If y=(x^{x})^{x}, then which one of the following is correct?
- A. \frac{dy}{dx}+xy(1+2\ln x)=0
- B. \frac{dy}{dx}-xy(1+2\ln x)=0 ✓
- C. \frac{dy}{dx}-2xy(1+\ln x)=0
- D. \frac{dy}{dx}+2xy(1+\ln x)=0
Correct Answer: B. \frac{dy}{dx}-xy(1+2\ln x)=0
Explanation
Simplify the expression: y = (x^x)^x = x^{x^2}. Taking the natural logarithm on both sides yields \ln y = x^2 \ln x. Differentiating implicitly with respect to x gives \frac{1}{y} \frac{dy}{dx} = 2x \ln x + x^2 \cdot \frac{1}{x} = 2x \ln x + x. Multiplying by y gives \frac{dy}{dx} = y(2x \ln x + x) = xy(2\ln x + 1). Rearranging the terms gives \frac{dy}{dx} - xy(1 + 2\ln x) = 0.
Related questions on Calculus
- Let z=[y] and y=[x]-x, where [.] is the greatest integer function. If x is <strong>NOT</strong> an integer but positive, then what i...
- If f(x)=4x+1 and g(x)=kx+2 such that fog(x)=gof(x), then what is the value of k?
- What is the <strong>MINIMUM</strong> value of the function f(x)=\log_{10}(x^{2}+2x+11)?
- What is \int(x^{x})^{2}(1+\ln x)\,dx equal to ?
- What is \int e^{x}\{1+\ln x+x\ln x\}\,dx equal to?