For some constant k, find dy/dx . (a) e^xy + cos(y^3) = kx

To find \(\frac{dy}{dx}\) for the equation

\[
e^{xy} + \cos(y^3) = kx
\]

we'll differentiate both sides of the equation with respect to \(x\), using implicit differentiation because \(y\) is a function of \(x\). 

### Step 1: Differentiate \( e^{xy} \)
Using the product rule and the chain rule:
\[
\frac{d}{dx}(e^{xy}) = e^{xy} \cdot \frac{d}{dx}(xy) = e^{xy} \cdot \left(y + x\frac{dy}{dx}\right)
\]

### Step 2: Differentiate \( \cos(y^3) \)
Using the chain rule:
\[
\frac{d}{dx}(\cos(y^3)) = -\sin(y^3) \cdot \frac{d}{dx}(y^3) = -\sin(y^3) \cdot 3y^2 \frac{dy}{dx}
\]

### Step 3: Differentiate the right-hand side \(kx\)
\[
\frac{d}{dx}(kx) = k
\]

### Step 4: Combine the results
Putting everything together, we have:
\[
e^{xy} \cdot \left(y + x \frac{dy}{dx}\right) - \sin(y^3) \cdot 3y^2 \frac{dy}{dx} = k
\]

### Step 5: Solve for \(\frac{dy}{dx}\)
Now, isolate \(\frac{dy}{dx}\). Expand the left side first:
\[
e^{xy} \cdot y + e^{xy} \cdot x \frac{dy}{dx} - \sin(y^3) \cdot 3y^2 \frac{dy}{dx} = k
\]
Group the \(\frac{dy}{dx}\) terms:
\[
e^{xy} \cdot x \frac{dy}{dx} - \sin(y^3) \cdot 3y^2 \frac{dy}{dx} = k - e^{xy} \cdot y
\]
Factor out \(\frac{dy}{dx}\):
\[
\frac{dy}{dx} \left(e^{xy} \cdot x - \sin(y^3) \cdot 3y^2\right) = k - e^{xy} \cdot y
\]
Finally, solve for \(\frac{dy}{dx}\):
\[
\frac{dy}{dx} = \frac{k - e^{xy} \cdot y}{e^{xy} \cdot x - \sin(y^3) \cdot 3y^2}
\]

This is the derivative \(\frac{dy}{dx}\).

---

Would you like further details on any step or clarification? Here are five related questions to deepen understanding:

1. How is implicit differentiation applied when both \(x\) and \(y\) are variables?
2. How does the chain rule affect the differentiation of \(y^3\) in \(\cos(y^3)\)?
3. What happens if \(k = 0\) in this equation?
4. How do product and chain rules combine when differentiating \(e^{xy}\)?
5. How can the implicit method extend to second-order derivatives?

**Tip:** Always apply the product rule when differentiating expressions involving the product of two functions of \(x\).

For some constant k, find dy/dx for e^xy + cos(y^3) = kx

Learn how to use implicit differentiation to find dy/dx for the equation e^xy + cos(y^3) = kx. This step-by-step solution applies the product and chain rules, making it ideal for college-level calculus students.

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