Karl's Calculus Tutor - Solution to Exercise 7.2-1

Solution to Exercise 7.2-1

Step 1: If it is true that f(x) = arctan(x), then it is also true that

   tan(f(x)) = x

This is because tan(x) and arctan(x) are inverse functions of each other, and the above equation follows by the definition of inverse functions. Now apply the chain rule to take the derivative of both sides of this equation. When you are done, click here.

Step 2: When you applied the chain rule, you should have gotten,

   (1 + tan²(f(x)) )f'(x)  =  1

assuming you observed the hint in the original text of the problem. But recall now that in step 1 you determined that tan(f(x)) = x. Use that to make a substitution into the above equation. When you are done, click here.

Step 3: When you substituted x for tan(f(x)), you should have gotten

   (1 + x²)f'(x)  =  1

Now simply solve for f'(x), and you will be done. When you are, click here.

Step 4: You should have gotten,

                1
   f'(x)  =        
             1 + x²

which is indeed the derivative of f(x) = arctan(x). Once again we see that the derivative of a so-called transcendental function is a simple algebraic function. In this case the derivative doesn't even involve a square root. This is yet another example of how calculus weaves seemingly separate threads, like what you learned in algebra and what you learned in trig, into the same cloth. And I'm sure you will not be surprised a semester or two from now when you learn that the arctangent function is closely related to the natural log function (recall what the derivative of the natural log is).

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