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

1 Answer
Dec 4, 2016

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

Explanation:

If an equation does not express #y# explicitly in terms of #x# as in #y=f(x)#, and instead we have #g(y)=f(x)#, then when we differentiate we apply the chain rule so that we differentiate #g(y)# wrt #y# rather than #x# as in:

# 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 #dy/dx# being a function of #x# and #y#, rather than #dy/dx# being a function of #x# alone as we usually get

So for # 2x + y^2 = 5 # we have:

# 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 #x# we get:

# (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 #