Math / Calculus

QuestionLet $H(t)=\frac{a}{t^{40}}-b \cdot e^{t}$ where $a$ and $b$ are both fixed constants. What is $H^{\prime}(t)$ ?

Studdy Solution

STEP 1

What is this asking? We need to find the derivative of a function $H(t)$ with respect to $t$ , which involves a power function and an exponential function. Watch out! Don't forget the chain rule and the rules for differentiating power and exponential functions!
Also, remember that $a$ and $b$ are constants.

STEP 2

1. Rewrite the function
2. Differentiate the first term
3. Differentiate the second term
4. Combine the results

STEP 3

Let's rewrite $H(t)$ to make it easier to differentiate.
We can rewrite the first term with a negative exponent: $H(t) = a \cdot t^{-40} - b \cdot e^t$ This makes it clear that we're dealing with a power function.

STEP 4

Now, let's differentiate the first term, $a \cdot t^{-40}$ , with respect to $t$ .
Using the power rule, we multiply by the exponent and then decrease the exponent by 1.
Remember that $a$ is a constant, so it just hangs out: $\frac{d}{dt}(a \cdot t^{-40}) = a \cdot (-40) \cdot t^{-40 - 1} = -40a \cdot t^{-41}$

STEP 5

Next, let's differentiate the second term, $-b \cdot e^t$ , with respect to $t$ .
The derivative of $e^t$ is just $e^t$ , and $b$ is a constant: $\frac{d}{dt}(-b \cdot e^t) = -b \cdot e^t$ So, the derivative of the second term is simply $-b \cdot e^t$ .

STEP 6

Finally, let's combine the derivatives of the two terms to find the derivative of the entire function $H(t)$ : $H'(t) = \frac{d}{dt}(a \cdot t^{-40}) + \frac{d}{dt}(-b \cdot e^t) = -40a \cdot t^{-41} - b \cdot e^t$

STEP 7

Therefore, the derivative of $H(t)$ is: $H'(t) = -40a \cdot t^{-41} - b \cdot e^t$

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