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diamondboy
 one year ago
Just a quick question
diamondboy
 one year ago
Just a quick question

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diamondboy
 one year ago
Best ResponseYou've already chosen the best response.0dw:1442283012866:dw

diamondboy
 one year ago
Best ResponseYou've already chosen the best response.0could you help with one more?

diamondboy
 one year ago
Best ResponseYou've already chosen the best response.0or maybe I should say more?

diamondboy
 one year ago
Best ResponseYou've already chosen the best response.0dw:1442283174794:dw

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2\[y=\ln(5x^3) =\ln(5)+3 \ln(x) \\ y'=0+3 \frac{1}{x}=\frac{3}{x} \\ \text{ yep or you could have done } \\ y=\ln(5x^3) \implies y'=(5x^3)' \frac{1}{5x^3}=15x^2 \frac{1}{5x^3} =\frac{3}{x}\]

diamondboy
 one year ago
Best ResponseYou've already chosen the best response.0yes... I did it d second way

diamondboy
 one year ago
Best ResponseYou've already chosen the best response.0so I need to understand something about this chain rule

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2\[\text{ Let } f(x)>0 \\ y=\ln(f(x)) \implies y'=f'(x) \cdot \frac{1}{f(x)} \]

diamondboy
 one year ago
Best ResponseYou've already chosen the best response.0Oh ok...now I see it

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2there was another way... to find y' for the earlier problem \[y=\ln(5x^3) \\ e^y=5x^3 \\ \text{ now differentiate both sides } \\ y'e^{y}=15x^2 \\ \text{ solve for } y' \\ y'=15x^2 \cdot \frac{1}{e^y} \\ \text{ but } e^y=5x^3 \\ \text{ so } y'=15x^2 \cdot \frac{1}{5x^3}\]

diamondboy
 one year ago
Best ResponseYou've already chosen the best response.0dw:1442283585590:dw is that true?

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2well what did you get for f'(x)?

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2ok so you differentiate 2*(2)=4 not 4

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2\[y=e^{2x} \\ y'=(2x)'e^{2x}=2 e^{2x} \\ y''=2(2x)'e^{2x}\]

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2ok so you differentiate again but the coefficient should be 2*(2)=4 not 4 *

diamondboy
 one year ago
Best ResponseYou've already chosen the best response.0ok so I used the product rule to solve for d second derivative

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2product rule is not needed

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2constant multiple rule and chain rule

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2though you could use product rule

diamondboy
 one year ago
Best ResponseYou've already chosen the best response.0well what do you think?

diamondboy
 one year ago
Best ResponseYou've already chosen the best response.0o I leave you without a medal?....anyways thanks

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2had internet connection failure

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2or at least openstudy internet connection failure :p

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2what do I think about what?

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2you could use product rule but I think it is shorter to use constant multiple rule is that what you are talking about?

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2\[y'=2e^{2x} \\ y''=2 (e^{2x})' \text{ by constant multiple rule } \\ \text{ or as I said you could have use product rule to get this same result } \\ y'' =(2)'e^{2x}+2(e^{2x})'=0e^{2x}2(e^{2x})'=2(e^{2x})'\]

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2where you already found the derivative of exp(2x) w.r.t . x

diamondboy
 one year ago
Best ResponseYou've already chosen the best response.0wait wat do u mean?

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2are you asking what I mean by finding the derivative of exp(2x) w.r.t x?

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2\[\frac{d}{dx}e^{2x}=(2)e^{2x}=2e^{2x}\]
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