How do you find #(d^2y)/(dx^2)# for #5x^2=5y^2+4#?

Question

Luke Alvarez · Accepted Answer

# d^(2y)/(dx^2) = -4/(5y^3) #

The equation does not express #y# explicitly in terms of #x# as in #y=f(x)#, instead we have #g(y)=f(x)#, so 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) # So for # 5x^2 = 5y^2 + 4 # we have: # 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 #dy/dx=x/y# so substituting gives: # :. 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; # 5x^2 = 5y^2 + 4 => 5(y^2 - x^2) = -4 => y^2 - x^2 = -4/5 # Hence, # d^(2y)/(dx^2) =(-4/5)/y^3 = -4/(5y^3) #

Adam Adkins · Answer

undefined To find $\frac{{d^2y}}{{dx^2}}$ for $5x^2 = 5y^2 + 4$, first, rewrite the equation in terms of $y$. Then differentiate implicitly twice with respect to $x$. After differentiation, solve for $\frac{{d^2y}}{{dx^2}}$. The solution involves the chain rule and implicit differentiation.