https://r2.erweima.ai/mathbot/image/1724793321142cnst68u9.jpg

The expression provided in the image is:

\[
\frac{6x^2 - 1}{2\sqrt{x}}
\]

To differentiate this expression with respect to \( x \), we'll apply the quotient rule. The quotient rule states:

\[
\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot u' - u \cdot v'}{v^2}
\]

Here:
- \( u = 6x^2 - 1 \)
- \( v = 2\sqrt{x} \)

First, we find the derivatives of \( u \) and \( v \):

1. \( u' = \frac{d}{dx}(6x^2 - 1) = 12x \)
2. \( v' = \frac{d}{dx}(2\sqrt{x}) \)

Since \( 2\sqrt{x} \) can be rewritten as \( 2x^{1/2} \):

\[
v' = 2 \cdot \frac{1}{2} \cdot x^{-1/2} = \frac{1}{\sqrt{x}}
\]

Now, applying the quotient rule:

\[
\frac{d}{dx}\left(\frac{6x^2 - 1}{2\sqrt{x}}\right) = \frac{2\sqrt{x} \cdot 12x - (6x^2 - 1) \cdot \frac{1}{\sqrt{x}}}{(2\sqrt{x})^2}
\]

Simplify the expression:

\[
= \frac{24x\sqrt{x} - \frac{6x^2 - 1}{\sqrt{x}}}{4x}
\]

This can be simplified further by combining like terms:

\[
= \frac{24x\sqrt{x} - \frac{6x^2}{\sqrt{x}} + \frac{1}{\sqrt{x}}}{4x}
\]

\[
= \frac{24x\sqrt{x} - 6x^{3/2} + x^{-1/2}}{4x}
\]

Dividing each term by \( 4x \):

\[
= \frac{24x\sqrt{x}}{4x} - \frac{6x^{3/2}}{4x} + \frac{x^{-1/2}}{4x}
\]

\[
= 6\sqrt{x} - \frac{3\sqrt{x}}{2} + \frac{1}{4x\sqrt{x}}
\]

This is the derivative of the given expression with respect to \( x \).

Would you like further details or have any questions?

Here are some related questions:
1. How does the quotient rule work in differentiation?
2. What is the derivative of \( \sqrt{x} \) with respect to \( x \)?
3. How can you simplify expressions involving radicals during differentiation?
4. What are the steps to applying the chain rule in differentiation?
5. Can the expression be simplified before applying the quotient rule?

**Tip:** When simplifying derivatives, always look for common factors in the numerator and denominator to reduce complexity.

Learn how to differentiate the expression 6x^2 - 1 / 2sqrt(x) using the quotient rule in calculus. This detailed step-by-step solution covers differentiation techniques involving radicals and algebraic expressions. Suitable for college-level students and anyone studying calculus.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation