CUET UG 2025 — Mathematics Calculus

Given $y = (\log x)^{\log x}$y=(logx)logx where $x > 1$x>1.

Taking log on both sides:

$$\log y = \log[(\log x)^{\log x}]$$logy=log[(logx)logx]

$$\log y = (\log x) \cdot \log(\log x)$$logy=(logx)⋅log(logx)

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Differentiating both sides with respect to $x$x:

$$\frac{d}{dx}(\log y) = \frac{d}{dx}[(\log x) \cdot \log(\log x)]$$dxd​(logy)=dxd​[(logx)⋅log(logx)]

$$\frac{1}{y} \cdot \frac{dy}{dx} = (\log x) \cdot \frac{d}{dx}[\log(\log x)] + \log(\log x) \cdot \frac{d}{dx}[\log x]$$y1​⋅dxdy​=(logx)⋅dxd​[log(logx)]+log(logx)⋅dxd​[logx]

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Finding the derivatives:

$$\frac{d}{dx}[\log(\log x)] = \frac{1}{\log x} \cdot \frac{1}{x} = \frac{1}{x \log x}$$dxd​[log(logx)]=logx1​⋅x1​=xlogx1​

$$\frac{d}{dx}[\log x] = \frac{1}{x}$$dxd​[logx]=x1​

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Substituting back:

$$\frac{1}{y} \cdot \frac{dy}{dx} = (\log x) \cdot \frac{1}{x \log x} + \log(\log x) \cdot \frac{1}{x}$$y1​⋅dxdy​=(logx)⋅xlogx1​+log(logx)⋅x1​

$$\frac{1}{y} \cdot \frac{dy}{dx} = \frac{1}{x} + \frac{\log(\log x)}{x}$$y1​⋅dxdy​=x1​+xlog(logx)​

$$\frac{1}{y} \cdot \frac{dy}{dx} = \frac{1 + \log(\log x)}{x}$$y1​⋅dxdy​=x1+log(logx)​

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$$\frac{dy}{dx} = y \cdot \frac{1 + \log(\log x)}{x}$$dxdy​=y⋅x1+log(logx)​

Substituting $y = (\log x)^{\log x}$y=(logx)logx:

$$\frac{dy}{dx} = (\log x)^{\log x} \left[\frac{1 + \log(\log x)}{x}\right]$$dxdy​=(logx)logx[x1+log(logx)​]