SOLUTION: Use logarithmic differentiation to solve:
f(x) =(x^(3/2)*e^(-(x^2)))/(1-e^(x))
Don't know if I did this correct.
ln(f(x)) = ln((x^(3/2)*e^(-(x^2)))/(1-e^(x)))
ln(f(x)) = l
Algebra.Com
Question 1002755: Use logarithmic differentiation to solve:
f(x) =(x^(3/2)*e^(-(x^2)))/(1-e^(x))
Don't know if I did this correct.
ln(f(x)) = ln((x^(3/2)*e^(-(x^2)))/(1-e^(x)))
ln(f(x)) = ln((x^(3/2)*e^(-(x^2)) - ln(1-e^(x))
ln(f(x)) = ln(x^(3/2))+ln(e^(-(x^2))) - ln(1-e^(x))
1/(f(x))*f'(x) = 3/2ln(x)-x^2-(x)(ln(1)
f'(x) = f(x)(3/2)ln(x)-x^2-x
Pretty sure this is incorrect
Thank you
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
Here is how I did it
f(x) =(x^(3/2)*e^(-x^2))/(1-e^x)
ln(f(x)) = ln[ (x^(3/2)*e^(-x^2))/(1-e^x) ]
ln(f(x)) = ln[ x^(3/2)*e^(-x^2) ] - ln(1-e^x)
ln(f(x)) = ln(x^(3/2))+ln(e^(-x^2)) - ln(1-e^x)
ln(f(x)) = (3/2)*ln(x)-x^2*ln(e) - ln(1-e^x)
ln(f(x)) = (3/2)*ln(x)-x^2*1 - ln(1-e^x)
ln(f(x)) = (3/2)*ln(x) - x^2 - ln(1-e^x)
1/(f(x))*f'(x) = (3/2)*(1/x) -2x - [1/(1-e^x)]*(-e^x)
1/(f(x))*f'(x) = 3/(2x) - 2x + (e^x)/(1-e^x)
f'(x) = f(x) * [ 3/(2x) - 2x + (e^x)/(1-e^x) ]
You were on the correct path. You just had a few parenthesis out of place (or one too many of one side). You also made a few differentiating errors.
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