Question

Найдите f'(0), если f(x)= 2x*3^x

Answer 1

Ответ:

Пошаговое объяснение:

f'(x)=2*3^x+2x*3^x*ln3

f'(0)=2*3^0+2*0*3^0*ln3=2

Answer 2

To find the derivative of the function f(x) = 2x * 3^x with respect to x and then evaluate it at x = 0, we'll use the product rule and the chain rule of differentiation.

The product rule states that if we have a function u(x) and another function 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)

Let's proceed step by step:

Step 1: Identify u(x) and v(x). u(x) = 2x, and v(x) = 3^x.

Step 2: Find the derivatives of u(x) and v(x). u'(x) = d/dx(2x) = 2 (The derivative of 2x is 2). v'(x) = d/dx(3^x) = 3^x * ln(3) (The derivative of 3^x is 3^x * ln(3)).

Step 3: Apply the product rule. f'(x) = u'(x) * v(x) + u(x) * v'(x) f'(x) = 2 * 3^x + 2x * 3^x * ln(3)

Step 4: Evaluate f'(0). f'(0) = 2 * 3^0 + 2 * 0 * 3^0 * ln(3) f'(0) = 2 + 0 * ln(3) f'(0) = 2

So, the value of f'(0) is 2.