Derivative Question: Why is dy/dx for cos(2x) not -sin(2x)?

  • Thread starter Miike012
  • Start date
In summary, the derivative of cos(2x) is not -sin(2x) because it is a composite function and requires the use of the chain rule. The derivative of cos(2x) is actually -2sin(2x).
  • #1
Miike012
1,009
0
Example Problem: Finding dy/dx for cos(2x).

knowing that the derivative of cos(x) is -sin(x)...

thus my thought is... when differentiating cos(2x)... why is the derivative not -sin(2x)?

I wasn't aware that I would have to treat that as a composite function...
 
Physics news on Phys.org
  • #2
Have you learned the chain rule yet?
 
  • #3
Yes I have... But I thought I didn't think I had to use the chain rule on this function.
 
  • #4
Miike012 said:
Example Problem: Finding dy/dx for cos(2x).

knowing that the derivative of cos(x) is -sin(x)...

thus my thought is... when differentiating cos(2x)... why is the derivative not -sin(2x)?

I wasn't aware that I would have to treat that as a composite function...

You have to use the chain rule. You are right in which cos(x) is -sin(x) but you also have 2x which the derivative is 2 so the final answer is -2sin(x)

Hope it make things more clear for you.
 
  • #5
McAfee said:
You have to use the chain rule. You are right in which cos(x) is -sin(x)
No, cos(x) IS NOT -sin(x), but d/dx(cos(x)) = -sin(x).
McAfee said:
but you also have 2x which the derivative is 2 so the final answer is -2sin(x)

Hope it make things more clear for you.
 
  • #6
Mark44 said:
No, cos(x) IS NOT -sin(x), but d/dx(cos(x)) = -sin(x).

You knew what I meant. :D
 
  • #7
McAfee said:
You knew what I meant. :D
Yes, I knew what you meant, so you should work a little harder to say what you mean.
 

Related to Derivative Question: Why is dy/dx for cos(2x) not -sin(2x)?

1. Why is the derivative of cos(2x) not -sin(2x)?

The derivative of a function is the rate of change of the function at a specific point. In the case of cos(2x), the derivative is not -sin(2x) because the derivative of a composite function is not equal to the derivative of the inner function multiplied by the derivative of the outer function. In other words, the derivative of cos(2x) is not equal to the derivative of 2x multiplied by the derivative of cos(x).

2. How do you find the derivative of cos(2x)?

To find the derivative of cos(2x), we use the chain rule. The chain rule states that the derivative of a composite function is equal to the derivative of the outer function multiplied by the derivative of the inner function. In this case, the outer function is cos(x) and the inner function is 2x. Therefore, the derivative of cos(2x) is equal to -sin(2x) multiplied by the derivative of 2x, which is 2. This gives us a final answer of -2sin(2x).

3. Can you explain why the derivative of cos(2x) is -2sin(2x)?

The derivative of a function can be thought of as the slope of the line tangent to the graph of the function at a specific point. To find the slope of this tangent line, we use the formula for the derivative, which in this case is -2sin(2x). This means that for every value of x, the slope of the tangent line to the graph of cos(2x) will be -2sin(2x).

4. How does the graph of the derivative of cos(2x) compare to the graph of -sin(2x)?

The graph of the derivative of cos(2x) will be a cosine function with a different amplitude and phase shift. However, the graph of -sin(2x) will be a sine function with the same amplitude and phase shift as the original function. This means that while the two functions will have the same general shape, they will differ in their exact values at each point.

5. Why is it important to know the derivative of cos(2x)?

The derivative of cos(2x) can be used in various applications, such as in physics, engineering, and economics. It can help us find the slope of a curve, which is important in understanding the rate of change of a particular quantity. It is also useful in finding maximum and minimum values of a function, as well as in optimization problems. Overall, understanding the derivative of cos(2x) can greatly enhance our understanding of various real-world phenomena.

Similar threads

Arc length of vector function - the integral seems impossible
    • Calculus and Beyond Homework Help
Replies
2
Views
849
A question about the derivation of upper bound integrals
    • Calculus and Beyond Homework Help
Replies
23
Views
977
Question on Two Differentiation Techniques
    • Calculus and Beyond Homework Help
Replies
25
Views
391
Integration of a trigonometic function
    • Calculus and Beyond Homework Help
Replies
14
Views
970
Differentiation of Log(cos(X)/x^2)^2
    • Calculus and Beyond Homework Help
Replies
6
Views
959
Finding the Second Derivative of a Function with Two Variables
    • Calculus and Beyond Homework Help
Replies
5
Views
2K
Calculating Area Under a Curve Using Change of Variables and Quotient Rule
    • Calculus and Beyond Homework Help
Replies
5
Views
1K
Comparing Hyperbolic and Cartesian Trig Properties
    • Calculus and Beyond Homework Help
Replies
1
Views
755
Evaluating scalar products of two functions
    • Calculus and Beyond Homework Help
Replies
1
Views
776
How Do You Correctly Apply Integration by Parts to ∫-e^(2x)*sin(e^x) dx?
    • Calculus and Beyond Homework Help
Replies
9
Views
1K

Hot Threads

Recent Insights

Back
Top