### SOLUTIONS TO DIFFERENTIATION OF INVERSE TRIGONOMETRIC FUNCTIONS

SOLUTION 11 : Differentiate arc . What conclusion can be drawn from your answer about function y ? What conclusion can be drawn about functions arc and ? First, differentiate, applying the chain rule to the inverse cotangent function. Then   = 0 .

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

i.e.,

arc for all admissable values of x .

In particular, if x = 1 , then

arc i.e., .

Thus, c = 0 and arc for all admissable values of x . We conclude that

arc .

Note that this final conclusion follows even more simply and directly from the definitions of these two inverse trigonometric functions.

SOLUTION 12 : Differentiate . Begin by applying the product rule to the first summand and the chain rule to the second summand. Then    .

SOLUTION 13 : Find an equation of the line tangent to the graph of at x=2 . If x = 2 , then , so that the line passes through the point . The slope of the tangent line follows from the derivative      (Recall that when dividing by a fraction, one must invert and multiply by the reciprocal. That is .)  .

The slope of the line tangent to the graph at x = 2 is .

Thus, an equation of the tangent line is or or .

SOLUTION 14 : Evaluate . Since and , it follows that takes the indeterminate form `` zero over zero.'' Thus, we can apply L'H pital's Rule. Begin by differentiating the numerator and denominator separately. DO NOT apply the quotient rule ! Then = = (Recall that when dividing by a fraction, one must invert and multiply by the reciprocal. That is .)

= = = .

SOLUTION 15 : A movie screen on the front wall in your classroom is 16 feet high and positioned 9 feet above your eye-level. How far away from the front of the room should you sit in order to have the ``best" view ? Begin by introducing variables x and . (See the diagram below.) From trigonometry it follows that ,

so that . so that .

It follows that  ,

that is, angle is explicitly represented as a function of distance x . Now find the value of x which maximizes the value of function . Begin by differentiating function and setting the derivative equal to zero. Then   . .

Now solve this equation for x . Then iff iff iff iff iff feet .

(Use the first or second derivative test (The first derivative test is easier.) to verify that this value of x determines a maximum value for .)

Thus, the ``best'' view is found x=15 feet from the front of the room.