Differentiation of logarithm: f(x)= (1-3 ln x + 2e^6x)^11

J

jenna89

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I need help on the following problem.
The directions are to find the derivative of the of the following:
f(x)= (1-3 ln x + 2e^6x)^11

This is what I came up with. Please let me know if this is accurate.
11(1-3 ln x + 2e ^6x)^10 (-3(1/x)+2(-6e^-6x)

I thank you in advance for your assistance.
 
Re: Differentiation of logarithm

I agree with all but the last term
f[x]= [ 1-3 ln x + 2 e^6x]^11
f ' [x] = 11 [ 1 -3 ln x =2 e^6x]^10 d[1- 3 lnx + 2 e^6x ] /x

d/dx 1=0 agree
d[3 lnx] / dx = 3/x agree
d[2e^6x]/dx = 2[6] e^6x disagree with your nswer

Arthur
 
Re: Differentiation of logarithm

your welcome
the derivative of e^u is e^u du/dx

d/dx 2e^6x= 2 e^6x[d/dx 6x]
d / dx 2 e^6x=2 e^6x[6] = 12 e^6x

Arthur
 
thank you.
Does that mean my final answer would be f'(x)=11(1-3 ln x +2e^-6x)^10 (-3(1/x)-12e^-6x)
 
NO!
the last term should be 12 e^6x. No minus sign
d/dx [1-3lnx+2e^(6x)]^11= 11[1-3lnx+2e^(6x) ]^10] [- 3/x +12e^(6x)]

Arthur
 
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