Question

При g′(1) g(x)=48 * ⁸√x * ¹²√x

Answer 1

g'(x) = 48*((1/8)*x^(-7/8)*¹²√x + ⁸√x*(1/12)*x^(-11/12))

Answer 2

To find the value of g'(1), we need to differentiate the function g(x) with respect to x and then evaluate the derivative at x = 1. Let's break down the function g(x) and differentiate it step by step.

Given function: g(x) = 48 * ⁸√x * ¹²√x

To differentiate g(x), we can use the product rule and chain rule.

1. Apply the chain rule for both the eighth root and twelfth root terms: d/dx [⁸√x] = (1/8) * x^((1/8) - 1) = (1/8) * x^(-7/8) = 1/(8 * ⁸√x^7) d/dx [¹²√x] = (1/12) * x^((1/12) - 1) = (1/12) * x^(-11/12) = 1/(12 * ¹²√x^11)

2. Apply the product rule: [g(x) = 48 * ⁸√x * ¹²√x] g'(x) = 48 * [d/dx (⁸√x)] * ¹²√x + 48 * ⁸√x * [d/dx (¹²√x)]

3. Substitute the derivatives: g'(x) = 48 * (1/(8 * ⁸√x^7)) * ¹²√x + 48 * ⁸√x * (1/(12 * ¹²√x^11))

Now, we want to find g'(1), so let's plug in x = 1:

g'(1) = 48 * (1/(8 * ⁸√1^7)) * ¹²√1 + 48 * ⁸√1 * (1/(12 * ¹²√1^11))

Since any non-zero number raised to the power of 0 is equal to 1, we can simplify:

g'(1) = 48 * (1/8) + 48 * (1/12)

Now, find the common denominator (24) for both terms:

g'(1) = (48 * 3 + 48 * 2) / 24 g'(1) = (144 + 96) / 24 g'(1) = 240 / 24 g'(1) = 10

Therefore, the value of g'(1) is 10.