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

##### 1 Answer

#### Explanation:

If an 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) #

This will typically result in

So for

# d/dx(2x) + d/dx(y^2) = d/dx(5) #

# :. 2 + dy/dxd/dy(y^2) = 0) #

# :. 2 + dy/dx2y = 0 #

# :. 1 + dy/dxy = 0 #

# :. dy/dx = -1/y # ..... [1]

Differentiating again wrt

# (d^2y)/dx^2 = d/dx(-1/y) #

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

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

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

And substituting from [1] we get:

# (d^2y)/dx^2 = (-1/y)(1/y^2) #

# :. (d^2y)/dx^2 = -1/y^3 #