What is the result of implicit differentiation of the equation 8x^4 + 3y^4 = 32?

What is the result of implicit differentiation of the equation 8x^4 + 3y^4 = 32?

• A. (-8x^3)/(3y^3)
• B. (8x^3)/(3y^3)
• C. (-3y^3)/(8x^3)
• D. (3y^3)/(8x^3)

Answer: (A)

To find the derivative using implicit differentiation from the equation \(8x^4 + 3y^4 = 32\), we start by differentiating both sides with respect to \(x\). Differentiating the left side: 1. The term \(8x^4\) differentiates to \(32x^3\). 2. The term \(3y^4\) requires the chain rule, which gives us \(12y^3 \frac{dy}{dx}\). So, the derivative of the left side is: \[ 32x^3 + 12y^3 \frac{dy}{dx} \] The right side, which is a constant (32), differentiates to 0. Setting these equal, we get: \[ 32x^3 + 12y^3 \frac{dy}{dx} = 0 \] Next, we solve for \(\frac{dy}{dx}\): \[ 12y^3 \frac{dy}{dx} = -32x^3 \] Then, dividing both sides by \(12y^3\) results in: \[ \frac{dy}{dx} = -\frac{32x

To find the derivative using implicit differentiation from the equation (8x^4 + 3y^4 = 32), we start by differentiating both sides with respect to (x).

Differentiating the left side:

1. The term (8x^4) differentiates to (32x^3).

2. The term (3y^4) requires the chain rule, which gives us (12y^3 \frac{dy}{dx}).

So, the derivative of the left side is:

[

32x^3 + 12y^3 \frac{dy}{dx}

]

The right side, which is a constant (32), differentiates to 0.

Setting these equal, we get:

[

32x^3 + 12y^3 \frac{dy}{dx} = 0

]

Next, we solve for (\frac{dy}{dx}):

[

12y^3 \frac{dy}{dx} = -32x^3

]

Then, dividing both sides by (12y^3) results in:

[

\frac{dy}{dx} = -\frac{32x