We are asked to evaluate the derivative $\frac{dy}{dx}$ of the function $y = (x+1)(x + \sqrt{x})$ at $x = 9$.

Step 1: Define the function

The function is given as: $y = (x+1)(x + \sqrt{x})$

Step 2: Use the product rule

We have a product of two functions, so we will apply the product rule, which states: $\frac{d}{dx} [u(x) \cdot v(x)] = u'(x) \cdot v(x) + u(x) \cdot v'(x)$ Here, let:

• $u(x) = x + 1$
• $v(x) = x + \sqrt{x}$

Step 3: Find the derivatives of $u(x)$ and $v(x)$

1. $u(x) = x + 1$, so $u'(x) = 1$.
2. $v(x) = x + \sqrt{x}$, so we need to find $v'(x)$.

To differentiate $v(x)$: $v(x) = x + x^{1/2}$ The derivative is: $v'(x) = 1 + \frac{1}{2}x^{-1/2} = 1 + \frac{1}{2\sqrt{x}}$

Step 4: Apply the product rule

Now apply the product rule: $\frac{dy}{dx} = u'(x)v(x) + u(x)v'(x)$ Substitute $u(x)$, $v(x)$, and their derivatives: $\frac{dy}{dx} = (1)(x + \sqrt{x}) + (x+1)\left(1 + \frac{1}{2\sqrt{x}}\right)$ Simplify: $\frac{dy}{dx} = x + \sqrt{x} + (x+1)\left(1 + \frac{1}{2\sqrt{x}}\right)$

Step 5: Evaluate at $x = 9$

Now substitute $x = 9$ into the expression: $\frac{dy}{dx} = 9 + \sqrt{9} + (9+1)\left(1 + \frac{1}{2\sqrt{9}}\right)$ Simplify: $\frac{dy}{dx} = 9 + 3 + 10\left(1 + \frac{1}{6}\right)$ $\frac{dy}{dx} = 12 + 10\left(\frac{7}{6}\right)$ $\frac{dy}{dx} = 12 + \frac{70}{6}$ $\frac{dy}{dx} = 12 + 11.67$ $\frac{dy}{dx} \approx 23.67$

So, the value of $\frac{dy}{dx}$ at $x = 9$ is approximately 23.67.

Would you like any additional explanation or details on this?

Related Questions:

1. What is the general formula for the product rule in calculus?
2. How do we differentiate functions with square roots?
3. What are other applications of the product rule in physics?
4. How does the chain rule relate to differentiating nested functions?
5. What happens if you differentiate a product of three functions?

Tip: When differentiating functions involving radicals like $\sqrt{x}$, express them as exponents (e.g., $x^{1/2}$) to make the differentiation easier.