SOLUTIONS TO DIFFERENTIATION OF INVERSE TRIGONOMETRIC FUNCTIONS

Jun 21st, 2015
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SOLUTIONS TO DIFFERENTIATION OF INVERSE TRIGONOMETRIC FUNCTIONS SOLUTION 1 : Differentiate . Apply the product rule. Then (Factor an x from each term.) . SOLUTION 2 : Differentiate . Apply the quotient rule. Then . 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.) . (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 . 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., . 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.) . If , then , and letting , it follows that . The following problems require use of th

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