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Homework Help: Finding derivative of log and lnx. Please help! Need help before test!

  1. Oct 1, 2011 #1
    I need help in finding dy/dx or the derivative of these function:
    1. y=sin^2(lnx)
    so d/dx[sin^2(lnx)] I tried following this model of sin^x=(sinx)^2 so d/dx[(sinlnx)^2)]. Then use the chain rule of dy/dx= f'g(g') which is 2(sinlnx)(cos1/x) but I don't think that is right so can you please tell me another way to do it?thank you!

    2. y=x[log2(x^2-2x)]^3
    d/dx[xlog2(x^2-2x)]^3=3log2(x^2-2x)^2---> Again I don't think this is right so please direct me to the right one.

    3. y=exp(sqrt(1+5x^3)). This one I don't even know how to begin. Exponent of what?

    4. y=ln(cose^x) using the rule of d/dx[lnu]=1/u x (du/dx) I get dy/dx=(1/cose^x)(-sin(e^x)(e^x)
    dy/dx=(-e^xsine^x)/(cose^x)---> (tane^x)(-e^x)=-e^xtane^x~Please tell me if I did anything wrong

    5. Finding the derivative by using implict differentation
    y=ln(xtany). Please help me get started

    6. Finding the derivative by using logarithmic differentation:
    y=5root(x-1/x+1) or (x-1/x+1)^1/5
    so dy/dx=y[(2x+2)/(5x-5)(x+1)^2]

    Is this right?
    ~Thank you in advance =D
    Last edited: Oct 1, 2011
  2. jcsd
  3. Oct 1, 2011 #2


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    You just need to slow down and evaluate these things one step at a time.
    You started off okay by rewriting it in an easier form to differentiate.[tex]y = (\sin \log x)^2[/tex]When you apply the chain rule, you get[tex]y' = 2(\sin \log x)^1 \frac{d}{dx}[\sin (\log x)][/tex]
    You evaluated the derivative of the inside, sin(log x), incorrectly. You can't just differentiate the sine and log separately like that. You need to apply the chain rule again because log x is inside of sin.
    You need to apply the product rule first. Here you have y = fg where f=x and g=[log 2(x2-2x)]3 (Or is it supposed to be g=[log2 (x2-2x)]3?)
    Exp(x) is just a different way of writing ex, so what you have is[tex]y=e^\sqrt{1+5x^3}[/tex]
    Looks good!
    I'd start by using the property [itex]\log ab = \log a + \log b[/itex].
    You're good up to[tex]\log y = \frac{1}{5}\log \frac{x-1}{x+1}[/tex]Again, use the fact that log (a/b) = log a - log b to get rid of the quotient first, and then differentiate. That makes things much simpler.
  4. Oct 2, 2011 #3
    Hello, thank you so much for your help.
    ~I re-try the first one and I got: 2(sinlnx)(1/x)(cos(lnx)). Can you tell me if this is the right answer?

    ~the second one I retry and got dy/dx=1 [log2 (x2-2x)]3 + [3log2(x^2-2x)]2 (x)
    so to simplified it is : [log2(x2-2x)]3+ x[3log2(x^2-2x)]2~Can you tell me if i'm right?

    ~The third one i retry it and got y'=(1/2)(1+5x^3)-1/2 * eāˆš1+5x^3
    from the rule of y=eu the dy/dx=eu * du/dx
  5. Oct 2, 2011 #4


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    This is correct. Good work!
    This one isn't. You didn't calculate g' correctly. You need to use the chain rule:
    [tex]\frac{d}{dx}[\log_2 (x^2-2x)]^3 = 3[\log_2 (x^2-2x)]^2 \frac{d}{dx} [\log_2 (x^2-2x)][/tex]You have the first part correct; you're missing the second part. You'll want to use the fact that log2 u = (ln u)/(ln 2).
    You're almost there. Same mistake as in the second problem. You need to keep using the chain rule. When you differentiate [itex]\sqrt{1+5x^3}[/itex], you need to multiply by the derivative of the inside.
  6. Oct 2, 2011 #5
    I retry the 2nd one and here is the final answer (with your instructions =D)

    dy/dx=1[log2(x^2-2x)]3 + 3[log2(x^2-2x)]2[(ln(x^2-2x)/ln2)(x)]
    dy/dx=[log2(x^2-2x)]3 + 3x[log2(x^2-2x)]2 ((ln(x^2-2x)/ln2)
    Is this right?:confused:

    and I'm sorry I typed the last one wrong. It suppose to be y'=(1/2)(1+5x^3)^-1/2(15x^2) * eāˆš1+5x^3
  7. Oct 2, 2011 #6


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    Nope, not yet. You need to keep applying the chain rule. The base of the log just gives you a constant factor out front:
    [tex]\frac{d}{dx} \log_2 (x^2-2x) = \frac{d}{dx} \frac{\ln (x^2-2x)}{\ln 2} = \frac{1}{\ln 2}\frac{d}{dx} \ln (x^2-2x)[/tex]but you still need to differentiate the natural log and then what's inside of it.
    Looks good!
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