Use implicit differentiation to find the second derivative y'' of the function given.
x^2 + 5y^3 = 8
x^2 + 5y^3 = 8
2x + 15y^2 y' = 0
y' = -(2x)/(15y^2)
y" = - [(2)(15y^2)-(2x)(30yy')]/225y^4
= (60xyy'-30y^2)/(225y^4)
= (2xyy'-y^2)/(15y^4)
= (2xy'-y)/(15y^3)
= 2(2x(-(2x)/(15y^2))-y)/(15y^3)
= -2(4x^2+15y^3)/(225y^5)
To find the second derivative using implicit differentiation, we need to differentiate both sides of the equation with respect to x.
Starting with the equation: x^2 + 5y^3 = 8
Differentiating both sides with respect to x:
d/dx(x^2) + d/dx(5y^3) = d/dx(8)
Using the power rule for differentiation:
2x + 15y^2 * dy/dx = 0
Now, let's solve for dy/dx:
15y^2 * dy/dx = -2x
dy/dx = -2x / (15y^2)
To find the second derivative, we differentiate dy/dx with respect to x:
d/dx(-2x / (15y^2)) = d/dx(-2x) / (15y^2) = -2 / (15y^2) * (d/dx(x)) = -2 / (15y^2)
Therefore, the second derivative (y'') of the function x^2 + 5y^3 = 8 is -2 / (15y^2).
To find the second derivative y'', we will need to use implicit differentiation twice.
Step 1: Take the derivative of both sides with respect to x using the chain rule.
d/dx(x^2) + d/dx(5y^3) = d/dx(8)
2x + 15y^2 * y' = 0
Step 2: Now take the derivative of both sides again with respect to x (using the product rule on the left side).
d/dx(2x + 15y^2 * y') = d/dx(0)
2 + (30y * y' + 15y^2 * y'') = 0
Step 3: Solve the equation we obtained in Step 2 for y''.
30yy' + 15y^2y'' = -2
15yy' + 15y^2y'' = -2
15y(y' + y'') = -2
y' + y'' = -2 / (15y)
Therefore, the second derivative of the given function is y'' = -2 / (15y) - y'.