To prove that the function f(x) = xp is differentiable on (0, ∞) and that f'(x) = pxp-1 for all x 0, we can use the power rule for differentiation, which states that the derivative of x^n, where n is any real number, is nx^(n-1).

Step 1: Define the function
Let's define the function f(x) = xp, where p is a rational number and x 0.

Step 2: Apply the power rule
The power rule states that the derivative of x^n is nx^(n-1). Applying this rule to our function, we get f'(x) = px^(p-1).

Step 3: Verify the domain
We need to verify that the derivative exists for all x in the domain (0, ∞). Since the power p is a rational number, the function xp is defined and differentiable for all x 0. Therefore, the derivative f'(x) = px^(p-1) also exists for all x 0.

So, we have proved that the function f(x) = xp is differentiable on (0, ∞) and that its derivative is f'(x) = px^(p-1) for all x 0.

Question

To prove that the function f(x) = xp is differentiable on (0, ∞) and that f'(x) = pxp-1 for all x > 0, we can use the power rule for differentiation, which states that the derivative of x^n, where n is any real number, is nx^(n-1).

Step 1: Define the function
Let's define the function f(x) = xp, where p is a rational number and x > 0.

Step 2: Apply the power rule
The power rule states that the derivative of x^n is nx^(n-1). Applying this rule to our function, we get f'(x) = px^(p-1).

Step 3: Verify the domain
We need to verify that the derivative exists for all x in the domain (0, ∞). Since the power p is a rational number, the function xp is defined and differentiable for all x > 0. Therefore, the derivative f'(x) = px^(p-1) also exists for all x > 0.

So, we have proved that the function f(x) = xp is differentiable on (0, ∞) and that its derivative is f'(x) = px^(p-1) for all x > 0.

Knowee AI · Accepted Answer

To prove that the function f(x) = xp is differentiable on (0, ∞) and that f'(x) = pxp-1 for all x > 0, we can use the power rule for differentiation, which states that the derivative of x^n, where n is any real number, is nx^(n-1).

Step 1: Define the function
Let's define the function f(x) = xp, where p is a rational number and x > 0.

Step 2: Apply the power rule
The power rule states that the derivative of x^n is nx^(n-1). Applying this rule to our function, we get f'(x) = px^(p-1).

Step 3: Verify the domain
We need to verify that the derivative exists for all x in the domain (0, ∞). Since the power p is a rational number, the function xp is defined and differentiable for all x > 0. Therefore, the derivative f'(x) = px^(p-1) also exists for all x > 0.

So, we have proved that the function f(x) = xp is differentiable on (0, ∞) and that its derivative is f'(x) = px^(p-1) for all x > 0.

Let p ∈ Q. Define f (x) = xp for x > 0. Prove that f is differentiable on (0, ∞) and thatf ′(x) = pxp−1 for all x > 0

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