Find the slope of the tangent line to the curve y=√1+x³ at the given point (2, 3).

To find the slope of the tangent line to the curve \( y = \sqrt{1 + x^3} \) at the point \( (2, 3) \), we need to compute the derivative of the function \( y = \sqrt{1 + x^3} \) and then evaluate the derivative at \( x = 2 \).

### Step 1: Differentiate \( y = \sqrt{1 + x^3} \)
We can rewrite the equation as:
\[
y = (1 + x^3)^{1/2}
\]
Now, we differentiate \( y \) with respect to \( x \) using the chain rule.

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

Simplifying:
\[
\frac{dy}{dx} = \frac{3x^2}{2\sqrt{1 + x^3}}
\]

### Step 2: Evaluate the derivative at \( x = 2 \)
Now, substitute \( x = 2 \) into the derivative to find the slope of the tangent line at this point.

\[
\frac{dy}{dx} \bigg|_{x=2} = \frac{3(2)^2}{2\sqrt{1 + (2)^3}} = \frac{3 \times 4}{2\sqrt{1 + 8}} = \frac{12}{2\sqrt{9}} = \frac{12}{6} = 2
\]

### Final Answer:
The slope of the tangent line to the curve at the point \( (2, 3) \) is \( 2 \).

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Do you want more details or have any questions? Here are 5 related questions:

1. How do you find the equation of the tangent line using the point-slope form?
2. What is the geometric interpretation of the derivative in terms of the slope of a curve?
3. How does the chain rule apply when differentiating composite functions?
4. What are other ways to find the slope of a tangent line without calculus?
5. How would the slope change if the function was \( y = \sqrt{1 + x^4} \) instead?

**Tip:** When differentiating square roots, rewriting them as fractional exponents makes it easier to apply the chain rule!

Learn how to find the slope of the tangent line to the curve y = √1 + x³ at the given point (2, 3) using calculus concepts such as derivatives and the chain rule. This step-by-step solution explains how to differentiate the function and evaluate the slope at the specified point. Suitable for students in Grades 11-12.

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