	AllExperts > Math for Kids
	Search

About Brian Davidson
Expertise
I can answer questions ranging from Pre-Algebra through AP Calculus BC (first year college calculus), as well as some questions in Discreet Mathematics (probability, matrix theory, graph theory, and combinatorics).

Experience
From my earliest years as a math student, I have dedicated myself to learning the concepts and discipline of mathematics. In addition to a rigorous mathematics course sequence throughout junior high and then high school, I have served for four years on my school's math team and won several first place awards in both state and national competitions. I earned a perfect score on the AP Calculus BC exam, and (in addition) have done in-depth course work in discrete mathematics and differential calculus. Of course, I more commonly find myself offering aid in less advanced forms of math, which is why "Math for Kids" is perfect for me, also given my education credentials and previous experience.

Education/Credentials
Although I am just a college student, my educational experiences are broad. I have served as a certified student tutor for mathematics and physics both inside and outside of high school (2005 to 2009), served as a student leader by helping teach an integrated science/math class at my high school, where I was able to develop my teaching style through lessons and group interaction. Although I have much to learn before beginning to teach professionally (as I plan to do), I am proud to have helped over 80 students achieve success in mathematics over a four year period, all while expanding my own knowledge and appreciation for the discipline. And although I have studied many advanced forms of mathematics, some of my most successful students came to me for help with pre-algebra, algebra, or elementary geometry.

Awards and Honors
Outstanding Student Teacher Award (2009), Senior Student Tutor/Seminar Lecturer (2008), 3 First-place mathematical olympiad finishes, 4-second place.

Past/Present Clients
I have tutored between a 80 and 90 fellow students (both same age and younger) during high school before my graduation, although for privacy reasons, I'd rather not give specific names in such an online context. If more information is needed, please feel free to contact me through private email at bldavidson1990@hotmail.com

You are here: Experts > Science > Math for Kids > Math for Kids > ap calculus ab

Math for Kids - ap calculus ab

Expert: Brian Davidson - 11/6/2009

Question
my problems (set 2):
For Exercises 9, 10, 11, 12, 13, 19, 21, 22, 23, and 24 find the derivative of each expression and simplify. NOTE: Differentiate with respect to x.
9. (1/a)((1/b)(x²)-(2/a)(x)-(d/x))
10. (-8x^-8)+(12√x)
11. (6x^-7)-(4√x)
12. (x^-5)+(1/x^8)
13. (√x)+(1/x³)
19. (x²+8x-4)((2x^-2)+(x^-4))
21. (x+1)³
22. (√x)+(3rd root of x)+(3rd root of (x²))
23. x(2x+7)(x-2)
24. (√x)((3rd root of x)+(5 root of x))

my problems (set 3):
2. What is the limit of (4-x²)/(x²-1) as x approaches ∞?
(A) 1, (B) 0, (C) -4, (D) -1, (E) ∞
4. What is the limit of (x/x) as x approaches 0?
(A) 1, (B) 0, (C) ∞, (D) -1, (E) nonexistent
6. What is the limit of (4-x²)/(4x²-x-2) as x approaches ∞?
(A) -2, (B) (-1/4), (C) 1, (D) 2, (E) nonexistent
7. Consider the function (this is a piecewise function) f(x)=(x²/x) if x≠0 or 3 if x=0. The function:
(A) is continuous everywhere
(B) is continuous except at x=0
(C) has a removable discontinuity at x=0
(D) has an infinite discontinuity at x=0
(E) has x=0 as a vertical asymptote
9. What is the limit of sin(x) as x approaches ∞?
(A) is nonexistent, (B) is infinity, (C) oscillates between -1 and 1 (D) is zero (E) is 1 or -1
10. Consider the function (this is a piecewise function) if f(x)=((x²-x)/(2x) for x≠0 or f(0)=k. If f is continuous at x=0, then what is k equal to?
(A) -1, (B) (-1/2), (C) 0, (D) (1/2), (E) 1
51. At which point(s) does the graph of the equation y=((1/3)(x³))-((3/2)(x²))+(2x) have a horizontal tangent line?
56. Find the value of a and b if the tangent to y=(a/x²)+b at (2,4) has slope m-sub-tan=-2.
58. Find k if the curve y=x²+k is tangent to the line y=2x.

Answer
Hi Asad,

Good questions. I am more than happy to answer all of them (that's what I am here for!), but please keep in mind that i can't possibly give a lot of in-depth explanation for each one when there are so many (and it would probably bore you even if I did).. I will do my very best to be complete, though..as always!

PART A
9. We combine the last two terms (since they both have an "x" to the first power) and differentiate just like we would with any normal polynomial. We get 2/(ab)*x-(2/a+d).

10. The first term isn't bad...It differentiates nicely into 64x^7. The second term involves a square root sign. Recall that rad(x) is equal to x^(1/2) by definition. Therefore, to find the derivative of the second term, we simply treat the exponent as 1/2. We get 12*1/2*x^(1/2-1) = 6x^(-1/2). This is equal to 6/rad(x). So the final answer (both terms included) is 64x^7 + 6/rad(x).

11. We follow the same process as in number 10. The answer is -42x^6-2/rad(x).

12. d/dx(x^-5)= -5x^(-6) (This is by the standard differentiation formula for an exponential term). And d/dx(x^-8) = -8x^-9. So the final answer is -5x^(-6)-8x^(-9).

13. Recall that the derivative of rad(x) is found by viewing rad(x) as x^(1/2) and using standard differentiation (1/2*x^(-1/2)) is the resulting derivative. And the derivative of 1/x^3 (which equals x^-3) is -3x^-4. So we have 1/(2*rad(x))-3x^-4.

19. This is kind of an ugly one. I wouldn't multiply it out, because that would take a long time. I'd use the product rule. We take d/dx(x^2+8x-4)*((2x^-2)+(x^-4)) + d/dx((2x^-2)+(x^-4))*(x^2+8x-4).   After computing, I get the final answer to be (2x+8)(2/x^2+1/x^4) + (-4/x^3+-4/x^5)(x^2+8x-4). I'll leave it to you to simplify, although I don't believe that simplifies much.

21. Normally, we've just considered functions like x^3. Now, we must differentiate (x+1)^3. It turns out (in this case) that the process is exactly the same. We treat (x+1) as if it were simply a single variable, and differentiate (basically just ignoring the 1).   So we have 3(x+1)^2.   That is the answer. Later in this course, you will learn about something called the "Chain rule" which applies to more complex functions. In this case, because the coefficient on "x" was 1, we could treat (x+1) like simply a variable when differentiating. If the expression would have been (2x+1)^3, however, the answer would NOT have been simply 3(2x+1)^2. You will get to that later in the course, I am sure.

22. The first two terms are rather straight forward. d/dx(radx)= 1/(2radx) and d/dx(x^1/3) = 1/(3x^-2/3). For the last term, we note that the cube root of x^2 is actually (by definition) x^(2)^(1/3) which is equal to x^(2/3) by exponent rules. So we have d/dx(x^2/3)= 2/3*x^(-1/3). Therefore, the final answer (putting them all together) is 1/(2radx)+1/(3x^-2/3)+2/3*x^(-1/3).

23. I'd just multiply it out...Don't worry about the extended product rule here..thats just making life harder than it needs to be. Multiplying through by x, we get (2x^2+7x)(x-2) which is equal to 2x^3+3x^2-14x. The derivative, therefore, is 6x^2+6x-14.

24. The answer is 5/6x^(-1/6)+1/5x^(-4/5). I'll leave it to you to write the solution for practice.

PART B
2. I am assuming that you are now familiar with L'Hopital's rule for determining the limits of expressions that approach infinity/infinity or 0/0. However, in case you are not, I will give you solutions that do NOT require L'Hopital's rule. If we look at (4-x²)/(x²-1), there is something very clever we can do to make the answer visible. Let's factor out an x^2 of both the numerator and denominator. In other words, let's divide both the numerator AND denominator by x^2 (this should NOT change the value of the expression, because we are doing it to both the numerator and denominator!). We then have (4/x^2-1)/(1-1/x^2). You are probably wondering why this is any easier to work with. Well, think about it this way... If "x" approaches infinity, what happens to both 4/x^2 and 1/x^2? Thats right..they approach zero! So if they approach zero in the new fraction above, our new fraction simply becomes (-1/1)=-1. This is the answer we are looking for.

4. The same process can be used here as in #2. We factor out an "x" from the numerator and denominator, leaving us with an equivalent new fraction, 1/1. As x goes to zero, 1/1 is still just 1/1! So the answer is 1. (Again, we are able to do this factoring out of variables because we do it to BOTH the numerator AND denominator, keeping the fraction's value identical).

6. The answer is -1/4. I'll leave it to you to figure out why (hint: just use the steps I highlighted above in #2 and 4, factoring out an x^2 of the numerator and denominator, then seeing how the expression behaves as x approaches infinity).

7. The function f(x) simplifies to just "x" ONLY if x does not equal zero.   But since we are given that x does not equal zero when this function is valid, we are allowed to simply divide by x! So f(x)=x, plain and simple. Therefore, f(1)=1, f(.5)=.5, and f(0.0001)=0.0001. Now this creates a problem with the other half of the piecewise function. Note that if x=0, then f(x)=3. But we know that when we get very close to x=0, f(x) does NOT approach "3"! (See the previous sentence). Therefore, we have a 'rogue' point, (0,3), which is not continuous with the rest of our function, which approaches 0 as x approaches 0. Therefore, the only POSSIBLE correct answers could be B or C. But for a removable discontinuity, the limit has to approach the point in question, which is does not here (it approaches 0 as x goes to 0, not 3), so the answer is "B"... It is continuous everywhere except at x=0,

9. This limit does not exist. From a visual standpoint, it makes sense...the sine curve oscillates between -1 and 1, never 'leveling off' or approaching a fixed y value. We could draw it to infinity, and it still wouldn't choose if it wanted to stay at -1 or 1, so the limit is nonexistent. Without getting into something called "epsilon/delta proofs", which can be very advanced, I don't see an easy way to find that answer purely mathematically (without making a visual argument). But that's the answer.

10. If the function is to be continuous, then the value of (x^2-x)/(2x) when x approaches 0 has to equal k.   But we can't just plug x=0 into (x^2-x)/(2x) because we will get an undefined answer. We must treat (x^2-x)/(2x) as a fraction that we want to find the limit of as x approaches 0 (using the method I described in the first few problems of part A). For practice, I'll leave it to you to find this limit and set it equal to "K" for the answer.

51. Horizontal tangents arise when the slope of a function is zero (by the definition of a horizontal tangent). So we must find where the given function's slope (derivative) is zero. Differentiating, we get x^2-3x+2. Setting this equal to 0, we find that x=1 or x=2. So horizontal tangents occur when x=1 or x=2. To find the points, we simply plug 1 and 2 in for "x" in the original function (to find the "y" value). The points are (1,5/6) and (2,2/3).

56. I am a little confused by the way you typed the question. What does m-sub-tan mean? I'll assume that you just meant m=-2. Differentiating, we find that dy/dx=2ax. So at (2,4), the slope will be the derivative at (2,4), which is 2(a)(2) =4a. Since 4a=-2, we find that a=-1/2.

58. The line y=2x has a slope of 2. Therefore, if y=x^2+k is to be tangent to y=2x, then the slopes must be equal. Differentiating x^2+k, we find that the slope is 2x. Therefore, since the slopes must be equal for a tangent point to take place, we know that 2x=2, so x=1. Therefore, by y=2x, y=2. And if y=2, then (using the first equation now), 2= (1)^2 + k. So k=1.

Hope that helps!

-Brian

Add to this Answer Ask a Question