To find the derivative of the given function $h(t) = \frac{6}{t} + \frac{4}{t^2}$, we rewrite and differentiate each term.

Step 1: Rewrite the function

Using properties of exponents, rewrite the terms: $h(t) = 6t^{-1} + 4t^{-2}.$

Step 2: Differentiate each term

Using the power rule $\frac{d}{dt}[t^n] = n t^{n-1}$, differentiate term by term:

1. For $6t^{-1}$:
   $\frac{d}{dt}[6t^{-1}] = 6(-1)t^{-2} = -\frac{6}{t^2}.$

2. For $4t^{-2}$:
   $\frac{d}{dt}[4t^{-2}] = 4(-2)t^{-3} = -\frac{8}{t^3}.$

Step 3: Combine results

Adding the derivatives together: $h'(t) = -\frac{6}{t^2} - \frac{8}{t^3}.$

Final Answer:

$h'(t) = -\frac{6}{t^2} - \frac{8}{t^3}.$

Would you like further clarification on any of these steps? Let me know!

Related Questions:

1. How do you apply the power rule for derivatives?
2. How do you differentiate functions involving negative exponents?
3. How would the derivative change if the powers of $t$ were different?
4. Can you simplify $h'(t)$ further or factor it?
5. How do you interpret the derivative $h'(t)$ graphically?

Tip:

When differentiating terms like $\frac{1}{t^n}$, first rewrite them as $t^{-n}$ to apply the power rule easily.