use implicit differentiation to find the 2nd derivative:
x^4+y^4=1
(pls do step by step, also when u solve for 1st deriv, you're to plug that answer for dy/dx when finding 2nd derivative)
Sure! Let's start by implicitly differentiating both sides of the given equation with respect to x.
Given: x^4 + y^4 = 1
1. Implicitly differentiate both sides with respect to x
d/dx (x^4) + d/dx (y^4) = d/dx (1)
2. Apply the power rule
4x^3 + 4y^3 * dy/dx = 0
since, d/dx (y^4) = 4y^3 * dy/dx (chain rule)
3. Solve for dy/dx
4y^3 * dy/dx = -4x^3
dy/dx = -4x^3 / 4y^3
dy/dx = -x^3 / y^3
Now, we have found the first derivative. Next, we will find the second derivative by differentiating dy/dx with respect to x again.
dy/dx = -x^3/y^3
4. Differentiate both sides with respect to x
d^2y/dx^2 = d/dx (-x^3/y^3).
5. Apply the quotient rule
d^2y/dx^2 = (y^3 * d/dx(-x^3) - (-x^3) * d/dx (y^3)) / (y^3)^2
6. Apply the chain rule
d^2y/dx^2 = (y^3 * (-3x^2) - (-x^3) * (3y^2 * dy/dx)) / (y^6)
Recall that we found dy/dx earlier:
dy/dx = -x^3 / y^3
7. Substitute this back into the expression for d^2y/dx^2
d^2y/dx^2 = (y^3 * (-3x^2) - (-x^3) * (3y^2 * (-x^3 / y^3))) / (y^6)
8. Simplify the expression
d^2y/dx^2 = (-3x^2y^3 + 3x^6) / y^6
So, the second derivative is:
d^2y/dx^2 = (-3x^2y^3 + 3x^6) / y^6