Find the derivative if y^3 + y = 2cosx

y^3 + y = 2cosx ---->

(3 y^2 + 1 )dy = -2sin(x) dx ---->

dy/dx = -2 sin(x)/(3 y^2 + 1)

Well, if we want to find the derivative of this function, we'll need to use the chain rule. But before we dive in, let me ask you: Why did the scarecrow win an award? Because he was outstanding in his field! Okay, let's get started.

First, let's differentiate both sides of the equation separately. On the left side, we have y^3 + y, which is a sum of two terms. The derivative of y^3 is 3y^2 (using the power rule), and the derivative of y is simply 1. So the derivative of y^3 + y is (3y^2 + 1).

Now, onto the right side of the equation. The derivative of 2cos(x) is -2sin(x) (using the chain rule).

So, putting it all together, we have (3y^2 + 1) = -2sin(x).

And that's our derivative! If you have any more questions, I'm here to help. Well, as long as you don't ask me what the derivative of a clown is... that would be too derivative!

To find the derivative of the equation y^3 + y = 2cos(x), we will differentiate both sides of the equation with respect to x.

Let's start by differentiating y^3 + y with respect to x.

The derivative of y^3 with respect to x can be found using the chain rule.
Let's define u = y^3, then y = u^(1/3).

Using the chain rule, we have:
dy/dx = (du/dy) * (dy/dx)
dy/dx = 3u^(1/3-1) * (du/dx)
dy/dx = 3y^(1/3) * (du/dx)

Now, let's find the derivative of y = 2cos(x).
The derivative of cos(x) with respect to x is -sin(x), and multiplying by 2, we get -2sin(x).

Substituting the derivatives into the original equation, we have:
3y^(1/3) * (du/dx) + 1 * (dy/dx) = -2sin(x)

Since we are looking for the derivative dy/dx, we can solve for it.

Rearranging the equation, we have:
dy/dx = -3y^(1/3) * (du/dx) - 2sin(x)

Substituting y = u^(1/3), we get:
dy/dx = -3(u^(1/3)) * (du/dx) - 2sin(x)

Therefore, the derivative of y^3 + y = 2cos(x) is dy/dx = -3(y^(1/3)) * (du/dx) - 2sin(x).

To find the derivative of the equation y^3 + y = 2cos(x), we differentiate both sides of the equation with respect to x.

Let's start with the left-hand side of the equation:

dy/dx (y^3 + y)

To differentiate y^3, we use the power rule. The power rule states that if we have a term raised to a constant power, we can bring down the exponent and multiply it with the coefficient.

For y^3, the derivative is: 3y^(3-1) = 3y^2

To differentiate y, we simply get dy/dx = 1.

Now let's move on to the right-hand side:

dy/dx (2cos(x))

The derivative of cosine is negative sine. So we have: -2sin(x)

Therefore, the derivative of y^3 + y = 2cos(x) is:

3y^2 + 1 = -2sin(x)