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|dw:1442283012866:dw|
yes for x>0
alright

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could you help with one more?
k
or maybe I should say more?
|dw:1442283174794:dw|
find y prime
is it 3/x?
\[y=\ln(5x^3) =\ln(5)+3 \ln(x) \\ y'=0+3 \frac{1}{x}=\frac{3}{x} \\ \text{ yep or you could have done } \\ y=\ln(5x^3) \implies y'=(5x^3)' \frac{1}{5x^3}=15x^2 \frac{1}{5x^3} =\frac{3}{x}\]
yes... I did it d second way
so I need to understand something about this chain rule
\[\text{ Let } f(x)>0 \\ y=\ln(f(x)) \implies y'=f'(x) \cdot \frac{1}{f(x)} \]
Oh ok...now I see it
there was another way... to find y' for the earlier problem \[y=\ln(5x^3) \\ e^y=5x^3 \\ \text{ now differentiate both sides } \\ y'e^{y}=15x^2 \\ \text{ solve for } y' \\ y'=15x^2 \cdot \frac{1}{e^y} \\ \text{ but } e^y=5x^3 \\ \text{ so } y'=15x^2 \cdot \frac{1}{5x^3}\]
ok
|dw:1442283585590:dw| is that true?
well what did you get for f'(x)?
e^(-2x)*-2
ok so you differentiate -2*(-2)=4 not -4
yes thats true
\[y=e^{-2x} \\ y'=(-2x)'e^{-2x}=-2 e^{-2x} \\ y''=-2(-2x)'e^{-2x}\]
ok so you differentiate again but the coefficient should be -2*(-2)=4 not -4 *
ok so I used the product rule to solve for d second derivative
product rule is not needed
yea I get that
constant multiple rule and chain rule
though you could use product rule
ok
well what do you think?
o I leave you without a medal?....anyways thanks
had internet connection failure
or at least openstudy internet connection failure :p
what do I think about what?
you could use product rule but I think it is shorter to use constant multiple rule is that what you are talking about?
\[y'=-2e^{-2x} \\ y''=-2 (e^{-2x})' \text{ by constant multiple rule } \\ \text{ or as I said you could have use product rule to get this same result } \\ y'' =(-2)'e^{-2x}+-2(e^{-2x})'=0e^{-2x}-2(e^{-2x})'=-2(e^{-2x})'\]
where you already found the derivative of exp(-2x) w.r.t . x
yep
wait wat do u mean?
are you asking what I mean by finding the derivative of exp(-2x) w.r.t x?
\[\frac{d}{dx}e^{-2x}=(-2)e^{-2x}=-2e^{-2x}\]
oh noo

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