Let and be functions, both differentiable in a neighborhood of 0, with and such that
Prove or disprove each the following statements.
- as .
- as .
Proof. Since as we know by the definition of that
Thus, for every there exists a such that
So, for we have
The final line follows since by hypothesis. Therefore,
By definition, we then have
Consider for and for . Then, for ,
For we have .
Since we have as . However, since
does not exist.
Prove the inequality
Proof. Define a function . Then, since for we have
Since and is increasing on (since its derivative is positive) we have
For the inequality on the left (which is much more subtle), we want to show
So, we consider the function
Now, if we can show that is decreasing on the whole interval then we will be done (since this would mean on the whole interval since if it were less than somewhere then it would have to increase to get back to on the right end of the interval).
To show is decreasing on the whole interval we will show that its derivative is negative. To that end, define a function
(This is the numerator in the expression we got for . Since we know the denominator of that expression is always positive, we are going to show this is always negative to conclude is always negative.) Now, and
for . Thus, is negative, and so is decreasing. Since we then have is negative on the whole interval. Therefore, on the interval. Hence, is decreasing. Hence, we indeed have
Derive the formulas for the derivative of a product and the derivative of a quotient from the corresponding formulas for the derivative of a sum and the derivative of a difference.
We know the derivative rules for sums and differences are:
To derive the derivative rule for products using logarithmic differentiation we let and compute
This is the usual rule for derivative of a product.
Similarly, for the derivative of a quotient, let and then compute,
Which is the usual rule for derivative of a quotient.
Given a function satisfying the properties:
Prove the following:
- The derivative exists for all .
- We must have .
This problem is quite similar to two previous exercises here and here (Section 6.17, Exercises #39 and #40).
- Proof. To show that the derivative exists for all we must show that the limit
exists for all . Using the given properties of we can evaluate this limit
Therefore, for all , so the derivative is defined everywhere
- Proof. From part (a) we know . By Section 6.17, Exercise #39 (linked above) we know that the only functions which satisfy this equation are for all or for some constant (where in the linked exercise). However, since the derivative of exists everywhere, and differentiability implies continuity, we know is continuous everywhere. Hence, . Then,
since , so . Therefore, we must have for some constant . Furthermore, we must have since . Thus,
Consider the function
Compute the derivative and use the sign of the derivative to prove the inequality
Proof. First, we compute the derivative
Therefore, for all . This implies is strictly increasing if , and in particular, is increasing for . This implies
Proof. First, we consider the derivatives of the left and right side of the given equation. (Treating as a function of and remembering to use the chain rule.) So, for the derivative on the left, we have
On the right we have,
Now, using the given equation we have