## theEric 3 years ago Hi! I'm looking at an example from my book and I'm hoping somebody can explain this. It's in the first chapter, but I don't want to skip out on anything. Thanks! Here is the issue: My book says that $$\dfrac{dp/dt}{p-900}=\dfrac{1}{2}$$ where $$p\neq 900$$. Theennn... "By the chain rule the left side of [that equation] is the derivative of $$\ln\left|p-900\right|$$ with respect to $$t$$, so we have $$\dfrac{d}{dt} \ln\left| p-900\right|=\dfrac{1}{2}$$" My issue is that I'm missing something. I see that $$\cancel{\dfrac{d}{dt}\dfrac{1}{p-900}= \ln \left| p-900\right|}$$.

1. theEric

I know the chain rule, like if $$f=f(g(t))$$ then $$\dfrac{df}{dt}=\dfrac{df}{dg}\dfrac{dg}{dt}$$. I think I'm just getting confused from looking at this too much...

2. theEric

I crossed that out because it's really $$\dfrac{1}{p-900}=\dfrac{d}{dt}\ln\left| p-900\right|$$

3. theEric

And so comes my problem... The book says that $$\dfrac{dp/dt}{p-900}=\dfrac{1}{2}$$ and $$\dfrac{d}{dt}\ln\left|p-900\right| = \dfrac{1}{2}$$ Which means that $$\dfrac{dp/dt}{p-900}=\dfrac{d}{dt}\ln\left|p-900\right|$$ And I think that $$\dfrac{1}{p-900}=\dfrac{d}{dt}\ln\left|p-900\right|$$ I can't be right unless $$\dfrac{dp/dt}{p-900}=\dfrac{1}{p-900}$$. This would imply that $$dp/dt=1$$, which isn't necessarily true for this problem, that I see.

4. theEric

I said that I think $$\dfrac{1}{p-900}=\dfrac{d}{dt}\ln\left|p-900\right|$$, but really $$\dfrac{1}{p-900}=\dfrac{d}{dp}\ln\left|p-900\right|$$.

5. theEric

So then.... $$\dfrac{dp/dt}{p-900}=\dfrac{dp}{dt}\dfrac{d}{dp}\ln\left|p-900\right|$$

6. theEric

Ohhh....

7. theEric

$$\dfrac{dp}{dt}\dfrac{d}{dp}\longrightarrow\dfrac{d}{dt}$$