Given y^4 -x^4=16 find and simplify d2y/dx2 using implicit differentiation.
I got the first derivative which was x^3/y^3 and no im am stuck on using implicit differentiation for the second derivative part
y^4 - x^4 = 16
4y^3 y' - 4x^3 = 0
y' = x^3/y^3
y'' = (3x^2*y^3 - x^3*3y^2*y')/y^6
= 3x^2(y-xy')/y^4
= 3x^2 (y-x*x^3/y^3)/y^4
= 3x^2 (y^4 - x^4)/y^7
= 48x^2/y^7
or,
y^3 y' = x^3
3y^2 y'^2 + y^3 y'' = 3x^2
3y^2 (x^6/y^6) + y^3 y'' = 3x^2
y^3 y'' = 3x^2 - 3x^6/y^4
y'' = 3x^2 (y^4 - x^4)/y^7
To find the second derivative, we need to differentiate the first derivative we found with respect to x, using implicit differentiation.
Let's start by differentiating the first derivative, which is x^3/y^3, with respect to x. But before we do that, let's rewrite the equation y^4 - x^4 = 16 as y^4 = x^4 + 16.
Now let's differentiate both sides of the equation y^4 = x^4 + 16 with respect to x, using the chain rule.
Differentiating y^4 with respect to x gives us 4y^3 * (dy/dx), and differentiating x^4 + 16 with respect to x gives us 4x^3.
So, we have 4y^3 * (dy/dx) = 4x^3.
Now, let's solve this equation for (dy/dx) by dividing both sides by 4y^3:
(dy/dx) = (x^3) / (y^3).
Now, to find the second derivative d^2y/dx^2, we need to differentiate this expression (dy/dx) with respect to x again.
But before we do that, let's rewrite (dy/dx) = (x^3) / (y^3) as y^3 * (dy/dx) = x^3.
Differentiating both sides with respect to x using the chain rule, we get:
3y^2 * (dy/dx) + y^3 * (d^2y/dx^2) = 3x^2.
Now, we rearrange the equation to solve for (d^2y/dx^2):
y^3 * (d^2y/dx^2) = 3x^2 - 3y^2 * (dy/dx).
Finally, we divide both sides of the equation by y^3 to isolate (d^2y/dx^2):
(d^2y/dx^2) = (3x^2 - 3y^2 * (dy/dx)) / y^3.
Now, we can substitute the expression for (dy/dx) we found earlier, which is (dy/dx) = (x^3) / (y^3), into this equation to simplify further:
(d^2y/dx^2) = (3x^2 - 3y^2 * (x^3) / (y^3)) / y^3.
Simplifying this expression further will give you the final result for the second derivative, (d^2y/dx^2), in terms of x and y.
To find the second derivative using implicit differentiation, we need to differentiate both sides of the equation with respect to x. Let's start with the first derivative you obtained:
dy/dx = x^3/y^3
To find the second derivative (d^2y/dx^2), we will differentiate this expression again with respect to x. However, since y is an implicit function of x, we need to apply the chain rule.
Differentiating both sides, we get:
d/dx(dy/dx) = d/dx(x^3/y^3)
Let's start by differentiating the left side:
d/dx(dy/dx)
To differentiate this expression, we will apply the chain rule. The derivative of dy/dx with respect to x is the second derivative d^2y/dx^2.
So, d/dx(dy/dx) = d^2y/dx^2
Now, let's differentiate the right side using the quotient rule:
d/dx(x^3/y^3) = (x^3)(d/dx(1/y^3)) - (1/y^3)(d/dx(x^3))
Now, let's differentiate each term on the right side.
d/dx(1/y^3)
= -1/y^4 * dy/dx
d/dx(x^3)
= 3x^2
Substituting these derivatives back into the equation, we have:
d^2y/dx^2 = (x^3)(-1/y^4 * dy/dx) - (1/y^3)(3x^2)
= -x^3/y^4 * dy/dx - 3x^2/y^3
Now, we can substitute the value of dy/dx that you found earlier:
= -x^3/y^4 * (x^3/y^3) - 3x^2/y^3
= -x^6/y^7 - 3x^2/y^3
Thus, the second derivative, d^2y/dx^2, is -x^6/y^7 - 3x^2/y^3.