The slope of the tangent line to the graph of the function 𝑓(𝑥)=𝑥2+5𝑥+3‾‾‾‾‾‾‾‾‾‾‾‾√𝑓(𝑥)=𝑥2+5𝑥+3 at 𝑥=1𝑥=1 equals just the final answer

To find the slope of the tangent line to the graph of the function \( f(x) = \sqrt{x^2 + 5x + 3} \) at \( x = 1 \), we need to compute the derivative of \( f(x) \) and then evaluate it at \( x = 1 \).

First, let's find \( f'(x) \):

\[
f(x) = \sqrt{x^2 + 5x + 3}
\]

Using the chain rule:

\[
f'(x) = \frac{d}{dx} \left( \sqrt{x^2 + 5x + 3} \right) = \frac{1}{2\sqrt{x^2 + 5x + 3}} \cdot \frac{d}{dx} (x^2 + 5x + 3)
\]

\[
f'(x) = \frac{1}{2\sqrt{x^2 + 5x + 3}} \cdot (2x + 5)
\]

\[
f'(x) = \frac{2x + 5}{2\sqrt{x^2 + 5x + 3}}
\]

Now, evaluate \( f'(x) \) at \( x = 1 \):

\[
f'(1) = \frac{2(1) + 5}{2\sqrt{(1)^2 + 5(1) + 3}} = \frac{2 + 5}{2\sqrt{1 + 5 + 3}} = \frac{7}{2\sqrt{9}} = \frac{7}{2 \cdot 3} = \frac{7}{6}
\]

Therefore, the slope of the tangent line to the graph of \( f(x) \) at \( x = 1 \) is:

\[
\boxed{\frac{7}{6}}
\]

Would you like more details or have any questions?

Here are 5 questions you can ask:
1. How do you use the chain rule to differentiate composite functions?
2. Can you explain the steps to differentiate a square root function?
3. How do you evaluate a derivative at a specific point?
4. What are some common rules for differentiation?
5. How do you find the equation of a tangent line to a curve?

**Tip**: When differentiating, always carefully apply the chain rule and simplify your expressions step by step.

Learn how to find the slope of the tangent line to the function f(x) = sqrt(x^2 + 5x + 3) at x = 1 with a step-by-step solution. This problem involves differentiation using the chain rule and evaluating a derivative at a specific point. Suitable for advanced high school students.

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