Sin x vs sin^2 x

  • Thread starter BoogieL80
  • Start date
  • #1
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Homework Statement


Find any critical numbers of the function


Homework Equations


sin^2 x + cos x


The Attempt at a Solution



I actually have a sort of silly question. Woud sin^2 in the equation be solved using the differeniation rule of d/dx[sin x] = cos x or d/dx [sin u] = (cos u)d/dx u?
 

Answers and Replies

  • #2
529
1
d/dx (sin^2 x)= d/dx (sin x * sin x)
= d/dx (sin x) * sin x + sin x * d/dx (sin x)
=2 d/dx (sin x) sin x
=2cos x sin x
 
  • #3
39
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So I follow that thinking, but that would be the same as d/dx[sin u] right?
 
  • #4
529
1
No- I think you're referring to a case like this:

A=d/dx sin(f(x))
substitue u=f(x)

A=du/dx * d/du sin(u)
=[d/dx f(x) ] cos u
=[d/dx f(x) ] cos(f(x))

Remember, sin^2(x) does not equal sin (sin (x)) or sin (x^2)
rather sin^2(x)= sin (x) * sin (x)
 
  • #5
39
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Thank you.
 

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