# 3.1 Defining the derivative  (Page 4/10)

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${v}_{\text{ave}}=\frac{s\left(t\right)-s\left(a\right)}{t-a}.$

As the values of $t$ approach $a,$ the values of ${v}_{\text{ave}}$ approach the value we call the instantaneous velocity    at $a.$ That is, instantaneous velocity at $a,$ denoted $v\left(a\right),$ is given by

$v\left(a\right)={s}^{\prime }\left(a\right)=\underset{t\to a}{\text{lim}}\frac{s\left(t\right)-s\left(a\right)}{t-a}.$

To better understand the relationship between average velocity and instantaneous velocity, see [link] . In this figure, the slope of the tangent line (shown in red) is the instantaneous velocity of the object at time $t=a$ whose position at time $t$ is given by the function $s\left(t\right).$ The slope of the secant line (shown in green) is the average velocity of the object over the time interval $\left[a,t\right].$

We can use [link] to calculate the instantaneous velocity, or we can estimate the velocity of a moving object by using a table of values. We can then confirm the estimate by using [link] .

## Estimating velocity

A lead weight on a spring is oscillating up and down. Its position at time $t$ with respect to a fixed horizontal line is given by $s\left(t\right)=\text{sin}\phantom{\rule{0.2em}{0ex}}t$ ( [link] ). Use a table of values to estimate $v\left(0\right).$ Check the estimate by using [link] .

We can estimate the instantaneous velocity at $t=0$ by computing a table of average velocities using values of $t$ approaching $0,$ as shown in [link] .

Average velocities using values of t Approaching 0
$t$ $\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}t-\text{sin}\phantom{\rule{0.1em}{0ex}}0}{t-0}=\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}t}{t}$
$-0.1$ $0.998334166$
$-0.01$ $0.9999833333$
$-0.001$ $0.999999833$
$0.001$ $0.999999833$
$0.01$ $0.9999833333$
$0.1$ $0.998334166$

From the table we see that the average velocity over the time interval $\left[-0.1,0\right]$ is $0.998334166,$ the average velocity over the time interval $\left[-0.01,0\right]$ is $0.9999833333,$ and so forth. Using this table of values, it appears that a good estimate is $v\left(0\right)=1.$

By using [link] , we can see that

$v\left(0\right)={s}^{\prime }\left(0\right)=\underset{t\to 0}{\text{lim}}\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}t-\text{sin}\phantom{\rule{0.1em}{0ex}}0}{t-0}=\underset{t\to 0}{\text{lim}}\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}t}{t}=1.$

Thus, in fact, $v\left(0\right)=1.$

A rock is dropped from a height of $64$ feet. Its height above ground at time $t$ seconds later is given by $s\left(t\right)=-16{t}^{2}+64,0\le t\le 2.$ Find its instantaneous velocity $1$ second after it is dropped, using [link] .

$-32$ ft/s

As we have seen throughout this section, the slope of a tangent line to a function and instantaneous velocity are related concepts. Each is calculated by computing a derivative and each measures the instantaneous rate of change of a function, or the rate of change of a function at any point along the function.

## Definition

The instantaneous rate of change    of a function $f\left(x\right)$ at a value $a$ is its derivative ${f}^{\prime }\left(a\right).$

## Chapter opener: estimating rate of change of velocity

Reaching a top speed of $270.49$ mph, the Hennessey Venom GT is one of the fastest cars in the world. In tests it went from $0$ to $60$ mph in $3.05$ seconds, from $0\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}100$ mph in $5.88$ seconds, from $0\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}200$ mph in $14.51$ seconds, and from $0\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}229.9$ mph in $19.96$ seconds. Use this data to draw a conclusion about the rate of change of velocity (that is, its acceleration ) as it approaches $229.9$ mph. Does the rate at which the car is accelerating appear to be increasing, decreasing, or constant?

First observe that $60$ mph = $88$ ft/s, $100$ mph $\approx 146.67$ ft/s, $200$ mph $\approx 293.33$ ft/s, and $229.9$ mph $\approx 337.19$ ft/s. We can summarize the information in a table.

$v\left(t\right)$ At different values of t
$t$ $v\left(t\right)$
$0$ $0$
$3.05$ $88$
$5.88$ $147.67$
$14.51$ $293.33$
$19.96$ $337.19$

Now compute the average acceleration of the car in feet per second on intervals of the form $\left[t,19.96\right]$ as $t$ approaches $19.96,$ as shown in the following table.

Average acceleration
$t$ $\frac{v\left(t\right)-v\left(19.96\right)}{t-19.96}=\frac{v\left(t\right)-337.19}{t-19.96}$
$0.0$ $16.89$
$3.05$ $14.74$
$5.88$ $13.46$
$14.51$ $8.05$

The rate at which the car is accelerating is decreasing as its velocity approaches $229.9$ mph $\text{(}337.19$ ft/s).

What is derivative of antilog x dx ?
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an area under a curve is continuous because you are looking at an area that covers a range of numbers, it is over an interval, such as 0 to 4
Lauren
using product rule x^3,x^5
Kabiru
may god be with you
Sunny
Luke 17:21 nor will they say, See here or See there For indeed, the kingdom of God is within you. You've never 'touched' anything. The e-energy field created by your body has pushed other electricfields. even our religions tell us we're the gods. We live in energies connecting us all. Doa/higgsfield
Scott
if you have any calculus questions many of us would be happy to help and you can always learn or even invent your own theories and proofs. math is the laws of logic and reality. its rules are permanent and absolute. you can absolutely learn calculus and through it better understand our existence.
Scott
ya doubtless
Bilal
help the integral of x^2/lnxdx
Levis
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Use integration by parts. Let u=lnx and dv=x2dx Then du=1xdx and v=13x3. ∫x2lnxdx=13x3lnx−∫(13x3⋅1x)dx ∫x2lnxdx=13x3lnx−∫13x2dx ∫x2lnxdx=13x3lnx−19x3+C
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itz 1/3 and 1/9
Bilal
now you can find the value of X from the above equation easily
Bilal
Pls i need more explanation on this calculus
usman
usman from where do you need help?
Levis
thanks Bilal
Levis
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yh
Gbesemete
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you just have to recognize the problem. there can be multiple ways to solve 1 problem. that's the hardest part about integration
Lauren
test
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Sunny
find integral of sin8xcos12xdx
don't share these childish questions
Bilal
well find the integral of x^x
Levis
bilal kumhar you are so biased if you are an expert what are you doing here lol😎😎😂😂 we are here to learn and beside there are many questions on this chat which you didn't attempt we are helping each other stop being naive and arrogance so give me the integral of x^x
Levis
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Bilal
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Levis
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Bilal
thank you Bilal Kumhar then how we may find definite integral let say x^x,3,5?
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Michael
welcome
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0 baby
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The denominator is the aggressive one
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or should I say any prime number greater then 11 ?
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just wondering
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I think as limit Approach infinity then X=0
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Fond the value of the six trigonometric function of an angle theta, which terminal side passes through the points(2x½-y)²,4
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that's means more power for all points
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0+0=0
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simply put the value of 0 in places of x.....
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the limit is 2x + 1
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the limit is 0
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limit s x
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The limit is 3
Levis
Leo we don't just do like that buddy!!! use first principle y+∆y=x+∆x ∆y=x+∆x-y ∆y=(x+∆x)^2+(x+∆x)-x^2+x on solving it become ∆y=3∆x+∆x^2 as ∆x_>0 limit=3 if you do by calculator say plugging any value of x=0.000005 which approach 0 you get 3
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write something lmit
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lnSecx+c
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