How to Find the Derivative of y=cosx Using the Limit Process?

  • Context: Undergrad 
  • Thread starter Thread starter lilxchristina
  • Start date Start date
  • Tags Tags
    Derivative
Click For Summary
SUMMARY

The derivative of y=cos(x) can be found using the limit process defined by the formula lim (h → 0) [f(x+h) - f(x)] / h. By applying this formula, the expression simplifies to [-sin(h)/h] * [(sin(h)/(cos(h)+1) * cos(x) + sin(x))]. The key identity to remember is lim (t → 0) [sin(t)/t] = 1, which is essential for evaluating the limit as h approaches 0. This process mirrors the derivation used for y=sin(x), confirming the relationship between the derivatives of sine and cosine functions.

PREREQUISITES
  • Understanding of basic calculus concepts, specifically limits.
  • Familiarity with trigonometric identities and their derivatives.
  • Knowledge of the limit definition of a derivative.
  • Proficiency in manipulating algebraic expressions involving trigonometric functions.
NEXT STEPS
  • Study the geometric proof of lim (t → 0) [sin(t)/t] = 1.
  • Explore the derivatives of other trigonometric functions, such as y=tan(x).
  • Learn about higher-order derivatives of trigonometric functions.
  • Investigate applications of derivatives in real-world scenarios, such as physics and engineering.
USEFUL FOR

Students studying calculus, mathematics educators, and anyone interested in understanding the derivation of trigonometric function derivatives using the limit process.

lilxchristina
Messages
4
Reaction score
0
how do you find the derivative of y=cosx by using the limit process of

limit as h --> 0 is f(x+h) - f(x) / all over h.

i did this with y=sinx, and the answer was cosx, but I'm having trouble figuring out y=cosx.


help?
 
Physics news on Phys.org
Well,
[tex]\frac{\cos(x+h)-cos(x)}{h}=\frac{\cos(h)-1}{h}\cos(x)-\frac{\sin(h)}{h}\sin(x)=-\frac{\sin(h)}{h}(\frac{\sin(h)}{\cos(h)+1}\cos(x)+\sin(x))[/tex]
using well-known identities. Can you finish this?
 
Last edited:
You need to know that

[tex]\lim_{t\to 0} \frac{\sin(t)}{t}=1.[/tex]

The most common elementary proof of this is geometric.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
Replies
4
Views
9K
  • · Replies 53 ·
2
Replies
53
Views
7K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K