Chapter 6 Solutions

These are explanations that I had emailed to various students while we were working on Chapter 6.  Use this information as you review old problems, or while you're studying for the final exam.

Remember, my Google Drive folder contains copies of worksheets and ANSWERS (on the last page of each file). CLICK HERE TO ACCESS.

Chp 6 Review (pg. 297):

R7fii) To write an equation for E(x), you need to integrate C(t). You can see this designated in the given diagram, where there’s an arrow pointing to the area under the curve. It also says in part i that E(x) can be expressed as the product of the concentration C(t) and the number of days t.

So, we need to integrate C(t) = 150e^-0.16t

The integral of the part with “e” will stay the same. But, we need to account for the derivative of the function inside the exponent of “e”. We need to divide the coefficient 150 by the derivative of the inside, which is -0.16. Here’s what we get:

E(x) = -937.5e^-0.16t + c

But, the key here is that we’re finding the integral of the curve under C(t) starting at x = 0. So, we should actually be finding a definite integral from x = 0 to x = x.

(integral sign from 0 to x) 150e^-0.16t

= -937.5e^-0.16t (from 0 to x)

= -937.5e^-0.16x - -937.5e^-0.16(0)

= -937.5e^-0.16x + 937.5

R7fiii) They just had you find the integral of C(t), and now they want you to find the derivative of that answer. So, we should right back at the C(t) equation as our answer here. The only difference is that the variable should now be “x”, because of how we found the integral in part ii from x = 0 to x = x. So, when we find the derivative of that integrated equation, it’ll have “x" as a variable instead of “t”. This is the second form of the Fundamental Theorem of Calculus.

E’(x) = 150e^-0.16x

R9biii) "I keep running into the same problem on a bunch of the review problems and I was wondering if you could help me. I use U Substitution and get to the point U^-1 and then don't know how to proceed from there because I would end up with U^0. This has happened a couple times, R9 biii most recently, and I was wondering how to take care of it."

Answer:  "Great question!  u^-1 is the same as 1/u, right?  Well, the integral of 1/u is…. ln u + c (with absolute value signs around the u).

You can't do the normal power rule method, because it does give you u^0.  That's why we use the natural logarithm instead!"

T3) Just to make sure I'm doing this right, 2nd FTC lets me change it to sin(x) and be done, right? Correct.

T6) Without first simplifying the equation for f(x):

f(x) = ln (x^3 * e^x)

Derivative of ln x is 1/x.  Here, when you multiply by the derivative of the inside, you have to use product rule.

f'(x) = 1 / (x^3 * e^x) * (3x^2 * e^x + x^3 * e^x)

Rewrite it in one fraction:

f'(x) =  (3x^2 * e^x + x^3 * e^x) / (x^3 * e^x)

Factor out a common factor of x^2 * e^x on the top:

f'(x) = x^2 * e^x * (3 + x) / (x^3 * e^x)

Cancel out x^2 and e^x:

f'(x) = (3 + x) / x

First simplify using ln properties:

f(x) = ln (x^3 * e^x)

f(x) = ln x^3 + ln e^x

f(x) = 3 ln x + x ln e

f(x) = 3 ln x + x

Now find the derivative:

f'(x) = 3 / x + 1

These two answers are the same.

First answer: f'(x) = (3 + x) / x

Second answer: f'(x) = 3 / x + 1

In the first answer, if you actually divide both terms in the numerator by the "x" in the denominator, you get:

f'(x) = (3 + x) / x

f'(x) = 3 / x + x / x

f'(x) = 3 / x + 1

T10) With this one, I just want to make sure that I can change ln(sin^2x+cos^2x) into ln(1) and then say the deriv is 0 because ln(1) is a constant.  Correct.

T11) Here, you're being asked to find the derivative of an integral.  According to the 2nd form of FTC, the derivative and integral will cancel out, leaving you with just the given integrand equation.  You need to plug in the upper bound for "t", then multiply by the derivative of the inside, or in other words, multiply by the derivative of the upper bound.

Step 1:  e^ln x * sin (ln x) * 1/x

Step 2:  e and ln cancel:

x sin (ln x) * 1/x

Step 3 (answer):  The x and 1/x cancel:

sin(ln x)

Sec 6.9 (pg. 294):

How do you integrate tan x?) 

(integral sign) tan x dx

You know that tan x = sin x / cos x, so now your integral looks like:

(integral sign) sin x / cos x * dx

Now, because we have one function divided by another function, we can try to use u-substitution to integrate, in hopes of ending up with an anti-derivative in terms of ln.

u = cos x
du = -sin x * dx
-du = sin x * dx

Substitute your u and -du back into the integral:

- (integral sign) 1 / u * du 

We know that the integral of 1 / u is ln |u| + c. So, our integral here is:

-ln |u| + c

Then plug back in for u:

-ln |cos x| + c 

This is the integral of tan x! A very similar method should be used to find the integral of cot x.

How do you integrate csc x?)  

(integral sign ) csc x * dx

We can't use the same trick as we did with tan x, because converting csc x to a different trig function won't result in one trig function over another trig function. (Because csc x = 1 / sin x). Instead, we're going to multiply the numerator and denominator by (csc x + cot x):

(integral sign) csc x * (csc x + cot x) / (csc x + cot x) * dx

Then, distribute the original csc x (which is in the numerator) into the parentheses that are also in the numerator:

(integral sign) (csc^2 x + csc x cot x) / (csc x + cot x) * dx

Now we can try u-substitution. Make the denominator equal to u:

u = csc x + cot x
du = (-csc x cot x - csc^2 x) * dx
-du = (csc x cot x + csc^2 x) * dx

Plug this u and -du back into the integral:

- (integral sign) du / u

This is the same thing as 1 / u * du, of which we know the integral is ln |u| + c. So, when we integrate, we get:

-ln |u| + c

= -ln |csc x + cot x| + c

This is the integral of csc x! A very similar method should be used to find the integral of sec x.

17) exp x means the same thing as e^x. I’ve been told that the reason exp e exists is to be able to use e^x when programming or coding, since you can’t actually type exponents in that type of situation.

77) Educreations video: Sec 6.9 #77 solution

Click here to view ENTIRE Sec 6.9 ANSWER KEY (stored in my Google Drive folder).
[If you get a file error, click the small arrowat the top of the screen to download the entire file.]

Sec 6.8 (pg. 288):

7) When you plug in x = 0 to this equation, it gives you 0 / 0:

lim        sin x        =    0
x—>0    x^2               0

This (the 0 / 0 answer) means we need to use l’Hospital’s Rule, where we find the derivative of the numerator and the denominator, then try to find the limit again:

lim        cos x    =    1 

x—>0    2x              0

This isn’t 0 / 0 or infinity / infinity, so we’re done. 1 / 0 is an undefined number, so you could answer “undefined” or “infinite limit” (because those two things mean the same thing).

9) You don’t have to do too much with the “+” in the lim of x—>0+ problems. The only reason it needs to be there is just in case the graph you’re taking the limit of doesn’t exist at x = 0. Then, by defining the limit as approaching from the right, you can still find the limit. This will apply in this problem.

When you plug in x = 0 to this equation, it gives you:

lim            ln x      =    infinity
x—>0+    1 / x            infinity

The numerator equals “infinity” because the graph of ln x approaches infinity as x approaches zero from the right. The denominator equals “infinity” because the graph of 1/x approaches infinity as x approaches zero from the right.

This infinity / infinity answer means we need to use l’Hospital’s Rule, where we find the derivative of the numerator and the denominator, then try to find the limit again.

The derivative of ln x is 1 / x, which is equal to x^-1.

The derivative of 1 / x is found more easily if we start by rolling “x” up to the numerator to make it x^-1, then find the derivative. The derivative ends up being: -x^-2

Let’s simplify this new equation before we actually find the limit:

lim        x^-1
x—>0    -x^-2

lim        -x^1
x—>0

lim        -x
x—>0

When you plug in x = 0, you find the final answer: 0

11) When you plug in x = 1 to this equation, it gives you 0 / 0:

lim        e^x - e        =    0
x—>1    5 ln x                0

This (the 0 / 0 answer) means we need to use l’Hospital’s Rule, where we find the derivative of the numerator and the denominator, then try to find the limit again:

lim        e^x        =    e
x—>1    5 / x            5

Sec 6.7 (pg. 282):

7) You could do this problem with product rule, because we don't need logarithms to find the derivative of e^x.

h(x) = x^3 * e^x

Product rule pieces:

u = x^3

u' = 3x^2

v = e^x

v' = e^x

Your answer:

h'(x) = 3x^2 * e^x + x^3 * e^x

Then you could factor out the common factors of x^2 and e^x:

h'(x) = x^2 * e^x (3 + x)

9) The long way to do this problem:

Use natural logs to simplify the equation:

r(t) = e^t * sin t

ln r(t) = ln e^t + ln (sin t)

ln r(t) = t + ln (sin t)

Now find the implicit derivative:

1/r(t) * r'(t) = 1 + 1/sin t * cos t

1/r(t) * r'(t) = 1 + cot t

r'(t) = (1 + cot t) * r(t)

r'(t) = (1 + cot t) * e^t * sin t

This answer is OK, but it can be simplified.  Start by distributing:

e^t * sin t + cot t * e^t * sin t

In that second part, cot t * sin t can be simplified, because cot = cos/sin.  So, cot t * sin t = cos t.

Your answer is now:

e^t * sin t + e^t * cos t

9) An alternate way to do this problem:

It's great to use logarithms, because it's an exponential equation, but you don't HAVE to when it's an equation with "e", because we know the derivative of e^x.  You could have done this problem with regular product rule:

Problem is: r(t) = e^t * sin t

Product rule pieces:

u = e^t

u' = e^t

v = sin t

v' = cos t

So your answer would be:

r'(t) = e^t * sin t + e^t * cos t

11) Starts with u = 3 * e^x * e^-x

The exponential part can be simplified before you do any derivative work.

e^x * e^-x = e^0 = 1

So, this whole equation is just:

u = 3

The derivative is u' = 0.

15) exp x means the same thing as e^x.  So, just think of it as y = e^x / ln x.  You'll need to use the quotient rule to find this derivative.

u = e^x v = ln x

u' = e^x v' = 1/x

y' = (e^x * ln x - e^x * 1/x) / (ln x)^2

49) Try simplifying this BEFORE you find the integral.  Use the logarithm properties you know.

3 ln x is the same as ln x^3

So it can be rewritten: e^(ln x^3)

Then e and ln cancel each other out, so it simplifies to just: x^3

Now integrate!

57) Find the antiderivatives first.  The antiderivative of e^x is e^x.  The antiderivative of -e^-x is e^-x (you can check this one by thinking: "What would the derivative of e^-x be? Would it give me the same thing that's listed in the integral?")

So, after integrating, your equation becomes: e^x + e^-x

Then, plug in the top bound and the lower bound, then subtract.

e^2 + e^-2 - (e^0 + e^0)

e^2 + e^-2 - (1 + 1)

e^2 + e^-2 - 2

= 5.524… which matches what you get from fnInt on your calculator (that's what "numerically integrating" means).

Sec 6.6 (pg. 276):

19) Educreations Video: Lava Flow problem

Tue 1/30 + Wed 1/31 Warm-up + answers (Logarithm property practice):

Sec 6.5 (pg. 269):

3) Find the derivative of g(x) = 4(7^x).

Use ln on both sides, then use ln properties to split it up.  Your goal here is to NOT have a variable in the exponent.

ln g(x) = ln (4(7^x))

ln g(x) = ln 4 + ln 7^x

ln g(x) = ln 4 + x ln 7

Now you can take the implicit derivative, because the variable is no longer in the exponent.  The "ln 4" is a constant, so that derivative is 0.  The "x ln 7" derivative is just "ln 7".  [This is like finding the derivative of 10x, which would just be 10.  The x "goes away", and the constant is still there.]

1/g(x) * g'(x) = ln 7

Now multiply both sides by g(x).

g'(x) = ln 7 * g(x)

Now substitute the equation back in for g(x).

g'(x) = ln 7 * 4(7^x)

In other words, to answer your question about the "4":  The "4" just ends up staying there, because it's a constant that is multiplying the function.  It doesn't go anywhere.  Another way to write the final answer:

g'(x) = 4 * 7^x * ln 7



7) In order to take the derivative of any exponential function, you have to use the logarithm power rule, where you "roll down the exponent", because we CANNOT have the variable in the exponent.

y = (cos x) ^ 0.7x

Take the ln of both sides, then "roll down" the exponent.

ln y = ln (cos x) ^ 0.7x

ln y = 0.7x * ln (cos x)

Now you can do the derivative!  Take the implicit derivative of each side.  Remember, when you take the derivative of ln (inside), it's 1/inside * derivative of inside.  Just like the chain rule in any other problem.

Notice that I'm using the product rule on the right side of the equation, because it's two functions multiplied together.

1/y * y' = 0.7 * ln (cos x) + 0.7x * 1/cos x * -sin x

Simplifying the right side a little bit, because sin/cos = tan.

1/y * y' = 0.7 ln (cos x) - 0.7x tan x

Now get y' by itself, since that's what you're looking for.

y' = (0.7 ln (cos x) - 0.7x tan x) * y

Substitute the original equation back in for y, because we know that's what y is equal to.

y' = (0.7 ln (cos x) - 0.7x tan x) * (cos x) ^ 0.7x

Sec 6.3 (pg. 260):

23) This shows two functions multiplied, so you will need to use product rule. Here’s the set up:

u = x^4
u’ = 4x^3

v = ln 3x
v’ = 1/3x * 3…which simplifies to...
v’ = 1/x

Here’s what your product rule will look like:

y’ = 4x^3 * ln 3x + x^4 * 1/x

In the second term, one of the x’s from x^4 can cancel with the one x in 1/x:

y’ = 4x^3 * ln 3x + x^3

You could leave it like this, but since there is a common factor of x^3 in both terms now, you could also factor it out:

y’ = x^3(4 ln 3x + 1)

29) "I checked my answers in the back of the book, but am stuck on #29 because I got it wrong. The answer on the back of the book is 1/3ln |x| + c, but I got 1/3 ln |3x| + c. What did I do wrong? Thank you!"

Your answer is OK, actually! Remember, to check our integrated answers, we should try taking the derivative of it (to see if we end up at our original given problem).

If you answer is 1/3 ln |3x| + c, the first step in doing the derivative would be to find the derivative of the outside function (ln):

1/3 * 1/3x …

Then multiply by the derivative of the inside function:

1/3 * 1/3x * 3….

Then find the derivative of +c, which is zero. Final derivative answer:

1/3 * 1/3x * 3

= 1/3 * 1/x

= 1/(3x)

Notice how the “3” from inside the ln of your answer ended up canceling out, because of the chain rule when we found the derivative? So, it doesn’t actually matter that the “3” is there. You could actually put ANY number inside the ln of your answer (but don’t actually put just any number there, that’d be weird), because it would cancel out when you find the derivative by using the chain rule.

Conclusion: You’re right. And the book is right. :)

Explanation of steps to integrate:

To integrate 1/(3x) dx, we could use u-substitution:

u = 3x
du = 3 dx

Before we plug these back into the equation, we need to solve for dx:

1/3 du = dx

Here’s what it looks like when we plug in our u’s:

1/3 (integral sign) 1/u du

The integral of 1/u is:

ln |u| + c

But we have a 1/3 on the front, and we need to substitute 3x back into the equation for u:

1/3 ln |3x| + c

31) Most of the time you are asked to integrate something that shows two functions divided, you should try using u-substitution (where u is equal to the denominator). Let’s try it here.

(integral sign) x^2 / (x^3 + 5) * dx

u = x^3 + 5
du = 3x^2 dx

This du equation almost matches the remaining part of our integral, but we should divide both sides of it by 3 to make it match exactly.

1/3 du = x^2 dx

Now it matches the x^2 dx part exactly, which means we can substitute 1/3 du back into the integral for that piece. We’ll also substitute in the “u” we picked originally:

1/3 (integral sign) 1/u * du

Here, remember that the integral of 1/x is ln | x | + c. So, this integral will be:

1/3 ln | u | + c

Then we can substitute back in for “u”:

1/3 ln | x^3 + 5 | + c
 
43) This explanation would be almost exactly the same as #31 explained above. Try making the denominator equal to “u”, then find “du”, then substitute both back into the integral, then integrate.

More solutions)

58) To write F as a function of h, just get F by itself.  In other words, subtract 30 from both sides to get:

F = k/h - 30

So, to finish this part of the problem, plug in F = 0 and h = 20 to get:

0 = k/20 - 30

30 = k/20

600 = k

Your equation should now be:

F = 600/h - 30

Final Exam Prep - 1st Semester

General final exam studying help: 

Our final will feel very much like a regular chapter test, except it will be on material from the entire semester. This could be good (easy stuff like limits from Chapter 1) or tough (difficult stuff like mean value theorem from Chapter 5). One good thing to keep in mind is that you’ve been consistently working on material from September and October within the context of the problems we’re learning right now. So, hopefully the older material won’t feel like you’ve totally forgotten it.

While you’re doing the review packets, keep this in mind…

You get the problem right on the first try, without looking at notes or anything.

• This means you understand how to do this one! Move on to the next one.

You get the problem wrong on the first try, but figure out how to do it after looking at the answer key.

• This means you need to keep working on this type of problem. Find another similar one to practice it, or have me or someone else make a new one up for you.

You get the problem wrong, and can’t figure out how to do it, even after looking at the answer key and your notes.

• Ask someone for help ASAP! That means calling or texting a friend, or emailing me this evening.

I hope this helps! Let me know what questions you have!

Final Exam Review Packet help:

11) You could use any definition of a derivative formula. Usually, people find it easiest to use the forward difference quotient:

lim    f(x + h) - f(x)
h->0        h

So, we need to find f(x) and f(x + h):

f(x) = x^3

f(x + h) = (x + h)^3

We can plug this in to the derivative formula, then begin simplifying:

lim    (x + h)^3 - x^3
h->0        h

Write out (x + h)^3 three times, then distribute. I would distribute two of the factors together, then distribute the third. Here are the steps:

(x + h)(x + h)(x + h)
(x + h)(x^2 + 2xh + h^2)
x^3 + 2x^2h + xh^2 + hx^2 + 2xh^2 + h^3
Then combine like terms: x^3 + 3x^2h + 3xh^2 + h^3

Here’s what your derivative formula looks like now:

lim    x^3 + 3x^2h + 3xh^2 + h^3 - x^3
h->0                h

The x^3 and -x^3 cancel out:

lim    3x^2h + 3xh^2 + h^3
h->0            h

Then, we can cancel an h from every term, because it’s common to all the terms on top and bottom:

lim    3x^2 + 3xh + h^2
h->0  

Now, we can finally find the limit as h approaches 0. Plug in 0 for each h:

3x^2 + 3x(0) + (0)^2

= 3x^2

Which is what we would get from doing the plain old power rule! :)

12) The limit of a sum property says that you can do this:

lim    x^2 + 6x + 10
x—>1

and get an answer of 17, rather than having to split it up like this:

lim    x^2    +    lim    6x    +    lim    10
x—>1               x—>1             x—>1

and get an answer of 1, an answer of 6, and an answer of 10, then add them together to get 17.

This is not a property that I’ll test you on specifically! The question is just meant to make you think about limits and how to find them. :)

17) This wants you to use the quotient rule to find the derivative of tan x. We can do this, because we know:

tan x = sin x / cos x

Here’s the set-up for the quotient rule:

u = sin x
u’ = cos x

v = cos x
v’ = -sin x

Quotient rule:

cos x*cos x - sin x*-sin x
        (cos x)^2

But, cos x*cos x is the same as saying (cos x)^2. The two negatives in the second term combine, so we can rewrite as:

(cos x)^2 + (sin x)^2
    (cos x)^2

Then, we need to recognize the numerator as one of our fundamental trig identities: (sin x)^2 + (cos x)^2 = 1. Now our equation simplifies to this:

      1        
(cos x)^2

Then the reciprocal of cos x is sec x, which gives us the answer of:

(sec x)^2

29) This one is tough to explain in writing, but here’s my best shot:

dx means the same thing as “change in x”, or “delta x”. Draw a Riemann Sum-like rectangle on the graph (however wide and wherever you want), and label the width of it dx. That’s part of what a Riemann Sum rectangle shows: its width shows a change in x from one x-value to the next x-value.

When you label a point (x, F) in that rectangle, that is supposed to help show you that the height of your rectangle is F.

Write an equation for the area of that rectangle:

height is F * width is dx

= F * dx

So, your equation is:

dW= F dx

The dW is the derivative of W. Think about what you would have to do to each side of your equation to make it an equation with regular W: you’d have to integrate each side (because an integral is the opposite of a derivative):

dW = F dx

(integral sign) dW = (integral sign) F dx

W = (integral sign) F dx

Look, it’s like a regular integral set up! You have an equation F with a dx next to it inside of the integral sign. This is because every single integral we work with is just the area of a rectangle, where the equation is the height of the rectangle and the width is dx.

31) It's not a proof that you need to know for the final exam tomorrow. Notice that it only wants you to “sketch a graph” to show that you understand the conclusion of the mean value theorem. So, your graph should be continuous and differentiable, with an interval from x = a to x = b with a secant line connecting a and b. Then, find an x = c (and label it) where there’s a parallel tangent line (and draw the tangent line). This would be a complete answer.

Bicycle #2) Remember, the sign of velocity indicates direction. So, the cyclist is changing directions whenever velocity changes from negative to positive or vice versa. This happens on our v(t) graph at t = 3 only.

Bicycle #3) In the intro to the problem, it says that “east is considered the positive direction”. So, the cyclist is traveling east whenever v(t) has positive values. This happens on our v(t) graph in the interval [0, 3). t = 0 is included because it says that the cyclist is starting at a velocity of approximately 42mph. t = 3 is NOT included, because the cyclist is traveling 0 mph at that moment (which means stopped, and not traveling anywhere).

Bicycle #5) Because the given graph is v(t), the derivative of it will be acceleration a(t).  This problem wants you to find the "average acceleration", which is the same as the average derivative, which is the same as the average slope.  In other words, you need to find the slope from x = 0 to x = 9.  Estimate the y-values for both points, then find the slope!

Bicycle #6) Because the given graph is v(t), the derivative (the slope) of it will be acceleration a(t).  So, this question is asking you to estimate the slope of the tangent line at x = 5.

Bicycle #8) This will also be an estimate.  This question is the same as asking: "Where is the instantaneous slope the same as the average slope?"  You found the average slope in Problem #5, so now you need to estimate where the instantaneous slope would be the same.  To do this, estimate the two locations where the tangent line would be parallel to the secant line from x = 0 to x = 9.

Here’s more of an explanation:

One of the things to keep in mind when you’re looking at a velocity graph or equation is this:

Derivative (slope) of displacement = velocity
Derivative (slope) of velocity = acceleration

So in this problem, when they’re asking about “instantaneous acceleration” and “average acceleration”, they mean “instantaneous slope” and “average slope” for this v(t) graph. In other words, they’re asking you to apply Mean Value Theorem! Average slope is finding the slope of the entire interval (the secant line slope) and instantaneous slope will be finding the derivative at a certain point.

To actually do this problem, where you’re not given an equation, you will need to draw lines and estimate. Start by drawing the secant line from x = 0 to x = 9. Then, estimate the other two points on your graph where the tangent line would be parallel to this secant line.

To draw the graph of a derivative) On your given graph, any max or min (i.e. a point with a slope of 0) will be a zero on the derivative graph.  Remember, the slopes on your given graph will be values on the derivative graph.  So, if you're given a segment with positive slopes, then your derivative graph will have positive y-values at those x-values.

Video: Final review #18 (Semester 1)