Derivation of f(x)=xcos(sen(x)) – 1

  • Thread starter schmitt
  • Start date
  • Tags
    Derivation
In summary, the conversation is about someone asking for help with differentiating a function and receiving advice on using the product rule and chain rule to find the first and second derivative. They also apologize for a mistake and a grammatical error.
  • #1
schmitt
3
0
Hi Could someone help me derivating this function:

f(x)=xcos(sen(x)) – 1
f'(x)= ?
f''(x)= ?

Thank you.
 
Physics news on Phys.org
  • #2
schmitt said:
Hi Could someone help me derivating this function:

f(x)=xcos(sen(x)) – 1
f'(x)= ?
f''(x)= ?

Thank you.

What kind of function is sen(x)?
Or did you mean sec(x)?
 
  • #3
If that sen(x) is indeed a sec(x) then just use the product rule between x and cos(sec(x)). The -1 becomes 0. When you get the new function derive it again and you'll have f''(x).

You might be asking for the answer in which case I won't be giving you because you'll never learn it yourself.

Woops. Just realized that mistake I made. That was embarrassing.
 
Last edited:
  • #4
I like Serena said:
What kind of function is sen(x)?
Or did you mean sec(x)?

sorry, its "sin(x)"
 
  • #5
You are going to need to use the product rule and the chain rule, in that order.

BTW, there is no such word as "derivating" in English, and I doubt that it is a word in any other language. To find the derivative of a function, we differentiate the function - we don't "derivate" it or "derive" it.

Starting with ax2 + bx + c = 0, I can derive the Quadratic Formula, but this has nothing to do with differentiation.
 
  • #6
Got it solved already, sorry for the bad english.
Topic can be deleted.
 
  • #7
We don't delete the topics so that other people with similar questions can learn from them.
 

1. What is the purpose of deriving the function f(x)=xcos(sen(x)) – 1?

The purpose of deriving a function is to find its derivative, which gives us information about the rate of change of the function at any given point. In this case, we want to find the slope of the tangent line to the curve of f(x) at any x-value.

2. What is the derivative of f(x)=xcos(sen(x)) – 1?

The derivative of f(x) is given by the product rule, which states that the derivative of a product of two functions is equal to the first function times the derivative of the second function, plus the second function times the derivative of the first function. Applying this rule, we get f'(x) = xcos(x)cos(x) - xsen(x)sen(x) - 1.

3. How do you solve for the derivative of f(x)=xcos(sen(x)) – 1?

To solve for the derivative of f(x), we first apply the product rule as described above. Then, we use the chain rule to find the derivative of cos(sen(x)) and sen(x). Finally, we simplify the expression to get the final answer of f'(x) = xcos(x)cos(x) - xsen(x)sen(x) - 1.

4. What is the significance of the derivative of f(x)=xcos(sen(x)) – 1?

The derivative of f(x) gives us information about the behavior of the function at any given point. It tells us the slope of the tangent line to the curve of f(x) at that point, which can be used to find the rate of change of the function. This information is useful in various fields such as physics, economics, and engineering.

5. Can the derivative of f(x)=xcos(sen(x)) – 1 have any real-world applications?

Yes, the derivative of f(x) has many real-world applications. For example, in physics, the derivative can be used to find the velocity and acceleration of an object in motion. In economics, it can be used to find the marginal cost and marginal revenue of a product. In engineering, it can be used to optimize designs and predict the behavior of systems.

Similar threads

  • Calculus and Beyond Homework Help
Replies
23
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
865
  • Calculus and Beyond Homework Help
Replies
6
Views
505
  • Calculus and Beyond Homework Help
Replies
9
Views
916
  • Calculus and Beyond Homework Help
Replies
16
Views
1K
  • Calculus and Beyond Homework Help
Replies
8
Views
343
  • Calculus and Beyond Homework Help
Replies
1
Views
781
  • Calculus and Beyond Homework Help
Replies
2
Views
924
  • Calculus and Beyond Homework Help
Replies
5
Views
859
  • Calculus and Beyond Homework Help
Replies
4
Views
518
Back
Top