INSTRUCTORS Raffi Hovasapian John Zhu

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• ## Related Books

 0 answersPost by Colby Trace on April 4, 2013I had same question^^ 2 answersLast reply by: Andrew MuTue Jul 9, 2013 3:51 PMPost by Redenna Chan Jing Si on May 11, 2012For 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!
• 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.