Given cosine, left parenthesis, x, y, cubed, right parenthesis, equals, x, cubedcos(xy
3
)=x
3
, find start fraction, d, y, divided by, d, x, end fraction
dx
dy
​
in terms of xx and y, .y.

To solve for \(\frac{dy}{dx}\) in terms of \(x\) and \(y\), we start with the given equation:

\[
\cos(xy^3) = x^3
\]

We need to differentiate both sides with respect to \(x\).

### Step 1: Differentiate the left-hand side

Using the chain rule, differentiate \(\cos(xy^3)\) with respect to \(x\):

\[
\frac{d}{dx} \left[\cos(xy^3)\right] = -\sin(xy^3) \cdot \frac{d}{dx}(xy^3)
\]

Now, apply the product rule to differentiate \(xy^3\) with respect to \(x\):

\[
\frac{d}{dx}(xy^3) = y^3 \cdot \frac{d}{dx}(x) + x \cdot \frac{d}{dx}(y^3)
\]

Since \(\frac{d}{dx}(x) = 1\) and using the chain rule on \(y^3\):

\[
\frac{d}{dx}(xy^3) = y^3 + 3xy^2 \cdot \frac{dy}{dx}
\]

Thus, the derivative of the left-hand side is:

\[
-\sin(xy^3) \cdot (y^3 + 3xy^2 \cdot \frac{dy}{dx})
\]

### Step 2: Differentiate the right-hand side

Differentiate \(x^3\) with respect to \(x\):

\[
\frac{d}{dx}(x^3) = 3x^2
\]

### Step 3: Set the derivatives equal

Now, equate the derivatives of both sides:

\[
-\sin(xy^3) \cdot (y^3 + 3xy^2 \cdot \frac{dy}{dx}) = 3x^2
\]

### Step 4: Solve for \(\frac{dy}{dx}\)

Distribute the \(-\sin(xy^3)\) on the left-hand side:

\[
-\sin(xy^3) \cdot y^3 - 3xy^2 \cdot \sin(xy^3) \cdot \frac{dy}{dx} = 3x^2
\]

Now, isolate the term involving \(\frac{dy}{dx}\):

\[
-3xy^2 \sin(xy^3) \cdot \frac{dy}{dx} = 3x^2 + \sin(xy^3) \cdot y^3
\]

Divide both sides by \(-3xy^2 \sin(xy^3)\) to solve for \(\frac{dy}{dx}\):

\[
\frac{dy}{dx} = \frac{3x^2 + \sin(xy^3) \cdot y^3}{-3xy^2 \sin(xy^3)}
\]

We can simplify this by factoring out \(\sin(xy^3)\):

\[
\frac{dy}{dx} = -\frac{x^2}{xy^2 \sin(xy^3)} - \frac{y^3}{3xy^2}
\]

Thus, the derivative \(\frac{dy}{dx}\) in terms of \(x\) and \(y\) is:

\[
\frac{dy}{dx} = -\frac{x^2}{xy^2 \sin(xy^3)} - \frac{y}{3x}
\]

Would you like more details or have any questions? 

Here are five related questions you might find interesting:
1. How would the solution change if the equation were \(\sin(xy^3) = x^3\)?
2. What if the given function were \(\cos(xy^2) = x^3\) instead of \(\cos(xy^3) = x^3\)?
3. How would you approach this problem if \(y\) were a function of another variable, say \(z\), instead of \(x\)?
4. How does the chain rule apply when differentiating functions of multiple variables?
5. What is the geometric interpretation of \(\frac{dy}{dx}\) in this context?

**Tip:** When differentiating implicit functions, always keep track of when to apply the chain rule and the product rule to avoid errors in the differentiation process.

Learn how to solve for dy/dx in terms of x and y from the equation cos(xy^3) = x^3 using differentiation. This problem involves applying the chain rule and product rule to find the derivative. Suitable for college-level mathematics students.

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