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Differentiating Logs (1 Viewer)
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Carrotsticks
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Question 1:
 \right ) \\\\ = 3x^2 \ln(x) + x^3 \times \frac{1}{x} \qquad $ by the Product Rule$ \\\\ = 3x^2 \ln(x) + x^2 \\\\ = x^2 \left ( 3 \ln(x) + 1 \right ) )
Question 2:
 \right ) \right ) \\\\ = \frac{1}{\ln(x)} \times \frac{1}{x} \qquad $ by the Chain Rule$ \\\\ = \frac{1}{x \ln(x)})
Question 3:
}{x-2} \right ) \\\\ = \frac{\frac{1}{x} \times (x-2) - \ln(x)}{(x-2)^2} \qquad $ by the Quotient Rule$ \\\\\ = \frac{\frac{x-2}{x} - \ln(x)}{(x-2)^2} \\\\ = \frac{x-2 - x\ln(x)}{x(x-2)^2})
Question 2:
Question 3:
Timske
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r) x^3Ln(x+1) use product rule , uv' + vu' = x^3/x+1 + 3x^2Ln(x+1)
= x^2[ x/x+1 + 3ln(x+1) ]
derivative of ln(x+1) = 1/x+1
<a href="http://www.codecogs.com/eqnedit.php?latex=\frac{d}{dx} ~ x^3\log_{e}(x@plus;1)~, ~ use ~product ~rule ~uv'~ @plus; ~vu'\\\\ = \frac{x^3}{x@plus;1} @plus; 3x^2\log_{e}(x@plus;1) \\\\ Factor ~ x^2 \\\\ \therefore x^2~(\frac{x}{x@plus;1} ~@plus; ~3\log_{e}(x@plus;1))" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\frac{d}{dx} ~ x^3\log_{e}(x+1)~, ~ use ~product ~rule ~uv'~ + ~vu'\\\\ = \frac{x^3}{x+1} + 3x^2\log_{e}(x+1) \\\\ Factor ~ x^2 \\\\ \therefore x^2~(\frac{x}{x+1} ~+ ~3\log_{e}(x+1))" title="\frac{d}{dx} ~ x^3\log_{e}(x+1)~, ~ use ~product ~rule ~uv'~ + ~vu'\\\\ = \frac{x^3}{x+1} + 3x^2\log_{e}(x+1) \\\\ Factor ~ x^2 \\\\ \therefore x^2~(\frac{x}{x+1} ~+ ~3\log_{e}(x+1))" /></a>
= x^2[ x/x+1 + 3ln(x+1) ]
derivative of ln(x+1) = 1/x+1
<a href="http://www.codecogs.com/eqnedit.php?latex=\frac{d}{dx} ~ x^3\log_{e}(x@plus;1)~, ~ use ~product ~rule ~uv'~ @plus; ~vu'\\\\ = \frac{x^3}{x@plus;1} @plus; 3x^2\log_{e}(x@plus;1) \\\\ Factor ~ x^2 \\\\ \therefore x^2~(\frac{x}{x@plus;1} ~@plus; ~3\log_{e}(x@plus;1))" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\frac{d}{dx} ~ x^3\log_{e}(x+1)~, ~ use ~product ~rule ~uv'~ + ~vu'\\\\ = \frac{x^3}{x+1} + 3x^2\log_{e}(x+1) \\\\ Factor ~ x^2 \\\\ \therefore x^2~(\frac{x}{x+1} ~+ ~3\log_{e}(x+1))" title="\frac{d}{dx} ~ x^3\log_{e}(x+1)~, ~ use ~product ~rule ~uv'~ + ~vu'\\\\ = \frac{x^3}{x+1} + 3x^2\log_{e}(x+1) \\\\ Factor ~ x^2 \\\\ \therefore x^2~(\frac{x}{x+1} ~+ ~3\log_{e}(x+1))" /></a>
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Carrotsticks
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Oh for some reason, I read the inside of the log function for Q1 to be x instead of x+1.
 \right ) \\\\ = 3x^2 \ln(x+1) + x^3 \times \frac{1}{x+1} \qquad $ by the Product Rule$ \\\\ = 3x^2 \ln(x+1) + \frac{x^3}{x+1} \\\\ = x^2 \left [ 3 \ln(x+1) + \frac{x}{x+1} \right ] )
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Carrotsticks
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Coookies
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Theres no x though
Carrotsticks
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Oh, so you just wanted the derivative of the constant:
If that was the question, then it would be 0, just like the derivative of any other constant.
Carrotsticks
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Now you are confusing me too! =p
You asked "could you write the derivative of loge?"
What exactly did you mean by that?
Carrotsticks
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We use the Chain Rule.
The derivative of:
is:
by the Chain Rule.
So when we differentiate ln(ln(x)), we have:
Coookies
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I'm really sorry about this, but where is the 1 from? The 1/lnx
Because isn't the derivative of lnx, 1/x?
Am I even making any sense? I don't think I even know what I'm talking about haha!
I think I'll ask my teacher tomorrow...
Because isn't the derivative of lnx, 1/x?
Am I even making any sense? I don't think I even know what I'm talking about haha!
I think I'll ask my teacher tomorrow...
Carrotsticks
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Indeed, explaining things via typing isn't the best way.
Yes, the derivative of ln(x) is 1/x, but we have a ln(x) WITHIN a ln(x) (ln-ception haha), so that changes things.
Coookies
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I have a few more questions lol!
9. Find the point of inflection on the curve y=xlogex-x^2
For this one, I just need help differentiating once (f'(x))
10. Find the stationary point on the curve y=lnx/x and determine its nature
9. Find the point of inflection on the curve y=xlogex-x^2
For this one, I just need help differentiating once (f'(x))
10. Find the stationary point on the curve y=lnx/x and determine its nature
Last edited:
Timske
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Do you have trouble differentiating* log?