Ask your own question, for FREE!
Mathematics 12 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

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Latest Questions
Countless7Echos: Ah trying out the whole T.V girl drawing :p (I love drawing eyes)
1 hour ago 8 Replies 5 Medals
kaelynw: starting to draw a hand
2 days ago 16 Replies 2 Medals
Twaylor: Rate it :D (Took 2 days)
2 days ago 7 Replies 0 Medals
XShawtyX: Art, Short Writing Assignment: Imagining Landscapes
4 hours ago 7 Replies 1 Medal
XShawtyX: Chemistry, Help ud83dude4fud83cudffe
3 days ago 13 Replies 1 Medal
kaelynw: tried a lil smt, the arm is off but i like the other stuff
3 days ago 27 Replies 3 Medals
kaelynw: art igg
3 days ago 14 Replies 1 Medal
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!