How do you find (d^2y)/(dx^2) for 2x-5y^2=3?

2 Answers
Jul 21, 2017

(d^2y)/dx^2=-1/(25y^3)

Explanation:

2x-5y^2=3 =>(d(2x))/dx-(d(5y^2))/dx=(d(3))/dx=>

2-10ydy/dx=0=>1-5ydy/dx=0 ..(1)

(d(1))/dx-(d(5ydy/dx))/dx=0=>-(d(5ydy/dx))/dx=0=>

(d(5y))/dxdy/dx+5y(d(dy/dx))/dx=0=>

5(dy/dx)^2+5y(d^2y)/dx^2=0

Let's substitude from the equation (1) that dy/dx=1/(5y)

So we get :

5(1/(5y))^2+5y(d^2y)/dx^2=0=>1/(5y^2)+5y(d^2y)/dx^2=0=>

5y(d^2y)/dx^2=-1/(5y^2)=>(d^2y)/dx^2=-1/(25y^3)

Jul 21, 2017

((d^2y)/(dx^2))^2 = 1/5 1/(2x-3)^3

Explanation:

Differentiate both sides of the equation with respect to x:

d/dx (2x-5y^2) = 0

2-10y dy/dx = 0

10y dy/dx = 2

(1)" "dy/dx =1/(5y)

and differentiating again:

(d^2y)/(dx^2) = -1/(5y^2)dy/dx

substitute now dy/dx from (1):

(2) " " (d^2y)/(dx^2) = -1/(25y^3)

We can also have an implicit equation for (d^2y)/(dx^2) by squaring both sides of (2):

((d^2y)/(dx^2))^2 = 1/(5^4y^6) = 1/5^4 (1/y^2)^3

and substituting y^2 = (2x-3)/5 from the original equation:

((d^2y)/(dx^2))^2 = 1/5 1/(2x-3)^3

Function:

graph{y^2 = (2x-3)/5 [-10, 10, -5, 5]}

Derivative:

graph{y^2 = 1/5 1/(2x-3)^3 [-10, 10, -5, 5]}