l'Hopitals Rule... need help with derivatives?


by m0286
Tags: derivatives, lhopitals, rule
m0286
m0286 is offline
#1
Jul6-06, 05:57 PM
P: 63
Question is:
Evaluate the following limits
lim x--> 0 4^x - e^2x / 2x
So i take derivatives
but thats where im confused.... Whats the derivate of 4^x... x4^-x???
and is the derivative of -e^2x -2e^x???
So then that leaves me with 4x^-x -2e^x /2which is 0... so do i do derivatives again??? It seems like the x's wont go away????
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StatusX
StatusX is offline
#2
Jul6-06, 06:04 PM
HW Helper
P: 2,566
You can use the chain rule and the fact that the derivative of e^x is e^x to answer both questions. That is, the derivative of e^f(x) is e^f(x)*f'(x). Try to rewrite 4^x as e^f(x). (use logs)
Jameson
Jameson is offline
#3
Jul7-06, 03:57 PM
P: 789
[tex]f(x)=4^x[/tex]
[tex]\ln(f(x))=x\ln(4)[/tex]
[tex]\frac{1}{f(x)}*f'(x)=\ln(4)[/tex]

You can take it from there.


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