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

$ \Rightarrow \,\frac{{{\text{dy}}}}{{{\text{dx}}}}\,\, = \,\,(1\, - \,{x^2})\, - \,2{x^2}\, = \,1\, - \,3{x^2}$

અને $\frac{{{{\text{d}}^{\text{2}}}y}}{{d{x^2}}}\,\, = \,\, - 6x$

હવે $\frac{{{\text{dy}}}}{{{\text{dx}}}}\, = \,\,0\,\, \Rightarrow \,x\,\, = \,\, \pm \,\,\frac{1}{{\sqrt 3 }}$

હવે ${\text{x}}\, = \,\,\frac{{\text{1}}}{{\sqrt {\text{3}} }}\,\,$ આગળ $\frac{{{{\text{d}}^{\text{2}}}y}}{{d{x^2}}}\,\, < \,0\,$ તેથી $\,{\text{x}}\, = \,\,\frac{{\text{1}}}{{\sqrt {\text{3}} }}$ આગળ ${\text{y}}$ મહતમ છે.