# How do you find #(d^2y)/(dx^2)# for #5x^2=5y^2+4#?

##### 1 Answer

# d^(2y)/(dx^2) = -4/(5y^3) #

#### Explanation:

The equation does not express

# g(y) = f(x) #

# :. d/dx g(y) = d/dx f(x) #

# :. dy/dx d/dy g(y) = f'(x) #

# :. g'(y) dy/dx= f'(x) #

So for

# 10x = 10y dy/dx #

# y dy/dx = x# ..... [1]

To find the second derivative we need to differentiate a second time again implicitly and using the product rule:

# y dy/dx = x#

# :. (y)(d/dxdy/dx) + (d/dxy)(dy/dx) = 1 #

# :. y(d^(2y)/(dx^2)) + (dy/dx)^2 = 1 #

From [1] we have

# :. y(d^(2y)/(dx^2)) + (x/y)^2 = 1 #

# :. yd^(2y)/(dx^2) = 1 - x^2/y^2#

# :. yd^(2y)/(dx^2) =(y^2-x^2)/y^2 #

so we have

# d^(2y)/(dx^2) =(y^2-x^2)/y^3 #

But from the original equation we have;

Hence,

# d^(2y)/(dx^2) =(-4/5)/y^3 = -4/(5y^3) #