Understanding the Derivative of sin^2x: Common Mistakes and Clarifications

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In summary, the discussion is about the difference between the derivative of two functions and how it can be different.
  • #1
Nerd
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Hi.

Please tell me where this reasoning is wrong, because I know it is but I can't see how.
f(x) = sinx
f ' (x) = cosx
f(60) = sin 60
f ' (60) = cos 60

but

g(x) = sin 2x
g'(x) = 2 cos 2x
set x = 30
then: g(30) = sin 60
g'(x) = 2 cos 60

but

f(60) = g(30)
so f ' (60) = g'(30)
i.e cos 60 = 2 cos 60
thus: 1 = 2

WHY! HOW!
 
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  • #2


The fact that two functions cross at a particular point does not mean that their derivative is the same.

You could do the same thing with much simpler functions eg f(x)=x, g(x)=2x these cross at x=0 but they obviously don't have the same gradient.
 
  • #3


Thank you. That is quite obvious, isn't it.
 
  • #4


Try this one:

x=0
x+1=1
(x+1)(x+1)=1(x+1)
x^2+2x+1=x+1
x^2+x=0
x(x+1)=0
x+1=0
x=-1

?
 
  • #5


darkside00 said:
Try this one:

x=0
x+1=1
(x+1)(x+1)=1(x+1)
x^2+2x+1=x+1
x^2+x=0
x(x+1)=0
x+1=0
x=-1

?

:wink: Ops!
 
  • #6


Where did it go wrong algebraically?
 
  • #7


darkside00 said:
Where did it go wrong algebraically?

I just told you! I boldfaced those parts.
 
  • #8


Actually, you could just as easily say that at...

x(x+1) = 0

You could just as easily divide by (x+1) to get...

x = 0.

After all, the first equation in this post says that x is EITHER 0 or -1.

And we know that it's zero.
 
  • #9


Char. Limit said:
Actually, you could just as easily say that at...

x(x+1) = 0

You could just as easily divide by (x+1) to get...

x = 0.

After all, the first equation in this post says that x is EITHER 0 or -1.

And we know that it's zero.

If you do that you might miss out a solution.

if ab=0, then either a=0 or b=0.

if you just divide by a, you'll get b=0 only.
 
  • #10


^What he said. Doing that causes a missing solution. By the end, you should get x= -1 or 0. However, -1 is an extraneous solution. Therefore, you're stuck with x=0. Check out http://en.wikipedia.org/wiki/Extraneous_solution if you're confused.
 
  • #11


Yeah, I know that...

My point was that he missed a solution.
 
  • #12


Moderator's note:

Please keep on topic with the thread. New topics should be started in a new thread.

(See post #1 if you don't know what the topic of this thread is.)
 

1. What is the derivative of sin^2x?

The derivative of sin^2x is 2sinxcosx.

2. How do you derive the formula for the derivative of sin^2x?

The formula for the derivative of sin^2x can be derived using the chain rule and the power rule. First, we rewrite sin^2x as (sinx)^2. Then, we apply the chain rule by multiplying the derivative of the outer function (2x) by the derivative of the inner function (sinx). This gives us 2sinxcosx.

3. What is a common mistake made when finding the derivative of sin^2x?

A common mistake is forgetting to use the chain rule when taking the derivative of sin^2x. It is important to remember that the power rule only applies to functions of the form x^n, so the chain rule must be used when the power is a function of x (in this case, sinx).

4. Can the derivative of sin^2x be simplified further?

Yes, the derivative of sin^2x, 2sinxcosx, can be simplified further using trigonometric identities. It can be rewritten as sin2x, which is equivalent to 2sinxcosx.

5. How does understanding the derivative of sin^2x relate to real-world applications?

The derivative of sin^2x is an important concept in calculus and is used in many real-world applications, such as physics, engineering, and economics. It can be used to analyze the rate of change of a system or to optimize functions. For example, in physics, it can be used to calculate the velocity of a moving object or the acceleration of a falling object.

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