How do you find the second derivative of #y^2 + x + sin y = 9#?

Question

Paisley Johnson · Accepted Answer

I get: #y'' = (-2+siny)/(2y+cosy)^3# #x+y^2+siny = 9# #d/dx(x+y^2+siny) = d/dx(9)# #1 + 2y dy/dx + cosy dy/dx = 0# #dy/dx = (-1)/(2y+cosy) = -(2y+cosy)^-1# #d/dx(dy/dx) = 1(2y+cosy)^-2*[d/dx(2y+cosy)]# # = (2y+cosy)^-2 [2 dy/dx - siny dy/dx]# # = (2y+cosy)^-2 [2 - siny] dy/dx# # = (2y+cosy)^-2 [2 - siny] [-(2y+cosy)^-1]# # = -(2-siny)(2y+cosy)^-3# # = (-2+siny)/(2y+cosy)^3#

Ezra Adams · Answer

The second derivative is #y''=(sin(y)-2)/(2y+cos(y))^3#

Implicit differentiation gives us

#d/dx(y^2)+d/dx(x)+d/dx(sin(y))=d/dx(9)# #2y dy/dx+1+cos(y) dy/dx=0# #2y dy/dx+cos(y) dy/dx=-1# #dy/dx(2y +cos(y))=-1#. Therefore,

#y'=dy/dx=(-1)/(2y +cos(y))#. This implies that #2y+cos(y)!=0#.

Taking the second implicit derivative gives us #d/dx[dy/dx(2y +cos(y))=-1]#

By application of the Product Rule, we have #d/dx(dy/dx)(2y +cos(y))+dy/dx(2dy/dx-sin(y) dy/dx)=0# #(d^2y)/dx^2(2y +cos(y))+2((dy)/dx)^2-sin(y) ((dy)/dx)^2=0# #(d^2y)/dx^2(2y +cos(y))=sin(y) ((dy)/dx)^2-2((dy)/dx)^2# #(d^2y)/dx^2=(sin(y) ((dy)/dx)^2-2((dy)/dx)^2)/((2y +cos(y)))# or more succinctly

#y''=(y')^2(sin(y)-2)/(2y+cos(y))# #y''=((-1)/(2y +cos(y)))^2(sin(y)-2)/(2y+cos(y))# #y''=1/(2y +cos(y))^2(sin(y)-2)/(2y+cos(y))# #y''=(sin(y)-2)/(2y+cos(y))^3#

Charles Anctil · Answer

undefined To find the second derivative of the given function, you need to follow these steps:

1. Differentiate the given equation implicitly with respect to $x$ to find the first derivative $\frac{dy}{dx}$.
2. Once you have the first derivative, differentiate it again with respect to $x$ to find the second derivative $\frac{d^2y}{dx^2}$.

Given equation: $y^2 + x + \sin(y) = 9$

1. Differentiate implicitly with respect to $x$:
$$ \frac{d}{dx}(y^2) + \frac{d}{dx}(x) + \frac{d}{dx}(\sin(y)) = \frac{d}{dx}(9) $$
$$ 2yy' + 1 + \cos(y) \cdot y' = 0 $$
$$ 2yy' + y' + \cos(y) = 0 $$
$$ y'(2y + 1) = -\cos(y) $$
$$ y' = \frac{-\cos(y)}{2y + 1} $$

2. Now, differentiate $y'$ with respect to $x$ to find the second derivative $\frac{d^2y}{dx^2}$:
$$ \frac{d}{dx}(y') = \frac{d}{dx}\left(\frac{-\cos(y)}{2y + 1}\right) $$
$$ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{-\cos(y)}{2y + 1}\right) $$
$$ \frac{d^2y}{dx^2} = \frac{-(-\sin(y)y' \cdot y' + \cos(y)(2yy'))}{(2y + 1)^2} $$

Substitute $y'$ from the first derivative:
$$ \frac{d^2y}{dx^2} = \frac{-(-\sin(y)\left(\frac{-\cos(y)}{2y + 1}\right) \cdot \left(\frac{-\cos(y)}{2y + 1}\right) + \cos(y)(2y\left(\frac{-\cos(y)}{2y + 1}\right) + 1))}{(2y + 1)^2} $$

Simplify this expression to get the second derivative.