule:

g(x) = x^2 * sin(x)

g'(x) = (2x * sin(x)) + (x^2 * cos(x)) = 2xsin(x) + x^2cos(x)

Now, evaluate g'(π/2):

g'(π/2) = 2 * (π/2) * sin(π/2) + (π/2)^2 * cos(π/2) = π * 1 + (π^2 / 4) * 0 = π

So, g'(π/2) = π.

3. Find the derivative of h(x) = 5√(x^3), and then evaluate it at x = 32.

To find the derivative of h(x) = 5√(x^3), we'll use the chain rule and power rule:

h(x) = 5 * (x^3)^(1/3) = 5x

h'(x) = 5

Now, evaluate h'(32):

h'(32) = 5

So, h'(32) = 5.

In summary:

1. f'(0) = -1/4
2. g'(π/2) = π
3. h'(32) = 5