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Lecture Comments (4)

0 answers

Post by Colby Trace on April 4, 2013

I had same question^^

2 answers

Last reply by: Andrew Mu
Tue Jul 9, 2013 3:51 PM

Post by Redenna Chan Jing Si on May 11, 2012

For the last example in this video, shouldn't the derivative of the numerator have an extra "6x^2-2" multiplied to it?

Derivatives of Inverse Trigonometric Functions

  • Memorize!
  • Chain rule often helpful
  • Other derivative rules still apply

Derivatives of Inverse Trigonometric Functions

Find f′(x) if f(x) = earcsinx
  • f′(x) = [d/dx] earcsinx
  • = earcsinx [d/dx] arcsinx
  • = earcsinx [1/(√{1 − x2})] =
[(earcsinx)/(√{1 − x2})]
Find f′(x) if f(x) = cos(arcsinx)
  • f′(x) = −sin(arcsinx) [d/dx] arcsinx
  • = [(−sin(arcsinx))/(√{1− x2})] =
[(−x)/(√{1 − x2})]
Find f′(x) if f(x) = [(ex)/arccosx]
  • f′(x) = [(arccosx [d/dx] ex − ex [d/dx] arccosx)/((arccosx)2)]
  • = [(ex arccosx − ex [(−1)/(√{1 − x2})])/((arccosx)2)]
[(ex arccosx + [(ex)/(√{1 − x2})])/((arccosx)2)]
Find f′(x) if f(x) = lnx (arctanx)
  • f′(x) = lnx [d/dx] arctanx + arctanx [d/dx] lnx
lnx ([1/(1 + x2)]) + arctanx ([1/x])
Find f′(x) if f(x) = √{x2 + cot−1 x}
  • u = x2 + cot−1 x , u′ = 2x − [1/(1 + x2)]
  • f′(x) = [1/2] u−[1/2] u′
  • = [1/(2 √u)] u′
[1/(2 √{x2 + cot−1 x})] (2x − [1/(1 + x2)])
Find f′(x) if f(x) = sec−1 (x3 − x)
  • u = x3 − x , u′ = 3x2 − 1
  • f′(x) = [1/(|u| √{u2 − 1 })] u′
[(3x2 − 1)/(|x3 − x| √{(x3 − x)2 − 1})]
Find dy given y = csc−1 ex
  • u = ex , u′ = ex
  • dy = [(−1)/(|u| √{u2 − 1})] u′
  • = [(−u′)/(|u| √{u2 − 1})]
  • = [(−ex)/(|ex| √{(ex)2 − 1})]
  • = [(−ex)/(ex √{e2x − 1})]
[(−1)/(√{e2x − 1})]
Find f′(x) if f(x) = ln(arcsin(x)arccos(x))
  • f(x) = lnarcsinx + lnarccosx
  • f′(x) = [1/arcsinx] [d/dx] arcsinx + [1/arccosx] [d/dx] arccosx
  • = [1/(√{1 − x2} arcsinx)] − [1/(√{1 − x2} arccosx)]
[1/(√{1 − x2})]([1/arcsinx] − [1/arccosx])
Find f′(x) if f(x) = [(cos−1 x )/cosx]
  • f′(x) = [(cosx [d/dx] cos−1 x + cos−1 x [d/dx] cosx)/(cos2 x)]
  • = [(cosx [(−1)/(√{1 − x2})] + cos−1 x (−sinx))/(cos2 x)]
[([(−cosx)/(√{1 − x2})] − sinx cos−1 x)/(cos2 x)]
Prove dy = [1/(√{1 − x2})] given y = sin−1 x
  • y = sin−1 x
  • siny = sin(sin−1 x)
  • siny = x
  • [d/dx] siny = [d/dx] x
  • cosy  dy = 1
  • dy = [1/cosy]
  • Using the following identity
  • cos2 y + sin2 y = 1
  • cos2 y = 1 − sin2 y
  • cosy = ±√{1 − sin2 y}
  • We have two potential solutions, but we know something about the range and domain of these functions. For y = sin−1 x, the range is [−[(π)/2], [(π)/2]]. Using what we know about y, we can determine acceptable values for cosy. With those inputs, cosy ≥ 0
  • cosy = √{1 − sin2 y} , x = siny
  • dy = [1/cosy]
  • dy = [1/(√{1 − sin2 y})]
[1/(√{1 − x2})]

*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 Inverse Trigonometric 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.

  • Intro 0:00
  • Types of Derivatives: Inverse Trigonometric Functions 0:06
    • 6 Fundamental Properties of Inverse Trigonometric Functions
  • Example 1 2:17
  • Example 2 3:41
  • Example 3 5:37
  • Example 4 7:24
  • Example 5 10:08