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INSTRUCTORS Raffi Hovasapian John Zhu
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Lecture Comments (5)

0 answers

Post by Lifan Yu on September 25, 2015

I think he's actually correct in the last example.

-8sin^2x+8cos^x = -8sin^2x-8cos^2x+16cos^x = -8+16cos^x. (Using pythagorean identity)

All the other terms match up.

0 answers

Post by John Zhu on August 12, 2013

There is definitely a mistake in there. Final answer is a bit long winded, but here it is:

30x^4-2x^3cos(x)+18x^2sin(x)-48xcos(x)+8cos^2(x)-8sin^2(x).

Thanks for spotting the error folks, a good example of how organization and sleep the night before the test helps boost scores!

0 answers

Post by Si Jia Wen on August 10, 2013

I think he made a mistake in the last example...

0 answers

Post by Cho HyuJang on November 10, 2012

Thaz the thing i wanna ask too? how did he get it?

0 answers

Post by Steve Denton on October 7, 2012

How did he simplify that to 16cos(sq)x + . . . . - 8 ???? The first and last terms came out of nowhere to me.

Product Rule

  • Organization is vital
  • Write more, more room to think, make less mistakes

Product Rule

Find the derivative of f(x) = g(x) * h(x) given g(x) = ln(x), g′(x) = [1/x] and h(x) = ex = h′(x)
  • f′(x) = g(x) h′(x) + h(x) g′(x)=
  • Remember, f′(x) is generally equivalent to [d/dx] f(x)
ln(x) ex + ex [1/x]
Find the derivative of y = x sin(x)
  • dy = x [d/dx] sin(x) + sin(x) [d/dx] x
  • = x cos(x) + sin(x) * 1 =
bsection*x cos(x) + sin(x)
Find the derivative of f(x) = sin(x) tan(x)
  • f′(x) = sin(x) [d/dx] tan(x) + tan(x) [d/dx] sin(x)
  • = sin(x) sec2(x) + tan(x) cos(x) =
sin(x) sec2(x) + sin(x)
Show that f′(x) = sin(2x) given f(x) = sin2(x)
  • f′(x) = [d/dx] sin2(x)
  • = [d/dx] (sin(x)sin(x))
  • = sin(x) [d/dx] sin(x) + sin(x) [d/dx] sin(x)
  • = sin(x) cos(x) + sin(x) cos(x)
  • = 2 sin(x) cos(x) =
= sin(2x)
Find the derivative of f(x) = (x3 + x + 3)(x2 + 1)
  • f′(x) = (x3 + x + 3) [d/dx] (x2 + 1) + (x2 + 1) [d/dx] (x3 + x + 3)
  • = (x3 + x + 3) 2x + (x2 + 1)(3x2 + 1)
  • = (2x4 + 2x2 + 6x) + (3x4 + 3x2 + x2 + 1) =
5x4 + 6x2 + 6x + 1
Find the derivative of f(x) = (x + 1)sec(x)
  • f′(x) = (x + 1) [d/dx] sec(x) + sec(x) [d/dx] (x + 1) =
(x + 1) sec(x) tan(x) + sec(x)
Find the derivative of f(x) = (x2 + 4)(x + 3)2
  • f′(x) = (x2 + 4) [d/dx] (x + 3)2 + (x + 3)2 [d/dx] (x2 + 4)
  • = (x2 + 4) [d/dx] (x + 3)2 + (x + 3)2 (2x)
  • = (x2 + 4) [d/dx] (x + 3)2 + (x2 + 6x + 9)(2x)
  • = (x2 + 4) [d/dx] (x + 3)2 + 2x3 + 12x2 + 18x
  • = (x2 + 4) ( (x + 3) [d/dx] (x + 3) + (x + 3) [d/dx] (x + 3) ) + 2x3 + 12x2 + 18x
  • = (x2 + 4) ( (x + 3) + (x + 3) ) + 2x3 + 12x2 + 18x
  • = (x2 + 4) (2x + 6) + 2x3 + 12x2 + 18x
  • = 2x3 + 8x + 6x2 + 24 + 2x3 + 12x2 + 18x
4x3 + 18x2 + 26x + 24
Find the derivative of f(x) = (7x + 53)2
  • f′(x) = (7x + 53) [d/dx] (7x + 53) + (7x + 53) [d/dx] (7x + 53)
14(7x + 53)
Given f(x) = [(x3)/3] + 2x2 + 3x, find the roots of f′(x)
  • The roots are given by values of x such that the function equals zero.
  • f′(x) = [d/dx] ([(x3)/3] + 2x2 + 3x)
  • = x2 + 4x + 3
  • f′(x) = 0 = x2 + 4x + 3
  • = (x + 1) (x + 3) = 0
  • x = −1, −3
  • A LOOK AHEAD: Below is the graph of f(x). Note the behavior of f(x) at the roots of its derivative.
x = −1, −3
Given f(x) = (x + 2)2 + 1, find the roots of f′(x)
  • f′(x) = [d/dx] ( (x + 2)2 + 1 )
  • = [d/dx] (x + 2)2
  • = (x + 2) [d/dx] (x + 2) + (x + 2) [d/dx] (x + 2)
  • = (x + 2) + (x + 2)
  • = 2(x + 2) = 0
  • x = −2
  • Graph of f(x)
x = −2

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

Product Rule

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Product Rule 0:07
    • Definition
    • Example 1
  • Example 2 2:11
  • Example 3 4:24
  • Example 4 5:24
  • Example 5 6:42
  • Example 6 7:51