# 3.1 Defining the derivative  (Page 6/10)

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$f\left(x\right)=4x+7;{x}_{1}=2,{x}_{2}=5$

$4$

$f\left(x\right)=8x-3;{x}_{1}=-1,{x}_{2}=3$

$f\left(x\right)={x}^{2}+2x+1;{x}_{1}=3,{x}_{2}=3.5$

$8.5$

$f\left(x\right)=\text{−}{x}^{2}+x+2;{x}_{1}=0.5,{x}_{2}=1.5$

$f\left(x\right)=\frac{4}{3x-1};{x}_{1}=1,{x}_{2}=3$

$-\frac{3}{4}$

$f\left(x\right)=\frac{x-7}{2x+1};{x}_{1}=-2,{x}_{2}=0$

$f\left(x\right)=\sqrt{x};{x}_{1}=1,{x}_{2}=16$

$0.2$

$f\left(x\right)=\sqrt{x-9};{x}_{1}=10,{x}_{2}=13$

$f\left(x\right)={x}^{1\text{/}3}+1;{x}_{1}=0,{x}_{2}=8$

$0.25$

$f\left(x\right)=6{x}^{2\text{/}3}+2{x}^{1\text{/}3};{x}_{1}=1,{x}_{2}=27$

For the following functions,

1. use [link] to find the slope of the tangent line ${m}_{\text{tan}}={f}^{\prime }\left(a\right),$ and
2. find the equation of the tangent line to $f$ at $x=a.$

$f\left(x\right)=3-4x,a=2$

a. $-4$ b. $y=3-4x$

$f\left(x\right)=\frac{x}{5}+6,a=-1$

$f\left(x\right)={x}^{2}+x,a=1$

a. $3$ b. $y=3x-1$

$f\left(x\right)=1-x-{x}^{2},a=0$

$f\left(x\right)=\frac{7}{x},a=3$

a. $\frac{-7}{9}$ b. $y=\frac{-7}{9}x+\frac{14}{3}$

$f\left(x\right)=\sqrt{x+8},a=1$

$f\left(x\right)=2-3{x}^{2},a=-2$

a. $12$ b. $y=12x+14$

$f\left(x\right)=\frac{-3}{x-1},a=4$

$f\left(x\right)=\frac{2}{x+3},a=-4$

a. $-2$ b. $y=-2x-10$

$f\left(x\right)=\frac{3}{{x}^{2}},a=3$

For the following functions $y=f\left(x\right),$ find ${f}^{\prime }\left(a\right)$ using [link] .

$f\left(x\right)=5x+4,a=-1$

$5$

$f\left(x\right)=-7x+1,a=3$

$f\left(x\right)={x}^{2}+9x,a=2$

$13$

$f\left(x\right)=3{x}^{2}-x+2,a=1$

$f\left(x\right)=\sqrt{x},a=4$

$\frac{1}{4}$

$f\left(x\right)=\sqrt{x-2},a=6$

$f\left(x\right)=\frac{1}{x},a=2$

$-\frac{1}{4}$

$f\left(x\right)=\frac{1}{x-3},a=-1$

$f\left(x\right)=\frac{1}{{x}^{3}},a=1$

$-3$

$f\left(x\right)=\frac{1}{\sqrt{x}},a=4$

For the following exercises, given the function $y=f\left(x\right),$

1. find the slope of the secant line $PQ$ for each point $Q\left(x,f\left(x\right)\right)$ with $x$ value given in the table.
2. Use the answers from a. to estimate the value of the slope of the tangent line at $P.$
3. Use the answer from b. to find the equation of the tangent line to $f$ at point $P.$

[T] $f\left(x\right)={x}^{2}+3x+4,P\left(1,8\right)$ (Round to $6$ decimal places.)

x Slope ${m}_{PQ}$ x Slope ${m}_{PQ}$
1.1 (i) 0.9 (vii)
1.01 (ii) 0.99 (viii)
1.001 (iii) 0.999 (ix)
1.0001 (iv) 0.9999 (x)
1.00001 (v) 0.99999 (xi)
1.000001 (vi) 0.999999 (xii)

a. $\text{(i)}\phantom{\rule{0.2em}{0ex}}5.100000,$ $\text{(ii)}\phantom{\rule{0.2em}{0ex}}5.010000,$ $\text{(iii)}\phantom{\rule{0.2em}{0ex}}5.001000,$ $\text{(iv)}\phantom{\rule{0.2em}{0ex}}5.000100,$ $\text{(v)}\phantom{\rule{0.2em}{0ex}}5.000010,$ $\text{(vi)}\phantom{\rule{0.2em}{0ex}}5.000001,$ $\text{(vii)}\phantom{\rule{0.2em}{0ex}}4.900000,$ $\text{(viii)}\phantom{\rule{0.2em}{0ex}}4.990000,$ $\text{(ix)}\phantom{\rule{0.2em}{0ex}}4.999000,$ $\text{(x)}\phantom{\rule{0.2em}{0ex}}4.999900,$ $\text{(xi)}\phantom{\rule{0.2em}{0ex}}4.999990,$ $\text{(x)}\phantom{\rule{0.2em}{0ex}}4.999999$ b. ${m}_{\text{tan}}=5$ c. $y=5x+3$

[T] $f\left(x\right)=\frac{x+1}{{x}^{2}-1},P\left(0,-1\right)$

x Slope ${m}_{PQ}$ x Slope ${m}_{PQ}$
0.1 (i) $-0.1$ (vii)
0.01 (ii) $-0.01$ (viii)
0.001 (iii) $-0.001$ (ix)
0.0001 (iv) $-0.0001$ (x)
0.00001 (v) $-0.00001$ (xi)
0.000001 (vi) $-0.000001$ (xii)

[T] $f\left(x\right)=10{e}^{0.5x},P\left(0,10\right)$ (Round to $4$ decimal places.)

x Slope ${m}_{PQ}$
$-0.1$ (i)
$-0.01$ (ii)
$-0.001$ (iii)
$-0.0001$ (iv)
$-0.00001$ (v)
−0.000001 (vi)

a. $\text{(i)}\phantom{\rule{0.2em}{0ex}}4.8771,$ $\text{(ii)}\phantom{\rule{0.2em}{0ex}}4.9875\phantom{\rule{0.2em}{0ex}}\text{(iii)}\phantom{\rule{0.2em}{0ex}}4.9988,$ $\text{(iv)}\phantom{\rule{0.2em}{0ex}}4.9999,$ $\text{(v)}\phantom{\rule{0.2em}{0ex}}4.9999,$ $\text{(vi)}\phantom{\rule{0.2em}{0ex}}4.9999$ b. ${m}_{\text{tan}}=5$ c. $y=5x+10$

[T] $f\left(x\right)=\text{tan}\phantom{\rule{0.1em}{0ex}}\left(x\right),P\left(\pi ,0\right)$

x Slope ${m}_{PQ}$
3.1 (i)
3.14 (ii)
3.141 (iii)
3.1415 (iv)
3.14159 (v)
3.141592 (vi)

[T] For the following position functions $y=s\left(t\right),$ an object is moving along a straight line, where $t$ is in seconds and $s$ is in meters. Find

1. the simplified expression for the average velocity from $t=2$ to $t=2+h;$
2. the average velocity between $t=2$ and $t=2+h,$ where $\text{(i)}\phantom{\rule{0.2em}{0ex}}h=0.1,$ $\text{(ii)}\phantom{\rule{0.2em}{0ex}}h=0.01,$ $\text{(iii)}\phantom{\rule{0.2em}{0ex}}h=0.001,$ and $\text{(iv)}\phantom{\rule{0.2em}{0ex}}h=0.0001;$ and
3. use the answer from a. to estimate the instantaneous velocity at $t=2$ second.

$s\left(t\right)=\frac{1}{3}t+5$

a. $\frac{1}{3};$ b. $\text{(i)}\phantom{\rule{0.2em}{0ex}}0.\stackrel{–}{3}$ m/s, $\text{(ii)}\phantom{\rule{0.2em}{0ex}}0.\stackrel{–}{3}$ m/s, $\text{(iii)}\phantom{\rule{0.2em}{0ex}}0.\stackrel{–}{3}$ m/s, $\text{(iv)}\phantom{\rule{0.2em}{0ex}}0.\stackrel{–}{3}$ m/s; c. $0.\stackrel{–}{3}=\frac{1}{3}$ m/s

$s\left(t\right)={t}^{2}-2t$

$s\left(t\right)=2{t}^{3}+3$

a. $2\left({h}^{2}+6h+12\right);$ b. $\text{(i)}\phantom{\rule{0.2em}{0ex}}25.22$ m/s, $\text{(ii)}\phantom{\rule{0.2em}{0ex}}24.12$ m/s, $\text{(iii)}\phantom{\rule{0.2em}{0ex}}24.01$ m/s, $\text{(iv)}\phantom{\rule{0.2em}{0ex}}24$ m/s; c. $24$ m/s

$s\left(t\right)=\frac{16}{{t}^{2}}-\frac{4}{t}$

Use the following graph to evaluate a. ${f}^{\prime }\left(1\right)$ and b. ${f}^{\prime }\left(6\right).$

a. $1.25;$ b. $0.5$

Use the following graph to evaluate a. ${f}^{\prime }\left(-3\right)$ and b. ${f}^{\prime }\left(1.5\right).$

For the following exercises, use the limit definition of derivative to show that the derivative does not exist at $x=a$ for each of the given functions.

$f\left(x\right)={x}^{1\text{/}3},x=0$

$\underset{x\to {0}^{-}}{\text{lim}}\frac{{x}^{1\text{/}3}-0}{x-0}=\underset{x\to {0}^{-}}{\text{lim}}\frac{1}{{x}^{2\text{/}3}}=\infty$

$f\left(x\right)={x}^{2\text{/}3},x=0$

$f\left(x\right)=\left\{\begin{array}{c}1,x<1\\ x,x\ge 1\end{array},x=1$

$\underset{x\to {1}^{-}}{\text{lim}}\frac{1-1}{x-1}=0\ne 1=\underset{x\to {1}^{+}}{\text{lim}}\frac{x-1}{x-1}$

$f\left(x\right)=\frac{|x|}{x},x=0$

[T] The position in feet of a race car along a straight track after $t$ seconds is modeled by the function $s\left(t\right)=8{t}^{2}-\frac{1}{16}{t}^{3}.$

1. Find the average velocity of the vehicle over the following time intervals to four decimal places:
1. [4, 4.1]
2. [4, 4.01]
3. [4, 4.001]
4. [4, 4.0001]
2. Use a. to draw a conclusion about the instantaneous velocity of the vehicle at $t=4$ seconds.

a. $\text{(i)}\phantom{\rule{0.2em}{0ex}}61.7244$ ft/s, $\text{(ii)}\phantom{\rule{0.2em}{0ex}}61.0725$ ft/s $\text{(iii)}\phantom{\rule{0.2em}{0ex}}61.0072$ ft/s $\text{(iv)}\phantom{\rule{0.2em}{0ex}}61.0007$ ft/s b. At $4$ seconds the race car is traveling at a rate/velocity of $61$ ft/s.

[T] The distance in feet that a ball rolls down an incline is modeled by the function $s\left(t\right)=14{t}^{2},$ where t is seconds after the ball begins rolling.

1. Find the average velocity of the ball over the following time intervals:
1. [5, 5.1]
2. [5, 5.01]
3. [5, 5.001]
4. [5, 5.0001]
2. Use the answers from a. to draw a conclusion about the instantaneous velocity of the ball at $t=5$ seconds.

Two vehicles start out traveling side by side along a straight road. Their position functions, shown in the following graph, are given by $s=f\left(t\right)$ and $s=g\left(t\right),$ where $s$ is measured in feet and $t$ is measured in seconds.

1. Which vehicle has traveled farther at $t=2$ seconds?
2. What is the approximate velocity of each vehicle at $t=3$ seconds?
3. Which vehicle is traveling faster at $t=4$ seconds?
4. What is true about the positions of the vehicles at $t=4$ seconds?

a. The vehicle represented by $f\left(t\right),$ because it has traveled $2$ feet, whereas $g\left(t\right)$ has traveled $1$ foot. b. The velocity of $f\left(t\right)$ is constant at $1$ ft/s, while the velocity of $g\left(t\right)$ is approximately $2$ ft/s. c. The vehicle represented by $g\left(t\right),$ with a velocity of approximately $4$ ft/s. d. Both have traveled $4$ feet in $4$ seconds.

[T] The total cost $C\left(x\right),$ in hundreds of dollars, to produce $x$ jars of mayonnaise is given by $C\left(x\right)=0.000003{x}^{3}+4x+300.$

1. Calculate the average cost per jar over the following intervals:
1. [100, 100.1]
2. [100, 100.01]
3. [100, 100.001]
4. [100, 100.0001]
2. Use the answers from a. to estimate the average cost to produce $100$ jars of mayonnaise.

[T] For the function $f\left(x\right)={x}^{3}-2{x}^{2}-11x+12,$ do the following.

1. Use a graphing calculator to graph f in an appropriate viewing window.
2. Use the ZOOM feature on the calculator to approximate the two values of $x=a$ for which ${m}_{\text{tan}}={f}^{\prime }\left(a\right)=0.$

a.

b. $a\approx -1.361,2.694$

[T] For the function $f\left(x\right)=\frac{x}{1+{x}^{2}},$ do the following.

1. Use a graphing calculator to graph $f$ in an appropriate viewing window.
2. Use the ZOOM feature on the calculator to approximate the values of $x=a$ for which ${m}_{\text{tan}}={f}^{\prime }\left(a\right)=0.$

Suppose that $N\left(x\right)$ computes the number of gallons of gas used by a vehicle traveling $x$ miles. Suppose the vehicle gets $30$ mpg.

1. Find a mathematical expression for $N\left(x\right).$
2. What is $N\left(100\text{)?}$ Explain the physical meaning.
3. What is ${N}^{\prime }\left(100\right)?$ Explain the physical meaning.

a. $N\left(x\right)=\frac{x}{30}$ b. $\sim 3.3$ gallons. When the vehicle travels $100$ miles, it has used $3.3$ gallons of gas. c. $\frac{1}{30}.$ The rate of gas consumption in gallons per mile that the vehicle is achieving after having traveled $100$ miles.

[T] For the function $f\left(x\right)={x}^{4}-5{x}^{2}+4,$ do the following.

1. Use a graphing calculator to graph $f$ in an appropriate viewing window.
2. Use the $\text{nDeriv}$ function, which numerically finds the derivative, on a graphing calculator to estimate ${f}^{\prime }\left(-2\right),{f}^{\prime }\left(-0.5\right),{f}^{\prime }\left(1.7\right),$ and ${f}^{\prime }\left(2.718\right).$

[T] For the function $f\left(x\right)=\frac{{x}^{2}}{{x}^{2}+1},$ do the following.

1. Use a graphing calculator to graph $f$ in an appropriate viewing window.
2. Use the $\text{nDeriv}$ function on a graphing calculator to find ${f}^{\prime }\left(-4\right),{f}^{\prime }\left(-2\right),{f}^{\prime }\left(2\right),$ and ${f}^{\prime }\left(4\right).$

a.

b. $-0.028,-0.16,0.16,0.028$

The f'(4)for f(x) =4^x
I need help under implicit differentiation
how to understand this
nhie
What is derivative of antilog x dx ?
what's the meaning of removable discontinuity
what's continuous
Brian
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
also find the value of "X" from the equation that follow (x-1/x)^4 +4(x^2-1/x^2) -6=0 please guy help
Levis
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
Bilal
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
integrate e^cosx
Uchenna
-sinx e^x
Leo
Do we ask only math question? or ANY of the question?
yh
Gbesemete
How do i differentiate between substitution method, partial fraction and algebraic function in integration?
usman
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
we asking the question cause only the question will tell us the right answer
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
Levis I am sorry
Bilal
Bilal it okay buddy honestly i am pleasured to meet you
Levis
x^x ... no anti derivative for this function... but we can find definte integral numerically.
Bilal
thank you Bilal Kumhar then how we may find definite integral let say x^x,3,5?
Levis
evaluate 5-×square divided by x+2 find x as limit approaches infinity
i have not understood
Leo
Michael
welcome
Sunny
I just dont get it at all...not understanding
Michagaye
0 baby
Sunny
The denominator is the aggressive one
Sunny
wouldn't be any prime number for x instead ?
Harold
or should I say any prime number greater then 11 ?
Harold
just wondering
Harold
I think as limit Approach infinity then X=0
Levis
ha hakdog hahhahahaha
ha hamburger
Leonito
Fond the value of the six trigonometric function of an angle theta, which terminal side passes through the points(2x½-y)²,4
What's f(x) ^x^x
What's F(x) =x^x^x
Emeka
are you asking for the derivative
Leo
that's means more power for all points
rd
Levis
iam sorry f(x)=x^x it means the output(range ) depends to input(domain) value of x by the power of x that is to say if x=2 then x^x would be 2^2=4 f(x) is the product of X to the power of X its derivatives is found by using product rule y=x^x introduce ln each side we have lny=lnx^x =lny=xlnx
Levis
the derivatives of f(x)=x^x IS (1+lnx)*x^x
Levis
So in that case what will be the answer?
Alice
nice explanation Levis, appreciated..
Thato
what is a maximax
A maxima in a curve refers to the maximum point said curve. The maxima is a point where the gradient of the curve is equal to 0 (dy/dx = 0) and its second derivative value is a negative (d²y/dx² = -ve).
Viewer
what is the limit of x^2+x when x approaches 0
it is 0 because 0 squared Is 0
Leo
0+0=0
Leo
simply put the value of 0 in places of x.....
Tonu
the limit is 2x + 1
Nicholas
the limit is 0
Muzamil
limit s x
Bilal
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
Levis
find derivatives 3√x²+√3x²
3 + 3=6
mujahid
How to do basic integrals
the formula is simple x^n+1/n+1 where n IS NOT EQUAL TO 1 And n stands for power eg integral of x^2 x^2+1/2+1 =X^3/3
Levis