Find the indicated derivative. d / d t ( − 5 t ) d/dt(-5t) d / d t ( − 5 t )

In this problem, we're asked to find the derivative d d t of negative five t. Now this d d t just means that we're differentiating with respect to the variable t, the same way that d d x means that we're differentiating with respect to the variable x. So we're gonna treat this the same way we would any derivative. Here, we're taking the derivative of negative 5, a constant, times t. Now I know that when faced with a derivative like this, I can pull my constant out to the front. So this is really negative 5 times the derivative d d t of t. Now using what we know, d d t of t is just going to be 1, so this becomes negative 5 times 1 or just a negative 5. So that's my final derivative here. Now, again, as you get further and further on in taking derivatives, you can recognize almost immediately that ddt of negative 5 t is just negative 5 without having to set up these additional rules. Now if you already recognize that here, great job. Let's keep getting some more practice. Thanks for watching.

3. Techniques of Differentiation

Basic Rules of Differentiation

Calculus

3. Techniques of Differentiation

Basic Rules of Differentiation

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Multiple Choice

Find the indicated derivative.
$d/dt(-5t)$

$0$

$-5$

$5$

$5+t$

Verified step by step guidance

Identify the function for which you need to find the derivative. In this case, the function is $-5t$.

Recall the basic rule for differentiating a constant multiplied by a variable: $\frac{d}{dt}(ct) = c$, where $c$ is a constant.

Apply this rule to the function $-5t$. Here, $-5$ is the constant $c$.

Differentiate $-5t$ with respect to $t$ using the rule: $\frac{d}{dt}(-5t) = -5$.

Conclude that the derivative of $-5t$ with respect to $t$ is $-5$.

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Bacterium population

When a bactericide was added to a nutrient broth in which bacteria were growing, the bacterium population continued to grow for a while, but then stopped growing and began to decline. The size of the population at time t (hours) was b = 10⁶ + 10⁴t − 10³t². Find the growth rates at

a. t = 0 hours.

b. t = 5 hours.

c. t = 10 hours.

Textbook Question

Vehicular stopping distance Based on data from the U.S. Bureau of Public Roads, a model for the total stopping distance of a moving car in terms of its speed is s = 1.1v + 0.054v², where s is measured in ft and v in mph. The linear term 1.1v models the distance the car travels during the time the driver perceives a need to stop until the brakes are applied, and the quadratic term 0.054v² models the additional braking distance once they are applied. Find ds/dv at v = 35 and v = 70 mph, and interpret the meaning of the derivative.

Textbook Question

Airplane takeoff Suppose that the distance an aircraft travels along a runway before takeoff is given by D = (10/9)t², where D is measured in meters from the starting point and t is measured in seconds from the time the brakes are released. The aircraft will become airborne when its speed reaches 200 km/h. How long will it take to become airborne, and what distance will it travel in that time?

Multiple Choice

Find the indicated derivative.

$f(x)=3x-4$