ln [(3x+1)/x^2+2)] please help!
- Thread starter Toshiba
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- Apr 12, 2005
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1) Check your notation. Do you REALLY mean ln[(3x+1)/x^2+2] or do you mean ln[(3x+1)/(x^2+2)]? It is VERY different.
2) You may wish to use rules of logarithms to simplify your life: ln[(3x+1)/(x^2+2)] = ln(3x+1) - ln(x^2+2)
Let's see what you get.
2) You may wish to use rules of logarithms to simplify your life: ln[(3x+1)/(x^2+2)] = ln(3x+1) - ln(x^2+2)
Let's see what you get.
Yes!! you are right I totaly forgot about that :s silly me so this is what I did:
ln(3x+1) - ln(x^2+2)
= 3/(3x+1) - 2x/(x^2+1)
=3(x^2+1) - 2x(3x+1)/ (3x+1)(x^2+2)
and then some simple multiplication.....
and i got the answer
)))
but i have a similar questions now and this dosent seem to work or any other thing I try i know this might be a bit silly because its similar but im going to go crazy if I cant figure it out can you help me?? the question is
y=ln(5x+1/x) this is exactly how it is......
and it ask agian to find dy/dx can you help me get on the right track???
Thank You
ln(3x+1) - ln(x^2+2)
= 3/(3x+1) - 2x/(x^2+1)
=3(x^2+1) - 2x(3x+1)/ (3x+1)(x^2+2)
and then some simple multiplication.....
and i got the answer
)))but i have a similar questions now and this dosent seem to work or any other thing I try i know this might be a bit silly because its similar but im going to go crazy if I cant figure it out can you help me?? the question is
y=ln(5x+1/x) this is exactly how it is......
and it ask agian to find dy/dx can you help me get on the right track???
Thank You
\(\displaystyle \L\\y=ln(5x+\frac{1}{x})\)
I reckon this is what they mean.
Chain rule, \(\displaystyle \L\\\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\).
Let \(\displaystyle \L\\u=5x+\frac{1}{x}\)
\(\displaystyle \L\\\frac{d}{du}[ln(u)]=\frac{1}{u}=\frac{dy}{du}\)
But \(\displaystyle \L\\\frac{du}{dx}=5-\frac{1}{x^{2}}\)
So:
\(\displaystyle \L\\\frac{5-\frac{1}{x^{2}}}{5x+\frac{1}{x}}=\frac{5x^{2}-1}{x(5x^{2}+1)}\)