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Homework Help: Some differentiation

  1. Jun 8, 2009 #1
    1. The problem statement, all variables and given/known data

    Differentiate with respect to X

    2. Relevant equations

    1) e^3x + ln2x

    2) (5+x^2)^3/2

    3. The attempt at a solution

    1) isnt it just normal differentiation? so

    3e^3x + 1/2x

    only thing i wasnt sure about was the differentiation of ln2x

    2) second one i let t=(5+x^2) differentiated that then differentiated t^3/2 to get 3/2t^1/2

    but this is where i got stuck as i wasnt sure where to go from here

    is the answer just

    3/2 (5+x^2)^1/2 or is it 3/2(2x)^1/2
  2. jcsd
  3. Jun 8, 2009 #2
    Both are examples of the chain rule.
    \frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}

    In 1) you've differentiated incorrectly the term ln(2x). The outer function is y=ln(u) but the inner function is u=2x.
    So from the result of differentiating ln(2x) should be (1/2x)*2 = 1/x

    In 2) the outer function is y=u^(3/2), but the inner function is u=5+x^2. Can you do it from here?
  4. Jun 9, 2009 #3
    so y=u^(3/2) differentiates to dy/dx= (3/2)(x^2)^(1/2)

  5. Jun 9, 2009 #4


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    No, no, no. y=(5+x^2)^(3/2), u=5+x^2. So y=u^(3/2), dy/dx=dy/du*du/dx. dy/du=d(u^(3/2))/du=(3/2)*u^(1/2)=(3/2)*(5+x^2)^(1/2). That's the part you've got. But you keep forgetting to multiply by the du/dx part. Look up the chain rule and work some more examples. This is a pretty important point.
  6. Jun 9, 2009 #5


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    By the way ln(2x)= ln(x)+ ln(2). That should be easier to differentiate.
  7. Jun 10, 2009 #6
    ahh ye course iv got it now.. it was just the final step i was missing
    ye i realised the ln(2x)=ln(x)+ln(2) is lost during differntiation because its a constant right?

    just lack of revision im afraid =]
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