Given, x 2 + y 2 + sin y = 4 After differentiating the above equation w.r.t.x we get 2x + 2y ${{dy} \over {dx}}$ + cos y ${{dy} \over {dx}}$ = 0 . . . . (1) $\Rightarrow$ 2x + (2y + cos y) ${{dy} \over {dx}}$ = 0 $\Rightarrow$ ${{dy} \over {dx}}$ = ${{ - 2x} \over {2y + \cos y}}$ At ($-$ 2, 0), ${\left( {{{dy} \over {dx}}} \right)_{\left( { - 2,0} \right)}}$ = ${{ - 2x - 2} \over {2 \times 0 + \cos 0}}$ $\Rightarrow$ ${\left( {{{dy} \over {dx}}} \right)_{\left( { - 2,0} \right)}}$ = ${4 \over {0 + 1}}$ $\Rightarrow$ ${\left( {{{dy} \over {dx}}} \right)_{\left( { - 2,0} \right)}}$ = 4 . . . . .(2) Again differentiating equation (1) w.r.t to x, we get 2 + 2 ${\left( {{{dy} \over {dx}}} \right)^2}$ + 2y${{{d^2}y} \over {d{x^2}}}$ $-$ sin y ${\left( {{{dy} \over {dx}}} \right)^2}$ + cos y ${{{d^2}y} \over {d{x^2}}}$ = 0 $\Rightarrow$ 2 + (2 $-$ sin y) ${\left( {{{dy} \over {dx}}} \right)^2}$ + (2y + cos y)${{{d^2}y} \over {d{x^2}}}$ = 0 $\Rightarrow$ (2y + cos y) ${{{d^2}y} \over {d{x^2}}}$ = $-$ 2 $-$ (2 $-$ sin y)${\left( {{{dy} \over {dx}}} \right)^2}$ $\Rightarrow$ ${{{d^2}y} \over {d{x^2}}}$ = ${{ - 2 - \left( {2 - \sin y} \right){{\left( {{{dy} \over {dx}}} \right)}^2}} \over {2y + \cos y}}$ So, at ($-$ 2, 0), ${{{d^2}y} \over {d{x^2}}}$ = ${{ - 2 - \left( {2 - 0} \right) \times {4^2}} \over {2 \times 0 + 1}}$ $\Rightarrow$ ${{{d^2}y} \over {d{x^2}}}$ = ${{ - 2 - 2 \times 16} \over 1}$ \Rightarrow $$$$\,\,\, ${{{d^2}y} \over {d{x^2}}}$ = $-$ 34