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OpenStudy (anonymous):
differentiate
y = (2x+5)^x
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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
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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
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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
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OpenStudy (anonymous):
yes
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