If
h(2) = 3 and h'(2) = −2,
find d/dx (h(x))/x)|x=2

We are tasked with finding \(\frac{d}{dx} \left( \frac{h(x)}{x} \right)\) evaluated at \(x = 2\), given that:

\[
h(2) = 3 \quad \text{and} \quad h'(2) = -2.
\]

To approach this, we will use the **quotient rule** for differentiation. The quotient rule states that for two differentiable functions \(f(x)\) and \(g(x)\):

\[
\frac{d}{dx} \left( \frac{f(x)}{g(x)} \right) = \frac{g(x) f'(x) - f(x) g'(x)}{(g(x))^2}.
\]

In our case, \(f(x) = h(x)\) and \(g(x) = x\). Let's apply the quotient rule:

\[
\frac{d}{dx} \left( \frac{h(x)}{x} \right) = \frac{x \cdot h'(x) - h(x) \cdot 1}{x^2}.
\]

Now, evaluate this expression at \(x = 2\):

\[
\frac{d}{dx} \left( \frac{h(x)}{x} \right)\bigg|_{x = 2} = \frac{2 \cdot h'(2) - h(2) \cdot 1}{2^2}.
\]

Substitute the known values \(h(2) = 3\) and \(h'(2) = -2\):

\[
= \frac{2 \cdot (-2) - 3 \cdot 1}{4}
= \frac{-4 - 3}{4}
= \frac{-7}{4}.
\]

Thus, the value of \(\frac{d}{dx} \left( \frac{h(x)}{x} \right)\) at \(x = 2\) is:

\[
\boxed{-\frac{7}{4}}.
\]

Would you like more details or have any further questions?

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Here are 5 related questions to expand your understanding:

1. How do you differentiate the product of two functions using the product rule?
2. What is the derivative of \(\frac{1}{x}\), and how does it relate to the quotient rule?
3. If \(h(x)\) were a specific polynomial, how would that change the derivative calculation?
4. How would the quotient rule change if \(g(x)\) were more complex than just \(x\)?
5. What are some practical applications of the quotient rule in physics or economics?

**Tip:** Always check if you can simplify the quotient before applying the quotient rule. This can sometimes make differentiation easier.

If h(2) = 3 and h'(2) = −2, find d/dx (h(x)/x)|x=2

Learn how to differentiate h(x)/x at x = 2 using the quotient rule. Given h(2) = 3 and h'(2) = -2, follow the step-by-step process to find the derivative. This problem covers key calculus concepts, including differentiation and the quotient rule. Suitable for high school students in Grades 11-12.

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