How do you find (d^2y)/(dx^2)d2ydx2 for 2x-5y^2=32x5y2=3?

2 Answers
Jul 21, 2017

(d^2y)/dx^2=-1/(25y^3)d2ydx2=125y3

Explanation:

2x-5y^2=3 =>(d(2x))/dx-(d(5y^2))/dx=(d(3))/dx=>2x5y2=3d(2x)dxd(5y2)dx=d(3)dx

2-10ydy/dx=0=>1-5ydy/dx=0 ..(1)210ydydx=015ydydx=0..(1)

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

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

5(dy/dx)^2+5y(d^2y)/dx^2=05(dydx)2+5yd2ydx2=0

Let's substitude from the equation (1)(1) that dy/dx=1/(5y)dydx=15y

So we get :

5(1/(5y))^2+5y(d^2y)/dx^2=0=>1/(5y^2)+5y(d^2y)/dx^2=0=>5(15y)2+5yd2ydx2=015y2+5yd2ydx2=0

5y(d^2y)/dx^2=-1/(5y^2)=>(d^2y)/dx^2=-1/(25y^3)5yd2ydx2=15y2d2ydx2=125y3

Jul 21, 2017

((d^2y)/(dx^2))^2 = 1/5 1/(2x-3)^3 (d2ydx2)2=151(2x3)3

Explanation:

Differentiate both sides of the equation with respect to xx:

d/dx (2x-5y^2) = 0ddx(2x5y2)=0

2-10y dy/dx = 0210ydydx=0

10y dy/dx = 210ydydx=2

(1)" "dy/dx =1/(5y)(1) dydx=15y

and differentiating again:

(d^2y)/(dx^2) = -1/(5y^2)dy/dxd2ydx2=15y2dydx

substitute now dy/dxdydx from (1)(1):

(2) " " (d^2y)/(dx^2) = -1/(25y^3)(2) d2ydx2=125y3

We can also have an implicit equation for (d^2y)/(dx^2)d2ydx2 by squaring both sides of (2)(2):

((d^2y)/(dx^2))^2 = 1/(5^4y^6) = 1/5^4 (1/y^2)^3(d2ydx2)2=154y6=154(1y2)3

and substituting y^2 = (2x-3)/5y2=2x35 from the original equation:

((d^2y)/(dx^2))^2 = 1/5 1/(2x-3)^3 (d2ydx2)2=151(2x3)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]}