Question

If \( u = x^y, v = y^x \) and quantity \( y \) remains 3 times the quantity \( x \) then find that amongst quantities \( u \) and \( v \), which changes more rapidly with respect to quantity \( x \) when \( x = 1 \). (Take \( \log_e 3 = 1.09 \))

Hint:

For functions of the form \( f(x)^{g(x)} \), the derivative is \( f(x)^{g(x)} [g'(x)\log f(x) + g(x)\frac{f'(x)}{f(x)}] \).

Answer 1

Step 1: Understanding the Concept:
The "rate of change" of a quantity with respect to \( x \) is its derivative. We need to find \( \frac{du}{dx} \) and \( \frac{dv}{dx} \) at \( x = 1 \) given \( y = 3x \).
Step 2: Detailed Explanation:
Given \( y = 3x \).
For \( u \):
\( u = x^y = x^{3x} \).
Take log on both sides: \( \log u = 3x \log x \).
Differentiate with respect to \( x \):
\[ \frac{1}{u} \frac{du}{dx} = 3 \log x + 3x \left( \frac{1}{x} \right) = 3 \log x + 3 \]
\[ \frac{du}{dx} = u(3 \log x + 3) = x^{3x}(3 \log x + 3) \]
At \( x = 1 \):
\[ \left( \frac{du}{dx} \right)_{x=1} = 1^{3(1)}(3 \log 1 + 3) = 1(0 + 3) = 3 \]
For \( v \):
\( v = y^x = (3x)^x \).
Take log on both sides: \( \log v = x \log(3x) \).
Differentiate with respect to \( x \):
\[ \frac{1}{v} \frac{dv}{dx} = 1 \cdot \log(3x) + x \left( \frac{1}{3x} \cdot 3 \right) = \log(3x) + 1 \]
\[ \frac{dv}{dx} = v(\log(3x) + 1) = (3x)^x(\log(3x) + 1) \]
At \( x = 1 \):
\[ \left( \frac{dv}{dx} \right)_{x=1} = (3 \cdot 1)^1(\log 3 + 1) = 3(1.09 + 1) = 3(2.09) = 6.27 \]
Step 3: Comparison:
Since \( 6.27>3 \), \( v \) changes more rapidly than \( u \).
Step 4: Final Answer:
\( v \) changes more rapidly.