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Find derivative of $$y= \frac{ax+b}{cx+d}$$

I found it to be $$\frac{dy}{dx}=\frac{a}{cx+d}-\frac{c(ax+b)}{(cx+d)^2}$$

Use it to evaluate:


I figured that here $y=\frac{x+1}{x+3}$ and $$\frac{dy}{dx}=\frac{1}{x+3}-\frac{(x+1)}{(x+3)^2}$$

and using the technique I learned from my last question I did this:


which I could then substitute back, having changed the limits by substituting $1$ into $y$ and then $0$ into $y$:




This gives me:


$$=2\left[y(\ln(y)-1)\right]_\frac{1}{3}^\frac{1}{2} = 2\left[\frac{1}{2}\left(\ln\left(\frac{1}{2}\right)-\frac{1}{2}\right)\right]-\frac{1}{3}\left[\ln\left(\frac{1}{3}\right)-\frac{1}{3}\right]\\$$


The problem is I am supposed to end up with something else. Can anyone spot any issues with this?

EDIT: This is the answer I am supposed to be getting:


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What's the answer supposed to be? – huon Aug 6 '12 at 13:28
If $dy/dx=2/(x+3)^2$, then $1/(x+3)^2=(1/2)dy/dx$. You've used $2dy/dx$ instead. – celtschk Aug 6 '12 at 13:33
  • You have:


But the integral is $I=\int_0^1{\dfrac{1}{(x+3)^2}}\ln\left(\dfrac{x+1}{x+3}\right)dx$ where ${\dfrac{1}{(x+3)^2}}$ is actually $\dfrac 12 \times \dfrac{2}{(x+3)^2}$. Therefore: $$I= \dfrac 12 \int_0^1{\frac{dy}{dx}}\ln(y)dx=\dfrac 12 \int_\frac{1}{3}^\frac{1}{2}{\ln(y)dy}$$

  • The other issue might be:

$$\dfrac 12 \int_\frac{1}{3}^\frac{1}{2}{\ln(y)dy}=\dfrac 12 \bigg[y(\ln(y)-1\bigg]_\frac{1}{3}^\frac{1}{2} =\\\frac 12 \left(\frac{1}{2}\left(\ln\left(\frac{1}{2}\right)-1\right)-\frac{1}{3}\left(\ln\left(\frac{1}{3}\right)-1\right)\right) \\=\frac 12 \left(-\frac 12 \ln 2 - \frac 12 +\frac 13 \ln 3+ \frac 13\right)\\=\frac 16 \ln 3 -\frac 14 \ln 2 -\frac 1{12}$$

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What do you mean the other issue might be? Once I did the $\frac{1}{2}$ I had the right answer. That what you've written needs correcting I think. – Magpie Aug 6 '12 at 14:22
@Magpie: Well, I thought you did not notice $\ln \frac 12 = - \ln 2$. – Gigili Aug 6 '12 at 14:24
Ok, I think I was ok with the logs but just to chekc I have put up what I did further up in the question. LEt me know if you see any problems with it.thanks – Magpie Aug 6 '12 at 14:45
@Magpie: No, that doesn't seem correct. You have $\int \ln x dx= x(\ln x -1)+C$. – Gigili Aug 6 '12 at 14:51
ok I have added a step to show what i was thinking, but I have to admit, I don't really get what you mean. – Magpie Aug 6 '12 at 16:56
up vote 1 down vote accepted

The solution I have now is:






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You should have been careful when multiplying by $2$, I think that you should have multiplied by $\frac{1}{2}$ in your formal calculation.

Also, you may need to rewrite your answer a bit to get it right, such as computing $\frac{2}{9} - 1 = -\frac{7}{9}$, which is implicitly required from you by any textbook, I presume.

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I posted the answer now, the reason I stopped there was that it was not going toward the answer needed. See above. – Magpie Aug 6 '12 at 13:37
also can you elaborate on why I should multiply but a half instead of 2? Thanks – Magpie Aug 6 '12 at 13:43
@Magpie: For your second question, see my comment to your post. In short, if $dy/dx=2f$, then $f=\frac12 dy/dx$, not $2dy/dx$. – celtschk Aug 6 '12 at 13:47
yeah that works. Thanks. – Magpie Aug 6 '12 at 14:00

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