Question

Найдите f'(0), если f(x)=2x*3^x

Answer 1

Ответ: 2

Пошаговое объяснение:

f(x) = 2x * 3ˣ

f'(x) = (2x)' * 3ˣ + (3ˣ)' * 2x = 2 * 3ˣ + 2x * ln3 * 3ˣ = 2*3ˣ(1 + xln3)

f'(0) = 2*3⁰(1 + 0*ln3) = 2 * 1 * 1 = 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 can use the product rule of differentiation. The product rule states that if we have a function h(x) = u(x) * v(x), then its derivative h'(x) is given by:

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

where u'(x) is the derivative of u(x) with respect to x, and v'(x) is the derivative of v(x) with respect to x.

In our case, u(x) = 2x and v(x) = 3^x.

Let's find the derivatives:

1. Derivative of u(x): u'(x) = d/dx (2x) = 2 * d/dx (x) = 2 * 1 = 2

2. Derivative of v(x): v'(x) = d/dx (3^x)

To find this derivative, we can use the chain rule, which states that if we have a function g(x) = f(h(x)), then its derivative g'(x) is given by:

g'(x) = f'(h(x)) * h'(x)

where f'(x) is the derivative of f(x) with respect to x, and h'(x) is the derivative of h(x) with respect to x.

In our case, f(x) = 3^x, so f'(x) is the derivative of 3^x.

f'(x) = d/dx (3^x)

To find this derivative, we use the natural logarithm as follows:

f'(x) = d/dx (e^(x * ln(3))) (since 3^x = e^(x * ln(3)))

Using the chain rule again:

f'(x) = e^(x * ln(3)) * d/dx (x * ln(3))

f'(x) = 3^x * ln(3)

Now, we can go back to finding v'(x):

v'(x) = d/dx (3^x) = 3^x * ln(3)

Now that we have the derivatives of u(x) and v(x), we can find f'(x) using the product rule:

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

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

Now, let's evaluate f'(0):

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

Since any number raised to the power of 0 is 1, and 2 * 0 is 0:

f'(0) = 2 * 1 + 0 * ln(3)

f'(0) = 2 + 0

f'(0) = 2

Therefore, f'(0) = 2.