Question

If x^2+y^2+Siny =4. Then the value of d^2y/dx^2 at the point (-2,0) is

Answer 1

The Question is
$x^{2} + y^{2} + siny =4$

Differentiating with respect to x

=> 2x + 2yy' + y'cosy = 0 ..................(1)

Differentiating again w.r.t. x
=> 2 + 2( yy'' + (y')²) + (y''cosy - y'siny)=0;

At (-2,0), x=-2 and y=0
using (1) y' at (-2,0) is 4
=> 2+ 2(16) + (y'')=0
=> y''= -34

Answer 2

The given equation is $x^{2} + y^{2} + sin y=4$.

Differentiating both sides with respect to x,
we get,

$2x + 2y \frac{dy}{dx} + cos y \frac{dy}{dx}= 0$
⇒$\frac{dy}{dx}= \frac{-2x}{2y + cos y}$

Differentiating that with respect to x, we get,
$\frac{ d^{2}y }{dx^{2}} = \frac{(2y+cos y)(-2) + 2x(2\frac{dy}{dx} - sin y \frac{dy}{dx})}{ (2y + cos y)^{2} }$
⇒$\frac{ d^{2}y }{dx^{2}} = \frac{(2y+cos y)(-4y+2cos y) - 8x^{2} -4 x^{2} sin y }{(2y+cos y)^3}$

At (-2, 0), the value of $\frac{d^2y}{dx^2} = \frac{(1)(2)-8(-2)^2-0}{1} = 2-8 = -6$

Hence, ANSWER = -6.

[ANSWERED]