Calculus 1 : How to find differential functions

Study concepts, example questions & explanations for Calculus 1

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Example Questions

Example Question #41 : How To Find Differential Functions

Given:

Evaluate the limit:

Possible Answers:

Correct answer:

Explanation:

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:

Possible Answers:

0

Correct answer:

Explanation:

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:

Possible Answers:

Correct answer:

Explanation:

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:

Possible Answers:

-

Correct answer:

Explanation:

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:

Possible Answers:

Correct answer:

Explanation:

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:

Possible Answers:

Correct answer:

Explanation:

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:

Possible Answers:

Correct answer:

Explanation:

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.

Possible Answers:

Correct answer:

Explanation:

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.

Possible Answers:

Correct answer:

Explanation:

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.

Possible Answers:

Correct answer:

Explanation:

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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