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For more information, please see full course syllabus of College Calculus: Level I

For more information, please see full course syllabus of College Calculus: Level I

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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) = e

^{arcsinx}- f′(x) = [d/dx] e
^{arcsinx} - = e
^{arcsinx}[d/dx] arcsinx - = e
^{arcsinx}[1/(√{1 − x^{2}})] =

[(e

^{arcsinx})/(√{1 − x^{2}})]Find f′(x) if f(x) = cos(arcsinx)

- f′(x) = −sin(arcsinx) [d/dx] arcsinx
- = [(−sin(arcsinx))/(√{1− x
^{2}})] =

[(−x)/(√{1 − x

^{2}})]Find f′(x) if f(x) = [(e

^{x})/arccosx]- f′(x) = [(arccosx [d/dx] e
^{x}− e^{x}[d/dx] arccosx)/((arccosx)^{2})] - = [(e
^{x}arccosx − e^{x}[(−1)/(√{1 − x^{2}})])/((arccosx)^{2})]

[(e

^{x}arccosx + [(e^{x})/(√{1 − x^{2}})])/((arccosx)^{2})]Find f′(x) if f(x) = lnx (arctanx)

- f′(x) = lnx [d/dx] arctanx + arctanx [d/dx] lnx

lnx ([1/(1 + x

^{2})]) + arctanx ([1/x])Find f′(x) if f(x) = √{x

^{2}+ cot^{−1}x}- u = x
^{2}+ cot^{−1}x , u′ = 2x − [1/(1 + x^{2})] - f′(x) = [1/2] u
^{−[1/2]}u′ - = [1/(2 √u)] u′

[1/(2 √{x

^{2}+ cot^{−1}x})] (2x − [1/(1 + x^{2})])Find f′(x) if f(x) = sec

^{−1}(x^{3}− x)- u = x
^{3}− x , u′ = 3x^{2}− 1 - f′(x) = [1/(|u| √{u
^{2}− 1 })] u′

[(3x

^{2}− 1)/(|x^{3}− x| √{(x^{3}− x)^{2}− 1})]Find dy given y = csc

^{−1}e^{x}- u = e
^{x}, u′ = e^{x} - dy = [(−1)/(|u| √{u
^{2}− 1})] u′ - = [(−u′)/(|u| √{u
^{2}− 1})] - = [(−e
^{x})/(|e^{x}| √{(e^{x})^{2}− 1})] - = [(−e
^{x})/(e^{x}√{e^{2x}− 1})]

[(−1)/(√{e

^{2x}− 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 − x
^{2}} arcsinx)] − [1/(√{1 − x^{2}} arccosx)]

[1/(√{1 − x

^{2}})]([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)/(cos^{2}x)] - = [(cosx [(−1)/(√{1 − x
^{2}})] + cos^{−1}x (−sinx))/(cos^{2}x)]

[([(−cosx)/(√{1 − x

^{2}})] − sinx cos^{−1}x)/(cos^{2}x)]Prove dy = [1/(√{1 − x

^{2}})] 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
- cos
^{2}y + sin^{2}y = 1 - cos
^{2}y = 1 − sin^{2}y - cosy = ±√{1 − sin
^{2}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 − sin
^{2}y} , x = siny - dy = [1/cosy]
- dy = [1/(√{1 − sin
^{2}y})]

[1/(√{1 − x

^{2}})]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

^{−1}x- 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

There are technically many inputs for sin(x) that give [(√2)/2], but remember it must remain in the range of 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) = 0Find 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

sin

^{−1}([1/2]) = [(π)/6]Find tan(cos

^{−1}[(√2)/2]) cos

tan(cos

= 1

^{−1}[(√2)/2] = [(π)/4]tan(cos

^{−1}[(√2)/2]) = tan[(π)/4]= 1

Given f(x) = sin

^{2}x and g(x) = sin^{−1}(sin3x), evaluate f(g(x))- g(x) = sin
^{−1}(sin3x) - = 3x

f(g(x)) = sin

SPECIAL NOTE: By convention sin

^{2}3xSPECIAL 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
^{−1}1 = 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.

Answer

### 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
- Six Basic Inverse Trigonometric Functions 0:10
- Lecture Example 1 1:11
- Lecture Example 2 8:53
- Lecture Example 3 12:37
- Additional Example 4
- Additional Example 5

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