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Mathematics 49 Online
OpenStudy (anonymous):

differentiate y = (2x+5)^x

OpenStudy (anonymous):

do know ln and e

OpenStudy (anonymous):

would it be e^(xln(2x+5))

OpenStudy (anonymous):

if so i didnt know how to go from there

OpenStudy (anonymous):

you can right like this \[\ln y=\ln (2x+5)^x\] \[\ln y=x \ln (2x+5)\]

OpenStudy (anonymous):

oh yeah

OpenStudy (anonymous):

i'm always forgetful

OpenStudy (anonymous):

from there it would be chain rule right??

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

okay i did chain rule and i got x/(2x+5) + ln(2x+5) this is wrong though

OpenStudy (anonymous):

that is correct but the thing is the other side of the equation is still ln y

OpenStudy (anonymous):

right but when i did wolfram they got (2x+5)^5 * ( 2x/(2x+5) + ln(2x+5) )

OpenStudy (anonymous):

using implicit differentiation \[f(x) = \ln y\] \[f(x)\prime =y \prime/y\] and \[y \prime /(2x+5)^x=x/(2x+5) + \ln(2x+5)\]

OpenStudy (anonymous):

so to get the derivative finally we have to multiply with (2x+5)^x

OpenStudy (anonymous):

shouldn't it be 2x/2x+5 instead of x/2x+5

OpenStudy (anonymous):

the first one ofcourse but did you get the implicit differentiation part

OpenStudy (anonymous):

yes

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