Find
y''
by implicit differentiation.
3x3 + 4y3 = 1
3x^3 + 4y^3 = 1
9x^2 + 12y^2 y' = 0
y' = -3x^2/4y^2
now, using the quotient rule,
y'' = (-24xy^2 + 3x^2(8yy'))/ 16y^4
= (-3xy^2 + 3x^2yy')/2y^4
= -3x (y - xy')/2y^3
= -3x(y - x*(-3x^2/4y^2))/2y^3
= -3x(4y^3 + 3x^3)/8y^5
or, you can use implicit differentiation twice:
9x^2 + 12y^2 y' = 0
18x + 24y (y')^2 + 12y^2 y'' = 0
3x + 4y y'^2 + 2y^2 y'' = 0
2y^2 y'' = -(3x + 4y (y')^2)
y'' = -(3x + 4y (y')^2)/2y^2
= -(3x + 4y * 9x^4/16y^4)/2y^2
= -3x(4y^3 + 3x^3)/8y^5
Alright, let's roll up our sleeves and differentiate away!
To find y'', we need to differentiate our equation twice with respect to x using the chain rule. Strap on your differentiation caps, because things are about to get spicy!
We start by differentiating both sides of the equation with respect to x:
d/dx [3x^3 + 4y^3] = d/dx [1]
Now, differentiating each term separately, we get:
9x^2 + 12y^2 * (dy/dx) = 0
Still with me? Great! Now we tackle the second derivative:
We differentiate once more with respect to x:
d/dx [9x^2 + 12y^2 * (dy/dx)] = d/dx [0]
Differentiating each term one more time, we get:
18x + 12y^2 * (d^2y/dx^2) + 24y * (dy/dx)^2 = 0
And there you have it! The second derivative (y'') of the equation 3x^3 + 4y^3 = 1, obtained through implicit differentiation, is:
18x + 12y^2 * (d^2y/dx^2) + 24y * (dy/dx)^2 = 0
Feel free to hang it on your fridge and show it off to your friends, or just gaze at it admiringly in your spare time!
To find y'', we need to first differentiate the given equation with respect to x using the chain rule. Let's go step by step:
Step 1: Differentiate both sides of the equation with respect to x:
d/dx (3x^3 + 4y^3) = d/dx (1)
Using the chain rule, the derivative of y^3 with respect to x will be 3y^2 * (dy/dx).
Step 2: Differentiate each term on the left-hand side:
d/dx (3x^3) + d/dx (4y^3) = 0
Differentiating 3x^3 gives us: 9x^2
Differentiating 4y^3 gives us: 12y^2 * (dy/dx)
The equation now becomes:
9x^2 + 12y^2 * (dy/dx) = 0
Step 3: Now, we need to differentiate the equation obtained in Step 2 with respect to x once again to find y''.
Differentiating both sides (using the product rule for the second term on the left side), we get:
d/dx (9x^2) + d/dx (12y^2 * (dy/dx)) = 0
Differentiating 9x^2 gives us: 18x
For the second term, we apply the product rule where u = 12y^2 and v = (dy/dx):
d/dx (u * v) = u' * v + u * v'
Differentiating u = 12y^2 gives us: 24y * (dy/dx)
Differentiating v = (dy/dx) gives us: (d^2y/dx^2)
Now we can write the equation as:
18x + 24y * (dy/dx) + 12y^2 * (d^2y/dx^2) = 0
Step 4: Simplify if necessary. In this case, the equation we obtained in Step 3 is the expression for y'' (second derivative):
12y^2 * (d^2y/dx^2) + 24y * (dy/dx) + 18x = 0
Therefore, y'' = -18x/(12y^2 + 24y * (dy/dx))
To find y'', we will perform implicit differentiation on the given equation.
First, let's differentiate both sides of the equation with respect to x:
d/dx (3x^3) + d/dx (4y^3) = d/dx (1)
Using the power rule, the derivatives of x^3 and y^3 are:
9x^2 + 12y^2 * (dy/dx) = 0
Now, to find y'', we need to differentiate the equation we just obtained with respect to x again:
d/dx (9x^2) + d/dx (12y^2 * (dy/dx)) = d/dx (0)
Using the power rule and the product rule, we have:
18x + 12(dy/dx)^2 + 12y^2 * d^2y/dx^2 = 0
Finally, rearrange the equation to solve for y'':
12y^2 * d^2y/dx^2 = -18x - 12(dy/dx)^2
Divide both sides by 12y^2:
d^2y/dx^2 = (-18x - 12(dy/dx)^2) / (12y^2)
So, the value of y'' is given by the equation:
d^2y/dx^2 = (-18x - 12(dy/dx)^2) / (12y^2)