If 2, minus, 5, y, cubed, minus, x, y, equals, x, squared2−5*y*3−*xy*=*x*2 then find start fraction, d, y, divided by, d, x, end fraction*dxdy*​ in terms of x*x* and y, .*y*.

We are given the equation:

\[
2 - 5y^3 - xy = x^2
\]

We need to find \(\frac{dy}{dx}\) in terms of \(x\) and \(y\). To do this, we'll differentiate both sides of the equation implicitly with respect to \(x\), treating \(y\) as a function of \(x\).

### Step 1: Differentiate each term with respect to \(x\)

- For the constant \(2\), its derivative is 0.

- For \(-5y^3\), use the chain rule:
  \[
  \frac{d}{dx}(-5y^3) = -5 \cdot 3y^2 \cdot \frac{dy}{dx} = -15y^2 \cdot \frac{dy}{dx}
  \]

- For \(-xy\), apply the product rule. The derivative of \(xy\) is:
  \[
  \frac{d}{dx}(-xy) = -\left(\frac{d}{dx}(x \cdot y)\right) = -(y + x \cdot \frac{dy}{dx})
  \]

- For \(x^2\), the derivative is:
  \[
  \frac{d}{dx}(x^2) = 2x
  \]

### Step 2: Write the differentiated equation

Now, differentiating both sides of the equation:

\[
0 - 15y^2 \cdot \frac{dy}{dx} - (y + x \cdot \frac{dy}{dx}) = 2x
\]

Simplify this to:

\[
-15y^2 \cdot \frac{dy}{dx} - y - x \cdot \frac{dy}{dx} = 2x
\]

### Step 3: Collect \(\frac{dy}{dx}\) terms

Move the terms involving \(\frac{dy}{dx}\) to one side of the equation:

\[
-15y^2 \cdot \frac{dy}{dx} - x \cdot \frac{dy}{dx} = 2x + y
\]

Factor out \(\frac{dy}{dx}\):

\[
\frac{dy}{dx}(-15y^2 - x) = 2x + y
\]

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

Now, divide both sides by \((-15y^2 - x)\):

\[
\frac{dy}{dx} = \frac{2x + y}{-15y^2 - x}
\]

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

\[
\frac{dy}{dx} = \frac{2x + y}{-15y^2 - x}
\]

---

Would you like more details on any of these steps or have any questions?

Here are 5 related questions:
1. How does implicit differentiation differ from explicit differentiation?
2. What is the product rule and how is it applied in calculus?
3. How can we interpret the physical meaning of \(\frac{dy}{dx}\) in a real-world context?
4. Could we solve for \(y\) explicitly from the original equation instead of using implicit differentiation?
5. What happens if we change the powers of \(y\) or \(x\) in the original equation?

**Tip**: When using the product rule in differentiation, remember that you must differentiate both the first and second functions in the product separately.

If 2 − 5*y^3 − xy = x^2 then find dy/dx in terms of x and y.

Learn how to solve the implicit differentiation problem 2 - 5*y^3 - xy = x^2 and find dy/dx in terms of x and y. This detailed solution covers calculus concepts such as the product rule, chain rule, and implicit differentiation. Suitable for Grades 11-12.

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