Home » Difference Quotient

# Determine whether given statements about a function differentiable at a point are true or false

Let be a function and a point such that exists. Determine which of the following are true or false.

1. .
2. .
3. .
4. .

1. True. Let . Then as implies as . Thus,

2. True. Let , then

3. False. First, from the definition of limit we can conclude

(by taking ). Then we have,

Thus, this is not true in general (since there are many function differentiable at a point such that the derivative at that point is nonzero). ( Note: This statement would be true were the denominator in the given equation instead of .)

4. False. Here we evaluate,

# Find values of constants so that the derivative of a function exists

Consider the function defined by

Find values for the constants and such that the derivative exists.

We know that exists if and only if

exists. This limit exists if and only if the two one-sided limits exits and are equal:

Using the definition of , we then evaluate these limits,

In simplifying the right hand side we used that . Furthermore, for the limit on the left to exist we must have (otherwise the limit will diverge as ). Now for the expression on the right, we claim the limit in the expression is 0. We can see this because

But this limit is the derivative of at . Since and , this term is indeed 0.
So, coming back to our original equations we then have,

Furthermore, since we already established that we have,

Therefore, the expressions for and we are asked to find are,

# Find values of constants so that the derivative of a function exists

Consider the function defined by

Find values for the constants and such that the derivative exists.

First, let’s assume otherwise everywhere and the values of and are arbitrary (since the value of does not depend on them).

Then, exists implies the one-sided limits exist and are equal:

Using the definition of we plug in the expressions for when approaches from the left and from the right,

The limit on the left exists if and only if (otherwise the limit diverges to infinity has ). In this case the limit is 0, and we have two equations,

Using the equation on the right we have, . Then plugging this value into the equation on the left we obtain

Finally, using this in our expression we solve for in terms of alone, . Therefore,

# Find values of constants so that the derivative of a function exists

Consider the function defined by

Find values for the constants and such that the derivative exists.

We know that the derivative exists if and only if

exists. Furthermore, this limit exists if and only if the one-sided limits both exist and are equal:

So, plugging in the formula for (which is if we approach from the right, and is if we approach from the left, and noting that from the definition of ) we have,

For the limit on the left to exist we must have (otherwise the limit will diverge as ). Furthermore, this limit must be 0 since is a constant (and the limit of as is 0). Therefore, we have , and we have the equation

Therefore, and are the values of the requested constants in terms of .

# Use the binomial theorem to prove the formula for the derivative of x^n

Consider the function for a positive integer . Prove that

Conclude that by considering the limit of this as .

Proof. We recall the binomial theorem,

So, then we calculate,

Taking the limit as of both sides of the equation we conclude,

# Evaluate the limit

Evaluate

First, we simplify the expression,

Then, we evaluate the limit,

# Prove difference quotient formulas for sine and cosine

For prove

Proof. We use the sine and cosine of sum and difference formulas in the following computations:

And,