All Calculus 1 Resources
Example Questions
Example Question #41 : How To Find Differential Functions
Given:
Evaluate the limit:
Recalling properties of exponents:
Limit of product is the product of the limits:
And from the Pre-Question Text:
So:
Example Question #42 : How To Find Differential Functions
Given:
Evaluate the limit:
0
To solve this problem we use the variable substitution:
Obtaining:
And from the Pre-Question Text:
So:
Example Question #43 : How To Find Differential Functions
Given:
Evaluate the limit:
To solve this problem we use the variable substitution:
Obtaining:
We then observe that:
Product of limits is the limit of the products:
And from the Pre-Question Text:
So:
Example Question #44 : How To Find Differential Functions
Given:
Evaluate the limit:
-
Multiplying by we obtain:
Limit of product is the product of limits:
And from the Pre-Question Text:
and
So:
Example Question #232 : Functions
Given:
Evaluate the limit:
First observe that
Multiplying by and we obtain:
Limit of product is the product of limits:
And from the Pre-Question Text:
So:
Example Question #233 : Functions
Given:
Evaluate the limit:
Multiplying by we obtain:
Limit of product is the product of limits:
And from the Pre-Question Text:
So:
Example Question #45 : How To Find Differential Functions
Given:
Evaluate the limit:
Multiplying by we obtain:
Limit of product is the product of limits:
And from the Pre-Question Text:
So:
Example Question #1262 : Calculus
Differentiate the function using known derivatives and applying the product, quotient, and chain rules.
Using the product rule, ,
,
we observe the following:
, which is our final answer.
Example Question #1261 : Calculus
Differentiate the function using known derivatives and applying the product, quotient, and chain rules.
We will need to use either the chain rule or product rule on both summands. The chain rule involves less simplifiction at the end. It states: , .
In this case, the outside function is, in fact, in both cases.
For the first summand:
And for the the second:
, which is our final answer.
[A function with a derivative of zero is a constant. You might recall the trig identity:
, which is another way to solve this problem.]
Example Question #1266 : Calculus
Differentiate the function using known derivatives and applying the product, quotient, and chain rules.
You could expand this problem out, but let's evaluate the derivative using a modified product rule:
.
Setting each of the multiplicands as , , and , respectively, we get:
, which is our final answer.
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