# Given that # x^2 - y^2 =25 # then show #(d^2y)/(dx^2) =-25/y^3 #?

##### 1 Answer

First a point on notation , we *solve* equations but we *derive* (or *show*) expressions.

So we are asked to show that given:

# x^2 - y^2 =25 # ..... [A]

Then

# (d^2y)/(dx^2) = f''(x,y) = -25/y^3 #

We proceed by differentiating the initial equation [A] implicitly:

# d/dx(x^2) - d/dx(y^2) = d/dx(25) #

# :. 2x - 2y dy/dx = 0 #

# :. dy/dx = (2x)/(2y) = x/y # ..... [B]

We now differentiate this second equation [B] implicitly and apply the quotient rule, giving:

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

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

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

# \ \ \ \ \ \ \ = ( y - x(x/y) ) / y^2 \ \ \ # using [B]

# \ \ \ \ \ \ \ = ( (y^2 - x^2) 1/y ) / y^2 #

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

# \ \ \ \ \ \ \ = -25 / y^3 \ \ \ # using [A] QED