Derivative defined as the limit of the difference quotient - AP Calculus AB

One should recognize the difference quotient as being equivalent to the derivative or instantaneous rate of change.

That is, for some continuous function, .

The function in our question is

We note from this that because the difference quotient is the derivative:

We use the power rule on each term of the function.

The first term

becomes

.

The second term

becomes

.

The final term, 7, is a constant, so its derivative is simply zero.

To get , we can use the power rule.

Since the exponent of the is , as , we lower the exponent by one and then multiply the coefficient by that original exponent:

Anything to the power is .

To solve this problem, we can use the power rule. That means we lower the exponent of the variable by one and multiply the variable by that original exponent.

Remember that anything to the zero power is one.

To solve this problem, we can use the power rule. That means we lower the exponent of the variable by one and multiply the variable by that original exponent.

We're going to treat as , as anything to the zero power is one.

That means this problem will look like this:

Notice that , as anything times zero is zero.

Remember, anything to the zero power is one.

To solve this equation, we can use the power rule. To use the power rule, we lower the exponent on the variable and multiply by that exponent.

We're going to treat as since anything to the zero power is one.

Notice that since anything times zero is zero.

To solve this equation, we can use the power rule. To use the power rule, we lower the exponent on the variable and multiply by that exponent.

We're going to treat as since anything to the zero power is one.

Notice that since anything times zero is zero.

That leaves us with .

Simplify.

As stated earlier, anything to the zero power is one, leaving us with:

To solve this equation, we can use the power rule. To use the power rule, we lower the exponent on the variable and multiply by that exponent.

We're going to treat as since anything to the zero power is one.

Notice that since anything times zero is zero.

Just like it was mentioned earlier, anything to the zero power is one.

To take the derivative of this equation, we can use the power rule. The power rule says that we lower the exponent of each variable by one and multiply that number by the original exponent.

Simplify.

Remember that anything to the zero power is equal to one.

To take the derivative of this equation, we can use the power rule. The power rule says that we lower the exponent of each variable by one and multiply that number by the original exponent.

We are going to treat as since anything to the zero power is one.

Notice that since anything times zero is zero.

Simplify.

As stated before, anything to the zero power is one.

To find the first derivative, we can use the power rule. We lower the exponent on all the variables by one and multiply by the original variable.

Anything to the zero power is one.

To find the first derivative, we can use the power rule. We lower the exponent on all the variables by one and multiply by the original variable.

We're going to treat as since anything to the zero power is one.

For this problem that would look like this:

Notice that since anything times zero is zero.

To find the first derivative, we can use the power rule. To do that, we lower the exponent on the variables by one and multiply by the original exponent.

We're going to treat as since anything to the zero power is one.

Notice that since anything times zero is zero.

This limit can't be evaluated by conventional limit laws. To see why the answer is , we have to recognize that the limit looks like

, with .

This new limit is a conventional expression for , with substituted in for

We can find with the power rule, and substituting gives .

To summarize, since

, we have

.

There are two commonly used formulations of the difference quotient used to compute derviatives. Both definitions are equivalent and should always give the same derivative for a given function .

The limit in this particular problem resembles the first difference quotient listed above.

By inspection, it's clear that the in our case must be . We are being asked to evaluate the limit by interpreting it as the definition of the derivative of a function. We must therefore identify the function, which is clearly , and then differentiate it using known rules of differentiation.

First rewrite the function to apply the rule for :

This means

The difference quotient of a function is the expression

.

If , this expression is

The derivative of a function can be defined according to the following:

Using this for our function - and not forgetting to write the limit for every step! - we get

The derivative of a function as defined by the limit of the difference quotient is

Evaluating the limit using our function - and always writing the limit symbol! - we get

To find the derivative using the limit of the difference quotient, we use the formula: