Use algebraic techniques to rewrite y=x√(5x12−3x52−6)
 as a sum or difference; then find y′
.



To rewrite \( y = x \sqrt{5x^{12} - 3x^{5/2} - 6} \) as a sum or difference, we'll first simplify the expression inside the square root, and then differentiate it using the chain rule and product rule.

### Step 1: Simplify the Expression

Given:

\[
y = x \sqrt{5x^{12} - 3x^{5/2} - 6}
\]

There isn't much simplification we can do to rewrite it as a sum or difference since the terms under the square root are already distinct powers of \(x\).

Thus, the simplified form of the expression is:

\[
y = x \sqrt{5x^{12} - 3x^{5/2} - 6}
\]

### Step 2: Differentiate \( y \)

To differentiate this expression, we apply the **product rule** because the function is a product of \(x\) and the square root term.

The product rule is:

\[
\frac{d}{dx}[u \cdot v] = u'v + uv'
\]

Here, \( u = x \) and \( v = \sqrt{5x^{12} - 3x^{5/2} - 6} \).

#### Differentiate \( u = x \):

\[
u' = 1
\]

#### Differentiate \( v = \sqrt{5x^{12} - 3x^{5/2} - 6} \):

To differentiate \( v \), we use the chain rule. Recall that \( v = (5x^{12} - 3x^{5/2} - 6)^{1/2} \), so:

\[
v' = \frac{1}{2}(5x^{12} - 3x^{5/2} - 6)^{-1/2} \cdot \frac{d}{dx}(5x^{12} - 3x^{5/2} - 6)
\]

Now, differentiate the inside expression \( 5x^{12} - 3x^{5/2} - 6 \):

\[
\frac{d}{dx}(5x^{12} - 3x^{5/2} - 6) = 60x^{11} - \frac{15}{2}x^{3/2}
\]

Thus, we get:

\[
v' = \frac{1}{2}(5x^{12} - 3x^{5/2} - 6)^{-1/2} \cdot \left( 60x^{11} - \frac{15}{2}x^{3/2} \right)
\]

### Step 3: Apply the Product Rule

Now we combine everything using the product rule:

\[
y' = u'v + uv'
\]

Substituting \( u' = 1 \), \( u = x \), \( v = \sqrt{5x^{12} - 3x^{5/2} - 6} \), and \( v' \) from the chain rule:

\[
y' = \sqrt{5x^{12} - 3x^{5/2} - 6} + x \cdot \frac{1}{2}(5x^{12} - 3x^{5/2} - 6)^{-1/2} \cdot \left( 60x^{11} - \frac{15}{2}x^{3/2} \right)
\]

This is the derivative \( y' \).

---

Would you like a more detailed breakdown or further assistance with this? 

Here are 5 related questions:
1. How do you apply the chain rule in calculus?
2. What are the key steps to simplify complex algebraic expressions?
3. Can you explain the product rule in differentiation with examples?
4. How do you differentiate expressions involving square roots?
5. How does the chain rule work with nested functions?

**Tip:** When differentiating a product of functions, always identify the two functions separately and apply the product rule carefully.

Use algebraic techniques to rewrite y=x√(5x12−3x52−6) as a sum or difference; then find y′.

Learn how to rewrite and differentiate the expression y = x√(5x^12 − 3x^(5/2) − 6) using algebraic techniques, the product rule, and the chain rule. This calculus problem involves simplifying square root functions and applying advanced differentiation methods, suitable for Grades 11-12 or college calculus students.

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