anonymous
  • anonymous
derivative of sqrt x + (1/4) sin (2x)^2
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
If the square is meant for the sin... y = sqrt x + (1/4)[sin 2x]^2 y' = (1/2)(1 / sqrt x) + (1/4)(2)(sin 2x)(cos 2x)(2) = (1 / 2 sqrt x) + (sin 2x)(cos 2x)
anonymous
  • anonymous
If the square is only on the 2x... y = sqrt x + (1/4)sin (2x)^2 = sqrt x + (1/4) sin (4x^2) y' = (1/2)(1 / sqrt x) + (1/4)[cos (4x^2)](4*2x) = (1 / 2 sqrt x) + (2x)cos (4x^2)
RadEn
  • RadEn
hint : derivative of sqrt(f(x)) = 1/(f'(x)*sqrt(f(x)) and derivative of sin(g(x))=g'(x)*cos(g(x))

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anonymous
  • anonymous
could you step by step for the solution (the square is only on the 2x)
anonymous
  • anonymous
If the square is only on the 2x... y = sqrt x + (1/4)sin (2x)^2 = sqrt x + (1/4) sin (4x^2) y' = (1/2)(1 / sqrt x) + (1/4)[cos (4x^2)](4*2x) = (1 / 2 sqrt x) + (2x)cos (4x^2)
RadEn
  • RadEn
@kelly226 is right
anonymous
  • anonymous
is the the (2x)^2 becoming (2x^2) product rule?
anonymous
  • anonymous
i dont think so...
RadEn
  • RadEn
its use chain rule... derivative of (2x)^2 is 2(2x)*2
anonymous
  • anonymous
wouldnt the derivative of (2x)^2 be 4x then according to chain rule= 2(2x)^2-1 (2)
RadEn
  • RadEn
2(2x)*2=8x not 4x
RadEn
  • RadEn
but, if we look the first question there is a coefficien of sin (2x)^2 (is 1/4), so multiply it by derivative of sin (2x)^2
RadEn
  • RadEn
therefore, 1/4*2(2x)*2*cos(2x)^2 = 2xcos(2x)^2
anonymous
  • anonymous
I understand the 2xcos part of the solution. But I dont understand why the original (2x)^2 remains
RadEn
  • RadEn
because derivative of sin(f(x) is f'(x)*cos(f(x))
anonymous
  • anonymous
oh okay thank you very much
RadEn
  • RadEn
wellcome :)

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