https://file.aiquickdraw.com/mathbot/file/17288617912275v0g3tez.jpg

We are given the equation:

\[
\sqrt{x + y} = x^4 + y^4
\]

We need to find \(\frac{dy}{dx}\) using implicit differentiation.

### Step 1: Differentiate both sides with respect to \(x\)

- The left side contains a square root term, so we can rewrite it as \((x + y)^{1/2}\) and apply the chain rule.
- The right side contains \(x^4 + y^4\), so we'll differentiate both terms separately.

#### Left-hand side:

\[
\frac{d}{dx}\left( (x + y)^{1/2} \right)
\]

Using the chain rule:

\[
\frac{1}{2}(x + y)^{-1/2} \cdot \left( \frac{d}{dx}(x + y) \right)
\]

Now, differentiate \(x + y\) with respect to \(x\):

\[
\frac{d}{dx}(x + y) = 1 + \frac{dy}{dx}
\]

So, the left-hand side becomes:

\[
\frac{1}{2}(x + y)^{-1/2} \cdot (1 + \frac{dy}{dx})
\]

#### Right-hand side:

\[
\frac{d}{dx}(x^4 + y^4)
\]

Differentiate both terms:

\[
\frac{d}{dx}(x^4) = 4x^3
\]
\[
\frac{d}{dx}(y^4) = 4y^3 \cdot \frac{dy}{dx}
\]

Thus, the right-hand side becomes:

\[
4x^3 + 4y^3 \cdot \frac{dy}{dx}
\]

### Step 2: Set up the full equation

Now we combine the derivatives:

\[
\frac{1}{2}(x + y)^{-1/2} \cdot (1 + \frac{dy}{dx}) = 4x^3 + 4y^3 \cdot \frac{dy}{dx}
\]

### Step 3: Solve for \(\frac{dy}{dx}\)

We now need to solve this equation for \(\frac{dy}{dx}\).

First, distribute the term on the left-hand side:

\[
\frac{1}{2}(x + y)^{-1/2} + \frac{1}{2}(x + y)^{-1/2} \cdot \frac{dy}{dx} = 4x^3 + 4y^3 \cdot \frac{dy}{dx}
\]

Now, move all terms involving \(\frac{dy}{dx}\) to one side:

\[
\frac{1}{2}(x + y)^{-1/2} \cdot \frac{dy}{dx} - 4y^3 \cdot \frac{dy}{dx} = 4x^3 - \frac{1}{2}(x + y)^{-1/2}
\]

Factor out \(\frac{dy}{dx}\) on the left-hand side:

\[
\frac{dy}{dx} \left( \frac{1}{2}(x + y)^{-1/2} - 4y^3 \right) = 4x^3 - \frac{1}{2}(x + y)^{-1/2}
\]

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

\[
\frac{dy}{dx} = \frac{4x^3 - \frac{1}{2}(x + y)^{-1/2}}{\frac{1}{2}(x + y)^{-1/2} - 4y^3}
\]

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

---

Would you like more details or further clarifications? 

Here are some related questions:

1. What does implicit differentiation mean, and when is it used?
2. How do you apply the chain rule in implicit differentiation?
3. Why do we need to move terms involving \(\frac{dy}{dx}\) to one side?
4. What happens if \(y\) is a constant with respect to \(x\)?
5. How can you interpret the result of \(\frac{dy}{dx}\) in this context?

**Tip**: When working with implicit differentiation, remember to always treat \(y\) as a function of \(x\), even if it isn't explicitly solved for.

Find dy/dx by implicit differentiation for the equation sqrt(x + y) = x^4 + y^4.

Learn how to find dy/dx using implicit differentiation for the equation sqrt(x + y) = x^4 + y^4. This step-by-step solution explains key calculus concepts such as the chain rule, power rule, and implicit differentiation, making it ideal for students in Grades 11-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation