QJEEM18Ma1bdf: JEE Main 2018 Mathematics - Class 12 JEEM18Ma1bdf Solution: If   x 2 + y 2 + sin y = 4, then the value of at the…

Given, x² + y² + 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