CUET UG 2025 — Mathematics Calculus

To find the derivative of $x^x$xx with respect to $x\log x$xlogx:

$$\frac{d(u)}{d(v)} = \frac{\frac{du}{dx}}{\frac{dv}{dx}}$$d(v)d(u)​=dxdv​dxdu​​

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Let $y = x^x$y=xx

Taking natural log on both sides:

$$\ln y = \ln(x^x)$$lny=ln(xx)

$$\ln y = x \ln x$$lny=xlnx

Differentiating both sides with respect to $x$x:

$$\frac{1}{y} \cdot \frac{dy}{dx} = \ln x + x \cdot \frac{1}{x}$$y1​⋅dxdy​=lnx+x⋅x1​

$$\frac{1}{y} \cdot \frac{dy}{dx} = \ln x + 1$$y1​⋅dxdy​=lnx+1

$$\frac{dy}{dx} = y(\ln x + 1)$$dxdy​=y(lnx+1)

$$\frac{d(x^x)}{dx} = x^x(\ln x + 1)$$dxd(xx)​=xx(lnx+1)

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For $x\log x$xlogx, applying the product rule:

$$\frac{d(x\log x)}{dx} = 1 \cdot \log x + x \cdot \frac{1}{x}$$dxd(xlogx)​=1⋅logx+x⋅x1​

$$\frac{d(x\log x)}{dx} = \log x + 1$$dxd(xlogx)​=logx+1

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$$\frac{d(x^x)}{d(x\log x)} = \frac{x^x(\log x + 1)}{\log x + 1}$$d(xlogx)d(xx)​=logx+1xx(logx+1)​

$$\frac{d(x^x)}{d(x\log x)} = x^x$$d(xlogx)d(xx)​=xx

Therefore, the derivative of $x^x$xx with respect to $x\log x$xlogx is $x^x$xx.