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Lecture Comments (8)

2 answers

Last reply by: Quinn Hollister
Mon May 25, 2015 6:03 PM

Post by Kenneth Cumins on May 5, 2014

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.

0 answers

Post by GEORGE KAMAU on June 3, 2013

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

0 answers

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

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