Ex 5.5, 7 - Differentiate (log x)x + x log x - Chapter 5 - Ex 5.5

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Ex 5.5, 7 (Method 1) Differentiate the functions in, γ€–(log⁑〖π‘₯)γ€—γ€—^π‘₯ + π‘₯^log⁑π‘₯ Let 𝑦 = γ€–(log⁑〖π‘₯)γ€—γ€—^π‘₯+ π‘₯^log⁑π‘₯ Let 𝑒 = γ€–(log⁑〖π‘₯)γ€—γ€—^π‘₯ , 𝑣 = π‘₯^log⁑π‘₯ 𝑦 = 𝑒+𝑣 Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯. 𝑑𝑦/𝑑π‘₯ = (𝑑 (𝑒 + 𝑣))/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = 𝑑𝑒/𝑑π‘₯ + 𝑑𝑣/𝑑π‘₯ 1/𝑒 (𝑑𝑒/𝑑π‘₯)" =" 1 . log (log" " π‘₯) + (1/(log" " π‘₯) .𝑑(log" " π‘₯)/𝑑π‘₯) Γ— π‘₯ 1/𝑒 (𝑑𝑒/𝑑π‘₯)" =" log (log" " π‘₯) + (1/(log" " π‘₯) . 1/π‘₯) Γ— π‘₯ 1/𝑒 (𝑑𝑒/𝑑π‘₯)" =" log (log" " π‘₯) + 1/log⁑π‘₯ Γ— π‘₯/π‘₯ 1/𝑒 (𝑑𝑒/𝑑π‘₯)" =" log (log" " π‘₯) + 1/log⁑π‘₯ 𝑑𝑒/𝑑π‘₯ " =" 𝑒 (log (log" " π‘₯)" + " 1/log⁑π‘₯ ) 𝑑𝑒/𝑑π‘₯ = (log⁑π‘₯ )^π‘₯ (log (log" " π‘₯)" + " 1/log⁑π‘₯ ) 𝑑𝑒/𝑑π‘₯ = (log⁑π‘₯ )^π‘₯ ((log⁑〖π‘₯ . γ€–log 〗⁑(log⁑π‘₯ ) +γ€— 1)/log⁑π‘₯ ) 𝑑𝑒/𝑑π‘₯ = (log⁑π‘₯ )^π‘₯/log⁑π‘₯ (log⁑〖π‘₯ . γ€–log 〗⁑(log⁑π‘₯ )+γ€— 1)

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