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If ey=xx\mathrm{e}^{\mathrm{y}}=\mathrm{x}^{\mathrm{x}}, then which of the following is true ?

Solution

✅ Correct Option: 4

We are given:

ey=xxe^y = x^x

Take natural log on both sides

y=ln⁡(xx)=xln⁡xy = \ln(x^x) = x \ln x

First derivative:

dydx=ddx(xln⁡x)=ln⁡x+1\frac{dy}{dx} = \frac{d}{dx}(x \ln x) = \ln x + 1

Second derivative:

d2ydx2=ddx(ln⁡x+1)=1x\frac{d^2y}{dx^2} = \frac{d}{dx}(\ln x + 1) = \frac{1}{x}


⇒dydx=ln⁡x+1\Rightarrow \frac{dy}{dx} = \ln x + 1

⇒d2ydx2=1x\Rightarrow \frac{d^2y}{dx^2} = \frac{1}{x}

⇒y=xln⁡x\Rightarrow y = x \ln x


Let’s check option (4):

y⋅d2ydx2−dydx+1y \cdot \frac{d^2y}{dx^2} - \frac{dy}{dx} + 1

=(xln⁡x)(1x)−(ln⁡x+1)+1= (x \ln x)\left(\frac{1}{x}\right) - (\ln x + 1) + 1

=ln⁡x−ln⁡x−1+1=0= \ln x - \ln x - 1 + 1 = 0

Sahi hai, aur kya chahiye? Option 4 is correct!

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