How do you find #(d^2y)/(dx^2)# for #5x^2=5y^2+4#?
1 Answer
# d^(2y)/(dx^2) = -4/(5y^3) #
Explanation:
The 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) #
So for
# 10x = 10y dy/dx #
# y dy/dx = x# ..... [1]
To find the second derivative we need to differentiate a second time again implicitly and using the product rule:
# y dy/dx = x#
# :. (y)(d/dxdy/dx) + (d/dxy)(dy/dx) = 1 #
# :. y(d^(2y)/(dx^2)) + (dy/dx)^2 = 1 #
From [1] we have
# :. y(d^(2y)/(dx^2)) + (x/y)^2 = 1 #
# :. yd^(2y)/(dx^2) = 1 - x^2/y^2#
# :. yd^(2y)/(dx^2) =(y^2-x^2)/y^2 #
so we have
# d^(2y)/(dx^2) =(y^2-x^2)/y^3 #
But from the original equation we have;
Hence,
# d^(2y)/(dx^2) =(-4/5)/y^3 = -4/(5y^3) #