Derivative of 101-x2 using Chain Rule - Calculus I Explained

  • Thread starter Thread starter UsernameValid
  • Start date Start date
  • Tags Tags
    Chain rule
Click For Summary

Homework Help Overview

The discussion revolves around finding the derivative of the function 101 - x² using calculus concepts, specifically the chain rule and power rule. Participants are exploring the correct application of these rules in the context of derivatives involving exponential functions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply the power rule and chain rule but questions the appearance of a term involving -20*ln(10) in their calculator's output. Other participants discuss the appropriate use of the power rule and the implications of variable exponents in the context of derivatives.

Discussion Status

Participants are actively engaging with the problem, with some providing insights into the correct application of derivative rules. There is an acknowledgment of misunderstandings regarding the use of the power rule and the nature of exponential functions, but no consensus has been reached on the final answer.

Contextual Notes

There is a mention of potential confusion regarding the application of derivative rules, particularly in relation to constants and variable exponents. The original poster also references a subsequent problem, indicating ongoing challenges in their understanding.

UsernameValid
Messages
12
Reaction score
0
Find the derivative of 101-x2.

1.) So, I used power rule to get: (1-x2)(10-x2).
2.) Then I did chain rule and multiplied my previous answer with (d/dx)(-x2).
3.) The answer I get is: (-2x)(1-x2)(10-x2).


When I use my calculator, the answer I get is: -20*ln(10)*10-x2. So, where does -20*ln(10) come from?
 
Physics news on Phys.org
The power rule applies when ##x## is the base, not in the exponent. What kinds of functions are the derivatives of the exponential function?
 
You have to use the formula for a constant to a variable exponent..

a^u = lna (u) (u')

So you use properties of exponents to get 10 times 10 to the power of negative x squared..

So that's a constant with a variable exponent..

So you get 10(ln10)(10^-x^2)(-2x)
^u ^u'
Then that simplifies to your answer, but you forgot the x.
 

Similar threads

Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
2
Views
2K
Replies
2
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 8 ·
Replies
8
Views
4K