How do you differentiate #f(x) = (e^(4-x))/(e^(1-x)-1)# using the quotient rule?

Question

William Abdalla · Accepted Answer

#f'(x) = e^(4-x) / (e^(1-x)-1)^2# #f(x)=u/v, f'(x)=(u'v-v'u)/v^2# So, we need #u'# and #v'#: #u' = (e^(4-x))' = e^(4-x) * (4-x)' = -e^(4-x)# #v' = (e^(1-x) - 1)' = e^(1-x) * (1-x)' = -e^(1-x)# #f'(x) = (-e^(4-x) * (e^(1-x) - 1) + e^(1-x) * e^(4-x)) / (e^(1-x)-1)^2# #f'(x) = (-e^(5-2x) + e^(4-x) + e^(5-2x)) / (e^(1-x)-1)^2# #f'(x) = e^(4-x) / (e^(1-x)-1)^2#

Adam Adkins · Answer

undefined To differentiate $ f(x) = \frac{e^{4-x}}{e^{1-x}-1} $ using the quotient rule, you would follow these steps:

1. Identify the numerator and denominator functions: $ u(x) = e^{4-x} $ and $ v(x) = e^{1-x}-1 $.
2. Use the quotient rule formula: $ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\cdot u' - u\cdot v'}{v^2} $.
3. Calculate the derivatives of $ u $ and $ v $: $ u' = -e^{4-x} $ and $ v' = -e^{1-x} $.
4. Substitute the derivatives and original functions into the quotient rule formula: $ f'(x) = \frac{(e^{1-x}-1)(-e^{4-x}) - (e^{4-x})(-e^{1-x})}{(e^{1-x}-1)^2} $.
5. Simplify the expression to get the final answer for $ f'(x) $.