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Homework Help: Derivative of square root

  1. Feb 4, 2013 #1
    1. The problem statement, all variables and given/known data

    Finding derivative of (sin(sqt3x+5))

    2. Relevant equations

    None given. Chain Rule

    3. The attempt at a solution

    The answer is: (cos(sqrt3x+5)) * 1/2(sqrt3x+5) * 3

    but I don't know how to get to the 3.

    I turned sin into cos and multiplied by the inside derivative giving the first two parts of the answer. How do I get 3? - is that from the inside derivative of 3x+5? How do you know which is outside or inside of a derivative
  2. jcsd
  3. Feb 4, 2013 #2
    Another problem:

    5√(x^2+1)^4 <-- That's all under the square root.

    i know the answer. But how do I figure out which is the outside then multiply that by the inside derivative?

    I understand how this becomes [(x^2+1)^4]^1/5 but then it turns into (x^2+1)^4/5 were the exponents multiplied 4/1 * 1/5 = 4/5?

    Then from there it becomes 4/5(x^2+1)^-1/5 * 2x
    Last edited: Feb 4, 2013
  4. Feb 4, 2013 #3


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    sin() is outside, sqrt() is inside of that. And 3x+5 is inside of sqrt(). You need two stages of chain rule.
  5. Feb 4, 2013 #4


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    5*sqrt() is outside. ()^4 is inside of that. And inside of that is x^2+1
  6. Feb 4, 2013 #5
    What about this problem..

    √x^3tan(x) <-- all under sqrt.

    so I did...

    1/2*(x^3tan(x))^-1/2 * 3x^2sec(x)^2

    but.. the answer says 1/2√x^3tan(x) * (3x^2tan(x) + x^3(sec(x))^2)

    I do know that √u = 1/2√u, but I wrote 1/2*(x^3tan(x))^-1/2 due to exponent rule?
  7. Feb 4, 2013 #6


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    That's just fine until you tried to find the derivative of x^3*tan(x). Now you need to use the product rule.
  8. Feb 4, 2013 #7
    Haha thank you! I stumbled for sure.
  9. Feb 4, 2013 #8
    Quick question:

    there is a problem i did which was (sin(x))^2

    = 2cos(x)) * 1 = 2cos(x) using chain rule, but for the problem I had to use the product rule in which I got...

    = 2(sin(x))(cos(x))

    do both answers equal each other? Trig identity unless I messed up somehow?
    Last edited: Feb 4, 2013
  10. Feb 4, 2013 #9


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    No. They aren't the same. The derivative of u^2 is 2*u*u'. The chain rule gives you 2*sin(x)*(sin(x))'=2*sin(x)*cos(x). NOT 2*cos(x). Same as the product rule.
  11. Feb 4, 2013 #10
    So the exponent 2 comes down multiplying with sin(x) then sin(x) is considered the inside derivative then you multiply that by cos(x)?

    I don't understand why it just doesn't become 2cos(x)
  12. Feb 5, 2013 #11


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    Yes! Yes! "the exponent 2 comes down multiplying with sin(x)" then you multiply by the derivative of sin(x). What happen to the sin(x) you multiplied by 2? Did you just drop it? This is just being sloppy.
  13. Feb 5, 2013 #12
    Haha alright it's just I have a hard time differentiating between inside and outside so 2*sin(x) with sin(x) turning into 2*cos(x) looks done, but I guess not. Thank you though!
  14. Feb 5, 2013 #13


    Staff: Mentor

    Someone reading your work would have no idea what you are doing, since you give no indication that you are taking a derivative.

    (sin(x))2 = sin(x) * sin(x), but
    sin(x) * sin(x) cos(x) sin(x)+ sin(x)cos(x)

    What you should be saying is that
    d/dx[sin(x) * sin(x)] = cos(x) * sin(x) + sin(x) * cos(x) = 2 * sin(x) * cos(x)

    You could have gotten this same result by using the chain rule when you differentiate sin2(x).
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