6. Application of derivatives — ગણિત ધોરણ 12 સાયન્સ — Question

હવે $\frac{{{\text{du}}}}{{{\text{dx}}}}\,\, = \,\,(2{x^2} - 2x - 1)\,2\sin x\cos x\, + \,\,(4x\, - \,2)\,{\sin ^2}x$

$ = \sin x\,[2(2{x^2} - 2x - 1)\,\cos x\, + \,(4x - 2)\,\sin x]$

$\frac{{du}}{{dx}}\,\, = \,\,0\,\, \Rightarrow \,\sin x\,\, = \,0\,\, \Rightarrow \,x\, = \,\,n\pi $

$\frac{{{d^2}u}}{{dx^2}}\,\, = \,\,\sin x\,\frac{d}{{dx}}\,[2(2{x^2} - 2x - 1)\,\cos x\, + \,\,(4x - 2)\sin x]\, + \,\,$

$\cos x[2\cos x(2{x^2} - 2x - 1) + (4x - 2)\,\sin x]$

${\text{x}}\, = \,\,{\text{n}}\pi $ આગળ $\frac{{{{\text{d}}^{\text{2}}}u}}{{d{x^2}}}\,\, = \,\,0\, + \,2\,{\cos ^2}n\pi \,(2{n^2}{\pi ^2} - 2n\pi  - 1) > 0$

આથી, $\,{\text{x}}\,\, = \,\,{\text{n}}\pi $ આગળ ${\text{u}}$ ની કિમત અને તેથી તેને અનુરૂપ ${\text{ y}}$ ની ન્યૂનતમ કે મહતમ કિમંત $ = \,\,{{\text{e}}^{\text{0}}}\,\, = \,\,1$