Question

प्रदत्त फलन का x के सापेक्ष अवकलन कीजिए-

(log x)^{log x}, x > 1

Answer 1

माना, y = (log x)^{log x}

दोनों ओर लघुगणक लेने पर,

${\displaystyle \frac{1}{y}\frac{dy}{dx}=(\sin x-\cos x)\frac{d}{dx} \log (\sin x-\cos x)+\log (\sin x-\cos x)\frac{d}{dx}(\sin x-\cos x)}$

${\displaystyle =(\sin x-\cos x)\times \frac{1}{\sin x-\cos x}\frac{d}{dx}(\sin x-\cos x)+\log (\sin x-\cos x)(\cos x+\sin x)}$

${\displaystyle =(\cos x+\sin x)+\log (\sin x-\cos x)(\cos x+\sin x)}$

${\displaystyle =(\cos x+\sin x)[1+\log (\sin x-\cos x)]}$

${\displaystyle \therefore \frac{dy}{dx}=y(\cos x+\sin x)[1+\log (\sin x-\cos x)]}$

${\displaystyle ={(\sin x-\cos x)}^{(\sin x-\cos x)} (\cos x+\sin x)[1+\log (\sin x-\cos x)]}$

Answer 2

माना, y = (log x)^{log x}

दोनों ओर लघुगणक लेने पर,

log y = log x log (log x)

दोनों पक्षों (1) का x के सापेक्ष अवकलन करने पर, हम पाते हैं,

${\displaystyle \frac{1}{y}\frac{dy}{dx}=\log x\cdot \frac{1}{\log x}\cdot \frac{1}{x}+\log \left(\log x\right)\cdot \frac{1}{x}}$

${\displaystyle =\frac{1}{x}\cdot [1+\log \left(\log x\right)]}$

${\displaystyle \frac{dy}{dx}={\left(\log x\right)}^{\log x}\cdot \frac{1}{x}\cdot [1+\log \left(\log x\right)],x>1}$