If x2 + y2 + sin y = 4, then the value of [d^2y dx^2] at the point (-2,0) is :

Question

If   x2 + y2 + sin y = 4, then the value of $${{{d^2}y} \over {d{x^2}}}$$ at the point (-2,0) is :

Accepted Answer

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}}}$=$-$

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