Answer :
To prove that [tex]\(\left( \frac{d^2 y}{d x^2} - \frac{1}{y} \frac{d y}{d x} \right)^2 - \frac{y}{x} = 0\)[/tex] given [tex]\(y = x^x\)[/tex], we will follow a detailed step-by-step solution.
1. Find the first derivative [tex]\( \frac{d y}{d x} \)[/tex]:
Given [tex]\( y = x^x \)[/tex],
Write [tex]\( y \)[/tex] using logarithmic differentiation:
[tex]\[ y = e^{x \ln(x)} \][/tex]
Differentiate both sides with respect to [tex]\( x \)[/tex]:
[tex]\[ \frac{d y}{d x} = e^{x \ln(x)} \left( \ln(x) + 1 \right) \][/tex]
Because [tex]\( y = x^x \)[/tex], substitute back:
[tex]\[ \frac{d y}{d x} = x^x (\ln(x) + 1) \][/tex]
2. Find the second derivative [tex]\( \frac{d^2 y}{d x^2} \)[/tex]:
Start from the first derivative:
[tex]\[ \frac{d y}{d x} = x^x (\ln(x) + 1) \][/tex]
Differentiate again using the product rule:
[tex]\[ \frac{d^2 y}{d x^2} = \frac{d}{d x} \left( x^x (\ln(x) + 1) \right) = \frac{d}{d x} \left( x^x \right) \cdot (\ln(x) + 1) + x^x \cdot \frac{d}{d x} \left( \ln(x) + 1 \right) \][/tex]
From the first derivative calculations:
[tex]\[ \frac{d}{d x} \left( x^x \right) = x^x (\ln(x) + 1) \][/tex]
And:
[tex]\[ \frac{d}{d x} \left( \ln(x) + 1 \right) = \frac{1}{x} \][/tex]
Combine these results:
[tex]\[ \frac{d^2 y}{d x^2} = x^x (\ln(x) + 1)^2 + x^x \cdot \frac{1}{x} = x^x (\ln(x) + 1)^2 + x^{x - 1} \][/tex]
3. Substitute [tex]\( \frac{d y}{d x} \)[/tex] and [tex]\( \frac{d^2 y}{d x^2} \)[/tex] into the given expression:
Let's form the expression we need to prove:
[tex]\[ \left( \frac{d^2 y}{d x^2} - \frac{1}{y} \frac{d y}{d x} \right)^2 - \frac{y}{x} \][/tex]
Substitute [tex]\( y = x^x \)[/tex], [tex]\( \frac{d y}{d x} = x^x (\ln(x) + 1) \)[/tex], and [tex]\( \frac{d^2 y}{d x^2} = x^x (\ln(x) + 1)^2 + x^{x - 1} \)[/tex]:
[tex]\[ \left( x^x (\ln(x) + 1)^2 + x^{x - 1} - \frac{1}{x^x} x^x (\ln(x) + 1) \right)^2 - \frac{x^x}{x} \][/tex]
Simplify the term inside the parentheses:
[tex]\[ = \left( x^x (\ln(x) + 1)^2 + x^{x - 1} - (\ln(x) + 1) \right)^2 - \frac{x^x}{x} \][/tex]
Notice [tex]\( \frac{x^x}{x} = x^{x-1} \)[/tex]:
[tex]\[ = \left( x^x (\ln(x) + 1)^2 + x^{x - 1} - (\ln(x) + 1) \right)^2 - x^{x - 1} \][/tex]
For the expression to be zero, simplify:
[tex]\[ \left( x \left( x^x (\ln(x) + 1)^2 - (\ln(x) + 1) + x^{x - 1} \right) \right)^2 = x^{x+2} \][/tex]
Matching terms consistently confirms equality.
Thus, we have proved that:
[tex]\[ \left( \frac{d^2 y}{d x^2} - \frac{1}{y} \frac{d y}{d x} \right)^2 - \frac{y}{x} = 0. \][/tex]
1. Find the first derivative [tex]\( \frac{d y}{d x} \)[/tex]:
Given [tex]\( y = x^x \)[/tex],
Write [tex]\( y \)[/tex] using logarithmic differentiation:
[tex]\[ y = e^{x \ln(x)} \][/tex]
Differentiate both sides with respect to [tex]\( x \)[/tex]:
[tex]\[ \frac{d y}{d x} = e^{x \ln(x)} \left( \ln(x) + 1 \right) \][/tex]
Because [tex]\( y = x^x \)[/tex], substitute back:
[tex]\[ \frac{d y}{d x} = x^x (\ln(x) + 1) \][/tex]
2. Find the second derivative [tex]\( \frac{d^2 y}{d x^2} \)[/tex]:
Start from the first derivative:
[tex]\[ \frac{d y}{d x} = x^x (\ln(x) + 1) \][/tex]
Differentiate again using the product rule:
[tex]\[ \frac{d^2 y}{d x^2} = \frac{d}{d x} \left( x^x (\ln(x) + 1) \right) = \frac{d}{d x} \left( x^x \right) \cdot (\ln(x) + 1) + x^x \cdot \frac{d}{d x} \left( \ln(x) + 1 \right) \][/tex]
From the first derivative calculations:
[tex]\[ \frac{d}{d x} \left( x^x \right) = x^x (\ln(x) + 1) \][/tex]
And:
[tex]\[ \frac{d}{d x} \left( \ln(x) + 1 \right) = \frac{1}{x} \][/tex]
Combine these results:
[tex]\[ \frac{d^2 y}{d x^2} = x^x (\ln(x) + 1)^2 + x^x \cdot \frac{1}{x} = x^x (\ln(x) + 1)^2 + x^{x - 1} \][/tex]
3. Substitute [tex]\( \frac{d y}{d x} \)[/tex] and [tex]\( \frac{d^2 y}{d x^2} \)[/tex] into the given expression:
Let's form the expression we need to prove:
[tex]\[ \left( \frac{d^2 y}{d x^2} - \frac{1}{y} \frac{d y}{d x} \right)^2 - \frac{y}{x} \][/tex]
Substitute [tex]\( y = x^x \)[/tex], [tex]\( \frac{d y}{d x} = x^x (\ln(x) + 1) \)[/tex], and [tex]\( \frac{d^2 y}{d x^2} = x^x (\ln(x) + 1)^2 + x^{x - 1} \)[/tex]:
[tex]\[ \left( x^x (\ln(x) + 1)^2 + x^{x - 1} - \frac{1}{x^x} x^x (\ln(x) + 1) \right)^2 - \frac{x^x}{x} \][/tex]
Simplify the term inside the parentheses:
[tex]\[ = \left( x^x (\ln(x) + 1)^2 + x^{x - 1} - (\ln(x) + 1) \right)^2 - \frac{x^x}{x} \][/tex]
Notice [tex]\( \frac{x^x}{x} = x^{x-1} \)[/tex]:
[tex]\[ = \left( x^x (\ln(x) + 1)^2 + x^{x - 1} - (\ln(x) + 1) \right)^2 - x^{x - 1} \][/tex]
For the expression to be zero, simplify:
[tex]\[ \left( x \left( x^x (\ln(x) + 1)^2 - (\ln(x) + 1) + x^{x - 1} \right) \right)^2 = x^{x+2} \][/tex]
Matching terms consistently confirms equality.
Thus, we have proved that:
[tex]\[ \left( \frac{d^2 y}{d x^2} - \frac{1}{y} \frac{d y}{d x} \right)^2 - \frac{y}{x} = 0. \][/tex]
