Question #ce2e0

1 Answer
Nov 12, 2016

Although we could solve for #y#, this looks like a classic implicit differentiation exercise.

Explanation:

I assume that you want the second derivative of #y# with respect to #x#. (Yes, there are other possibilities.)

#x^2+y^2=90#

Find #dy/dx#

#d/dx(x^2+y^2) = d/dx(90)#

#2x+2y dy/dx = 0#, so

#dy/dx = -x/y#

Differentiate again:

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

Simplify a bit,

# = (-y+xdy/dx)/y^2#

Replace #dy/dx# with its equivalent #-x/y#

# = (-y+x(-x/y))/y^2#

Simplify again

# = (-y-x^2/y)/y^2#

Clear the fraction in the nmumerator

# = ((-y-x^2/y)*y)/(y^2*y)#

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

Factor out a #-1)

# = (-(y^2+x^2))/y^3#

Use the original rfelationship #x^2+y^2 = 90# to finish simplifying.

# = (-90)/y^3#