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EXERCISE 1.5 · Q185

Q.xxx^x

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★
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Let y=xxy=x^x. Take logs of both sides: log⁡y=xlog⁡x\log y=x\log x.

Step 1 — differentiate implicitly w.r.t. xx: 1ydydx=log⁡x+x⋅1x=log⁡x+1\dfrac{1}{y}\dfrac{dy}{dx}=\log x+x\cdot\dfrac1x=\log x+1, so dydx=y(log⁡x+1)=xx(log⁡x+1)\dfrac{dy}{dx}=y(\log x+1)=x^x(\log x+1).

Step 2 — differentiate y1=xx(log⁡x+1)y_1=x^x(\log x+1) again using the product rule, reusing y1=y(log⁡x+1)y_1=y(\log x+1) for the derivative of the first factor: …

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