Register to reply 
L'Hopitals Rule... need help with derivatives? 
Share this thread: 
#1
Jul606, 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???? 


#2
Jul606, 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)



#3
Jul706, 03:57 PM

P: 788

[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. 


Register to reply 
Related Discussions  
Limits question L'Hopitals rule  Calculus & Beyond Homework  3  
L'Hopitals Rule and Standard Limits  Calculus & Beyond Homework  11  
L'Hopitals Rule  Calculus & Beyond Homework  5  
Limits without L'Hopitals rule  Introductory Physics Homework  8  
Using L'Hopitals rule  Introductory Physics Homework  17 