### SOLUTIONS TO DIFFERENTIATION OF INVERSE TRIGONOMETRIC FUNCTIONS

SOLUTION 1 : Differentiate . Apply the product rule. Then

(Factor an x from each term.)

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SOLUTION 2 : Differentiate . Apply the quotient rule. Then

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SOLUTION 3 : Differentiate arcarc . Apply the product rule. Then

arcarcarcarc

arcarc

= ( arcarc .

SOLUTION 4 : Let arc . Solve f'(x) = 0 for x . Begin by differentiating f . Then

(Get a common denominator and subtract fractions.)

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(It is a fact that if , then A = 0 .) Thus,

2(x - 2)(x+2) = 0 .

(It is a fact that if AB = 0 , then A = 0 or B=0 .) It follows that

x-2 = 0 or x+2 = 0 ,

that is, the only solutions to f'(x) = 0 are

x = 2 or x = -2 .

SOLUTION 5 : Let . Show that f'(x) = 0 . Conclude that . Begin by differentiating f . Then

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If f'(x) = 0 for all admissable values of x , then f must be a constant function, i.e.,

for all admissable values of x ,

i.e.,

for all admissable values of x .

In particular, if x = 0 , then

i.e.,

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Thus, and for all admissable values of x .

SOLUTION 6 : Evaluate . It may not be obvious, but this problem can be viewed as a derivative problem. Recall that

(Since h approaches 0 from either side of 0, h can be either a positve or a negative number. In addition, is equivalent to . This explains the following equivalent variations in the limit definition of the derivative.)

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If , then , and letting , it follows that

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The following problems require use of the chain rule.

SOLUTION 7 : Differentiate . Use the product rule first. Then

(Apply the chain rule in the first summand.)

(Factor out . Then get a common denominator and add.)

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SOLUTION 8 : Differentiate . Apply the chain rule twice. Then

(Recall that .)

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SOLUTION 9 : Differentiate . Apply the chain rule twice. Then

(Recall that .)

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SOLUTION 10 : Determine the equation of the line tangent to the graph of at x = e . If x = e , then , so that the line passes through the point . The slope of the tangent line follows from the derivative (Apply the chain rule.)

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The slope of the line tangent to the graph at x = e is

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Thus, an equation of the tangent line is

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