If #(dx)/(dy)=u# and #(du)/(dy)=v#, what is #(d^2y)/(dx^2)#?

2 Answers
Shwetank Mauria
Feb 1, 2018

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

Explanation:

As #(dx)/(dy)=u#, #(dy)/(dx)=1/u#

Further, #(d^2x)/(dy^2)=d/(dy)(dx)/(dy)=(du)/(dy)=v#

therefore #(dy)/(du)=1/v# and #(dy)/(dx)=(dy)/(du)*(du)/(dx)#

i.e. #1/u=1/v*(du)/(dx)# and #(du)/(dx)=v/u#

As #(dy)/(dx)=1/u#

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

= #-1/u^2(du)/(dx)=-1/u^2*v/u=-v/u^3#

Steve M
Feb 1, 2018

The quoted result is incorrect:

We have the corrected result:

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

Explanation:

The quoted result is incorrect:

If we start with a basic property of the chain rule, we can write

# dy/dx dx/dy = 1 => dy/dx = 1/(dx/dy) #

Leading to:

# dy/dx = 1/u #

Now, we have:

# dx/dy = u #

And then if we differentiate #(dy/dx)# wrt #x# and apply the chain rule we can write:

# (d^2y)/(dx^2) = d/dx (dy/dx) #
# \ \ \ \ \ \ \ = (dy/dx) d/dy (dy/dx) \ \ \ \ \ \ # (chain rule)
# \ \ \ \ \ \ \ = 1/u \ d/dy (1/u) #
# \ \ \ \ \ \ \ = 1/u \ (du)/(dy) \ d/(du) (1/u) \ \ \ \ \ \ # (chain rule)
# \ \ \ \ \ \ \ = 1/u \ (du)/(dy) \ (-1/u^2) \ \ \ \ \ \ (d/dx x^n=nx^(n-1))#
# \ \ \ \ \ \ \ = -1/u^3 \ (du)/(dy) #

# \ \ \ \ \ \ \ = -1/u^3 \ (d)/(dy)(u) #

# \ \ \ \ \ \ \ = -1/u^3 \ (d)/(dy) (dx/dy) \ \ \ \ \ \ # (using def'n of #u#)

# \ \ \ \ \ \ \ = -1/u^3 \ (d^2x)/(dy^2) #
# \ \ \ \ \ \ \ = -1/u^3 \ v \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ # (using def'n of #v#)

Hence we have the corrected result:

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

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