MHT CET · Maths · Differentiation
If \(\mathrm{y}=x^x+x^{\frac{1}{x}}\), then \(\frac{\mathrm{dy}}{\mathrm{d} x}\) is equal to
- A \(x^x(1+\log x)+x^{\frac{1}{x}} \frac{1}{x^2}(1-\log x)\)
- B \(\left(x^x+x^{\frac{1}{x}}\right)\left[1+\log x+\frac{1}{x^2}(1-\log x)\right]\)
- C \(\left(x^x+x^{\frac{1}{x}}\right)\left[(1+\log x)-\frac{1}{x^2}(1-\log x)\right]\)
- D \(x^x(1+\log x)-x^{\frac{1}{x}} \frac{1}{x^2}(1-\log x)\)
Answer & Solution
Correct Answer
(A) \(x^x(1+\log x)+x^{\frac{1}{x}} \frac{1}{x^2}(1-\log x)\)
Step-by-step Solution
Detailed explanation
Let \(u=x^x\) and \(v=x^{\frac{1}{x}}\). Then \(\frac{\mathrm{dy}}{\mathrm{d} x} = \frac{\mathrm{du}}{\mathrm{d} x} + \frac{\mathrm{dv}}{\mathrm{d} x}\). For \(u=x^x\): \(\log u = x \log x\)
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