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Post by Kenneth Cumins on May 5 at 09:43:19 PM

Nearly every video in the Calculus AB playlist has some type of mathematical error. That's pretty unprofessional for a website that charges such a large amount of money per month.

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Post by GEORGE KAMAU on June 3, 2013

how did 2XSQUARED+9 COME TO PLAY? THOUR OUR du/dx=9xsquared+2..

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Post by Riley Argue on April 25, 2012

there is an error

2 answers

Last reply by: Ash Niazi
Fri Dec 28, 2012 2:55 PM

Post by Daniel Gesla on April 9, 2012

@ 2:29 in this video how did you come up with 2x^2 + 9 for du/dx. I thought du/dx was 9x^2 + 2

Derivatives of Logarithmic Functions

  • All other derivative rules still apply
  • Substitute term inside log with “u” to attain ln(u).

Derivatives of Logarithmic Functions

Find dy if y = ln(g(x))

  • y = ln(g(x))
  • ey = eln(g(x))
  • ey = g(x)
  • [d/dx] ey = [d/dx] g(x)
  • ey dy = [d/dx] g(x)
  • g(x) dy = [d/dx] g(x) =

[d/dx] ln(g(x)) = [1/g(x)] [d/dx] g(x)

Find f′(x) if f(x) = ex ln x

  • f′(x) = [d/dx] ex lnx
  • = ex [d/dx] lnx + lnx [d/dx] ex
  • = ex [1/x] [d/dx] x + lnx ex
  • = [(ex)/x] + ex lnx =

ex([1/x] + lnx)

Find dy if y = eln(x2)

  • dy = [d/dx] eln(x2)
  • u = lnx2, u′ = [1/(x2)] [d/dx] x2 = [2x/(x2)] = [2/x]
  • Alternatively, u = lnx2 = 2 lnx, u′ = [2/x]
  • dy = [d/dx] eu
  • = u′eu
  • = [2/x] eln(x2)
  • = [2/x] x2
  • = 2x
  • It would have been much easier to just simplify in the first place.
  • y = eln(x2)
  • = x2
  • dy = 2x

dy = 2x

Find dy if y = xx

  • The power rule does not apply here because x is not a constant. Other methods must be used.
  • y = xx
  • lny = lnxx
  • lny = x lnx
  • [d/dx] lny = [d/dx] x lnx
  • [1/y] dy = x [d/dx] lnx + lnx [d/dx] x
  • [1/(xx)] dy = x ([1/x]) + lnx
  • [1/(xx)] dy = 1 + lnx

dy = xx (lnx + 1)

Find [(d2 y)/(dy2)] if y = x2 ln x

  • dy = [d/dx] x2 lnx
  • = x2 [d/dx] lnx + lnx [d/dx] x2
  • = x2 [1/x] + lnx (2x)
  • = x + 2x lnx
  • [(d2 y)/(dx2)] = [d/dx] (x + 2x lnx)
  • = 1 + 2x [d/dx] lnx + lnx [d/dx] 2x
  • = 1 + [2x/x] + 2 lnx
  • = 1 + 2 + 2 lnx =

3 + 2 lnx

Find f′(x) if f(x) = [1/ln(3x)]

  • f′(x) = [d/dx] [1/ln(3x)]
  • = [(ln(3x) [d/dx] 1 − 1 [d/dx] ln(3x))/((ln(3x))2)]
  • = [(0 − [3/3x])/((ln(3x))2)] =

[(−1)/(x (ln(3x))2)]

Find f′(x) if f(x) = ln([(ex)/x])

  • f(x) = ln([(ex)/x])
  • = lnex − lnx
  • = x − lnx
  • f′(x) = [d/dx] (x − lnx) =

1 − [1/x]

Find f′(x) if f(x) = ln(lnx)

  • f′(x) = [d/dx] ln( lnx)
  • = [1/lnx] [d/dx] lnx
  • = [1/lnx] [1/x] =

[1/x lnx]

Find f′(x) if f(x) = [x/sin(lnx )]

  • f′(x) = [(sin(lnx) [d/dx] x − x [d/dx] sin(lnx))/(sin2(lnx))]
  • = [(sin(lnx) − x (cos(lnx) [d/dx] lnx))/(sin2(lnx))]
  • = [(sin(lnx) − x (cos(lnx) [1/x]))/(sin2(lnx))]
  • = [(sin(lnx) − cos(lnx))/(sin2(lnx))]
  • = [(1 − cot(lnx))/sin(lnx)] =

csc(lnx) (1 − cot(lnx))

Find f′(x) if f(x) = ln(x csc(x))

  • f(x) = ln(x cscx)
  • = lnx + ln(cscx)
  • f′(x) = [d/dx] lnx + [d/dx] ln(cscx)
  • = [1/x] + [1/cscx] [d/dx] cscx
  • = [1/x] + [1/cscx] (−cotx cscx) =

[1/x] − cotx

*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

Derivatives of Logarithmic Functions

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.