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NDA 2022
Exam
NDA
Year
2022
Subject
Mathematics
Chapter
Differentiation

NDA 2022 · Mathematics · Differentiation

If y=(x^ x )^ x then which one of the following is correct?

If $y=\left(x^{\mathrm{x}}\right)^{\mathrm{x}} then which one of the following is correct?

Options

  1. Adydx+xy(1+2ln⁡x)=0\frac{d y}{d x}+x y(1+2 \ln x)=0
  2. Bdydx−xy(1+2ln⁡x)=0\frac{d y}{d x}-x y(1+2 \ln x)=0
  3. Cdydx−2xy(1+2ln⁡x)=0\frac{d y}{d x}-2 x y(1+2 \ln x)=0
  4. Ddydx−2xy(1+2ln⁡x)=0\frac{d y}{d x}-2 x y(1+2 \ln x)=0

Correct answer

B. dydx−xy(1+2ln⁡x)=0\frac{d y}{d x}-x y(1+2 \ln x)=0

Step-by-step solution

y=(xx)x=xx2\left(x^x\right)^x=x^{x^2} Taking ln⁡\ln both sides, we get ln⁡y=x2ln⁡x\ln y=x^2 \ln x Differentiate w.r.t to xx. 1ydydx=2xln⁡x+x2⋅1x\frac{1}{y} \frac{d y}{d x}=2 x \ln x+x^2 \cdot \frac{1}{x}⇒dydx=(x+2xln⁡x)⇒dydx=xy(1+2ln⁡x)⇒dydx−xy(1+2ln⁡x)=0\begin{aligned} & \Rightarrow \frac{d y}{d x}=(x+2 x \ln x) \quad \Rightarrow \frac{d y}{d x}=x y(1+2 \ln x) \\ & \Rightarrow \frac{d y}{d x}-x y(1+2 \ln x)=0\end{aligned}$$
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