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### Inverse Trigonometric Functions

• Check with your instructor as to whether or not you should memorize these formulas!
• One type of problem here simply incorporates inverse trigonometric functions into differentiation problems involving, for example, the Chain Rule.
• You can use implicit differentiation to derive these formulas.
• Check with your instructor as to whether or not you can use the “triangle approach” to derive these formulas.

### 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})]
Given f([(π)/6]) = [1/2] with f(x) = sinx and g(x) = sin−1 x, find g([1/2])
• An inverse function will "reverse" the action done by the original function. Kind of like working backwards.
• g(x) is the inverse of f(x)
g([1/2]) = [(π)/6]
Find the domain of f(x) = sin−1x
• The only valid inputs for x are given by the range of its inverse, sinx, [-1, 1]
Domain [-1, 1]
Find sin−1([(√2)/2])
• sin([(π)/4]) = [(√2)/2]
sin−1([(√2)/2]) = [(π)/4]
There are technically many inputs for sin(x) that give [(√2)/2], but remember it must remain in the range of sin−1 x, which is [−[(π)/2], [(π)/2]].
Find cos−1(1)
• Range of cos−1 x is [0,π].
• cos(0) = 1
cos−1(1) = 0
Find sin(cos−1[(√2)/2])
• cos([(π)/4]) = [(√2)/2]
• cos−1([(√2)/2]) = [(π)/4]
• sin(cos−1[(√2)/2]) = sin([(π)/4])
= [(√2)/2]
Find sin−1[1/2]
sin([(π)/6]) = [1/2]
sin−1([1/2]) = [(π)/6]
Find tan(cos−1[(√2)/2])
cos−1[(√2)/2] = [(π)/4]
tan(cos−1[(√2)/2]) = tan[(π)/4]
= 1
Given f(x) = sin2 x and g(x) = sin−1 (sin3x), evaluate f(g(x))
• g(x) = sin−1 (sin3x)
• = 3x
f(g(x)) = sin23x
SPECIAL NOTE: By convention sin−1 x is the inverse of sinx. (sinx)−1 is cscx.
Find at least one solution for θ given sin(θ) = 0
• Take the inverse sin of both sides
• sin(θ) = 0
• sin−1(sin(θ)) = sin−1(0)
θ = 0
Graph f(x) = 2 sin−1(3x + 1)
• We know the input for inverse sin must be within [-1, 1]. Therefore, −1 ≤ 3x + 1 ≤ 1. So [(−2)/3] ≤ x ≤ 0. Domain [[(−2)/3], 0]. Plugging in some values...
• f([(−2)/3]) = 2 sin−1(−1) = 2 [(− π)/2] = −π
• f([(−1)/3]) = 2 sin−1(0) = 0
• f(0) = 2 sin−11 = 2 [(π)/2] = π
• Increasing function with Range [−π, π]

*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.