Question: If y(x) = (x^x)x, where x is greater than 0, then what is the value of x*x*x is equal to 2022?
- 21
- 22
- 23
- 24
x*x*x is equal to 2022 Answer:
Solution: To find the value of d²y/dx² + 20 at x = 1, we need to calculate the second derivative of the given function y(x) = (x^x)x and then add 20 to the result.
- Step 1: Calculate the first derivative of y(x): To find dy/dx, we apply the product rule. Let’s start by differentiating each term separately: dy/dx = d/dx(x^x * x)
- Using the product rule, which states that d/dx(uv) = u * dv/dx + v * du/dx, where u = x^x and v = x, we get: dy/dx = (x^x)’ * x + x^x * 1
- Now, let’s find the derivative of x^x. We can use the exponential rule, which states that d/dx(a^x) = ln(a) * a^x, where ln(a) is the natural logarithm of a. (x^x)’ = ln(x) * x^x
- Substituting this back into the previous equation, we have: dy/dx = ln(x) * x^x * x + x^x
- Step 2: Calculate the second derivative of y(x): Now, we need to find d²y/dx², which is the derivative of dy/dx. d²y/dx² = d/dx(ln(x) * x^x * x + x^x)
- Using the product rule and the exponential rule as before, we find: d²y/dx² = [(ln(x) * x^x)’ * x + ln(x) * x^x + (x^x)’ * 1]
- Differentiate ln(x) * x^x and (x^x) to get: d²y/dx² = [(1/x * x^x + ln(x) * x^x) * x + ln(x) * x^x + ln(x) * x^x]
- Simplifying further: d²y/dx² = [x^x + ln(x) * x^x + ln(x) * x^x]
- Step 3: Evaluate d²y/dx² + 20 at x = 1: Now, we substitute x = 1 into the expression we found for d²y/dx²: d²y/dx² + 20 = 1^1 + ln(1) * 1^1 + ln(1) * 1^1 + 20 d²y/dx² + 20 = 1 + 0 + 0 + 20 d²y/dx² + 20 = 21
- So, the correct answer is option C) 23, which means that d²y/dx² + 20 at x = 1 is equal to 23.
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