Question

If ${x^2} + {y^2} + \sin y = 4,$ then the value of $\dfrac{{{d^2}y}}{{d{x^2}}}$ at point (-2, 0) is
$
  {\text{(A) }} - 34 \\
  {\text{(B) }} - 32 \\
  {\text{(C) }} - 2 \\
  {\text{(D) 4}} \\
 $

Answer 1

Hint: First we will differentiate the given equation. Later we will apply the quotient rule to get the desired answer. Finally we will find the value of $\dfrac{{{d^2}y}}{{d{x^2}}}$ at (-2,0).

Complete step-by-step answer:
From the problem, the equation given can be written as,
${x^2} + {y^2} + \sin y = 4,$
Differentiating the above equation with respect to $x$, we get the following expression,
$ \Rightarrow 2x + 2y\dfrac{{dy}}{{dx}} + \cos y\dfrac{{dy}}{{dx}} = 0$
This can be rearranged in the following way,
 $
   \Rightarrow 2x + 2y\dfrac{{dy}}{{dx}} + \cos y\dfrac{{dy}}{{dx}} = 0 \\
   \Rightarrow 2x + \dfrac{{dy}}{{dx}}(2y + \cos y) = 0 \\
   \Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{{2x}}{{2y + \cos y}}{\text{ }}...........................{\text{ (1)}}
 $
As it is difficult to find out the second derivative, we will use the Quotient Rule, which says
When $t = \dfrac{u}{v},$ where $u$ and $v$ are the functions in terms of $x$, then
$\Rightarrow$$\dfrac{{dt}}{{dx}} = \dfrac{{u'v - v'u}}{{{v^2}}}$ ………………………….. (2)
Where,$u' = \dfrac{{du}}{{dx}}{\text{ and }}v' = \dfrac{{dv}}{{dx}}$
Now assume that $\dfrac{{dy}}{{dx}} = t$, then
$\Rightarrow$$\dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{dt}}{{dx}}$ ………………………………….. (3)
Now, from eq. (1), for applying the Quotient Rule, we have
$u = 2x,{\text{ hence, }}u' = \dfrac{{du}}{{dx}} = 2$ …………………………….. (4)
And $v = 2y + \cos y,{\text{ hence, }}v' = \dfrac{{dv}}{{dx}} = 2\dfrac{{dy}}{{dx}} - \sin y\dfrac{{dy}}{{dx}}$ …………………. (5)
 Now from eq. 2, 3, 4, and 5 we get
$\Rightarrow$\[\dfrac{{{d^2}y}}{{d{x^2}}} = - \dfrac{{2(2y + \cos y) - 2x.\left( {2\dfrac{{dy}}{{dx}} - \sin y\dfrac{{dy}}{{dx}}} \right)}}{{{{(2y + \cos y)}^2}}}\] ………………….. (6)
Now, to find the value of $\dfrac{{{d^2}y}}{{d{x^2}}}$ at point (-2, 0), we have to calculate the value of $\dfrac{{dy}}{{dx}}$ from eq. (1). So the value of $\dfrac{{dy}}{{dx}}$ at point (-2, 0) is
$
   \Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{{2( - 2)}}{{2 \times 0 + \cos 0}} \\
   \Rightarrow \dfrac{{dy}}{{dx}} = 4
 $
Now, the value of $\dfrac{{{d^2}y}}{{d{x^2}}}$ at point (-2, 0) is from eq. (6)
\[
  \dfrac{{{d^2}y}}{{d{x^2}}} = - \dfrac{{2(2 \times 0 + \cos 0) - 2( - 2).\left( {2 \times 4 - \sin 0 \times 4} \right)}}{{{{(2 \times 0 + \cos 0)}^2}}} \\
  \dfrac{{{d^2}y}}{{d{x^2}}} = - (2 + 32) = - 34
 \]

So the correct option to the problem (A).

Note: Differentiation of these types of algebraic trigonometric equations, first order can be easy but when we have to find the second order derivative, the situation becomes complex. To overcome this situation, we can use the Quotient Rule.
The Quotient Rule can be written as,
When $t = \dfrac{u}{v},$ where $u$ and $v$ are the functions in terms of $x$, then
$\dfrac{{dt}}{{dx}} = \dfrac{{u'v - v'u}}{{{v^2}}}$ . In this equation,$u' = \dfrac{{du}}{{dx}}{\text{ and }}v' = \dfrac{{dv}}{{dx}}$