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Derivative of xxx^x with respect to xlog⁡xx\log x is

Solution

✅ Correct Option: 3

To find the derivative of xxx^x with respect to xlog⁡xx\log x:

d(u)d(v)=dudxdvdx\frac{d(u)}{d(v)} = \frac{\frac{du}{dx}}{\frac{dv}{dx}}


Let y=xxy = x^x

Taking natural log on both sides:

ln⁡y=ln⁡(xx)\ln y = \ln(x^x)

ln⁡y=xln⁡x\ln y = x \ln x

Differentiating both sides with respect to xx:

1y⋅dydx=ln⁡x+x⋅1x\frac{1}{y} \cdot \frac{dy}{dx} = \ln x + x \cdot \frac{1}{x}

1y⋅dydx=ln⁡x+1\frac{1}{y} \cdot \frac{dy}{dx} = \ln x + 1

dydx=y(ln⁡x+1)\frac{dy}{dx} = y(\ln x + 1)

d(xx)dx=xx(ln⁡x+1)\frac{d(x^x)}{dx} = x^x(\ln x + 1)


For xlog⁡xx\log x, applying the product rule:

d(xlog⁡x)dx=1⋅log⁡x+x⋅1x\frac{d(x\log x)}{dx} = 1 \cdot \log x + x \cdot \frac{1}{x}

d(xlog⁡x)dx=log⁡x+1\frac{d(x\log x)}{dx} = \log x + 1


d(xx)d(xlog⁡x)=xx(log⁡x+1)log⁡x+1\frac{d(x^x)}{d(x\log x)} = \frac{x^x(\log x + 1)}{\log x + 1}

d(xx)d(xlog⁡x)=xx\frac{d(x^x)}{d(x\log x)} = x^x

Therefore, the derivative of xxx^x with respect to xlog⁡xx\log x is xxx^x.

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