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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
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You are here: Experts > Science > Math for Kids > Math for Kids > ap calculus ab
Expert: Brian Davidson - 11/3/2009
Question
my problems (set 1):
In Exercises 4, 6, 10, 15, 17, 18 use the result of these problems to find dy/dx or its derivative.
4. y=(1/2)((x^4)+7).
6. y=(radical[2])(x)+(1/(radical[2])).
10. y=(1/a)(x²+(1/b)(x)+c) (a, b, c constant).
15. y=(3x²+6)(2x-(1/4)).
17. y=(x³+7x²-8)((2x^-3)+(x^-4)).
18. y=((1/x)+(1/x²))(3x³+27).
For Exercise 43, find dy/dx or its derivative using the extension of the product rule.
43. Find d/dx [(2x+1)(1+(1/x))((x^-3)+7)].
Answer
Thanks for your question. These questions, I believe, can all be answered with the use of either the product or quotient rule. Since this is AP Calculus, I'll assume you are familiar with the "general forms" of these rules/formulas.
4. For this problem, we could use the quotient rule, but that would be unnecessary since there is no variable in the denominator. Instead, we see that the expression simplifies to 1/2x^4 + 7/2. Taking the derivative of each term, we find that dy/dx = d/dx(1/2x^4) + d/dx(7/2) = 2x^3.
6. In this problem, the only term with a variable (x) is the first term. Therefore, just as in problem 4, the other term will disappear when we take the derivative. The derivative of the expression will be d/dx((rad2)*x)) +d/dx(1/rad2)) = rad(2)
10. We multiply through by 1/a to simplify. This gives us: y = (1/a)*x^2 + (1/ab)*x + c/a. Taking the derivative of each term separately, we find dy/dx, which is simply 2/a(x) + 1/(ab).
15. We can either simply by multiplying out the separate terms, or use the product rule--the answer will be the same either way. I will use the product rule for simplicity. To find the derivative (by the product rule), I take d/dx(3x^2+6)*(2x-1/4) + d/dx(2x-1/4)(3x^2+6). Since the derivative of (3x^2+6) is 6x, and the derivative of (2x-1/4) is 2, we are left with: 6x(2x-1/4) + 2(3x^2+6). This is the answer.
17. This looks ugly (and it is, in fact, uglier than the previous problem), but its solution can be found the same way. We take: d/dx(x^3+7x^2-8)*(2x^-3 +x^-4) + d/dx(2x^-3 +x^-4)*(x^3+7x^2-8). The derivative of (x^3+7x^2-8) is 3x^2+14x and the derivative of (2x^-3 +x^-4) is -6x^-4-4x^-5. Multiplying out as described above, we find the answer: (3x^2+14x)(2x^-3 +x^-4) + (-6x^-4-4x^-5)(x^3+7x^2-8).
18. I would multiply this one out before differentiating... Some nice things will happen. The answer is 6x+3-27/x^2-54/x^3. I'll leave the written solution for you to practice with.
43. There are several ways to attack this problem. I will go about it in the way that is recommended (extension of the product rule), although I think this is a rather inefficient method for this particular problem.
Recall that in the product rule, we start out with two expressions with "x". Call these two expressions f(x) and g(x). Then, if we are trying to find the derivative of f(x)*g(x), the answer is f(x)'*g(x) + g(x)'*f(x) where an apostrophe (') denotes the derivative of one of these functions. In problem number three, they refer to an "extension" of the product rule. What they are saying is this... Suppose instead of just f(x) and g(x) being multiplied, we have THREE functions of "x" being multiplied. Suppose we are trying to find the derivative of f(x)*g(x)*q(x) where "q(x)" is that third function. Although the product rule taught in most textbooks only deals with two functions of "x" being multiplied, the rule is the SAME with three. The derivative is simply:
f(x)'*g(x)*q(x) + g(x)'*f(x)*q(x) + q(x)'*f(x)*g(x) (They each take their turn being differentiated). This "extension" of the product rule works for ANY number of multiplied functions (not just 2 or 3), but the proof is very in-depth, so I'll leave it out.
Therefore, in problem number 43, we shall call (2x+1) our f(x), (1+1/x) our g(x), and (x^-3 +7) our q(x). Then, we simply multiply them out as described above, taking the derivatives of each. The answer is 14-7/x^2-4/x^3-9/x^4-4/x^5.
Hope that helps,
Brian
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