Find the derivative of the function
Using the chain rule and the formula for the derivative of arcsine we have
This is valid for all such that .
Find the derivative of the function
Using the chain rule and the formula for the derivative of arcsine we have
This is valid for all such that .
Find the derivative of the function
By definition of arcsecant we know
We computed the derivative of in this exercise (Section 6.22, Exercise #4) so,
valid for .
Find the derivative of the function
Using the formula for the derivative of and the chain rule we have
Since the domain of is , this formula is valid for .
Find the derivative of the function
Using the chain rule and the derivative of arcsine we have
This is valid for (since the domain of is ).
Why is this not a violation of the zero-derivative theorem (Theorem 5.2 in Apostol)?
Then, since , we have
Next, let . Then, Again, using that , we have
Hence, there is no constant such that for all
This is not a violation of the zero-derivative theorem since the function is constant on every open interval on which it is defined. Since it isn’t defined at , any open subinterval must be a subinterval of only positive or only negative reals. The function is constant on any of these subintervals.
Establish the following formula for the derivative of is correct,
For let
Then we know
Therefore, by Theorem 6.7 (p. 252 of Apostol) we have,
Where we use the trig identity in the final line. Then, since we have,
Establish the following formula for the derivative of is correct,
For let
Then we know
Therefore, by Theorem 6.7 (p. 252 of Apostol) we have,
Where we use the trig identity in the final line. Then, since we have,
Establish the following formula for the derivative of is correct,
For let
Then we know
Therefore, by Theorem 6.7 (p. 252 of Apostol) we have,
Using a trig identity for tangent and cosecant we have
since . Therefore we conclude,
Establish the following formula for the derivative of is correct,
For let
Then we know
Therefore, by Theorem 6.7 (p. 252 of Apostol) we have,
Using a trig identity for tangent and secant we have
since . Therefore we conclude,
Establish the following formula for the derivative of is correct,
For let
Then we know
Therefore, by Theorem 6.7 (p. 252 of Apostol) we have,
Using the pythagorean identity for sine and cosine we have
since . Furthermore, since on the range of (i.e., for ) we must take the positive square root. Therefore we conclude,