How do you differentiate #f(x) =1/(e^(3-x)+1)# using the quotient rule?

Question

Christopher Allers · Accepted Answer

#f'(x)=(e^(x+3))/(e^x+e^3)^2#

According to the quotient rule,

#f'(x)=((e^(3-x)+1)d/dx(1)-1d/dx(e^(3-x)+1))/(e^(3-x)+1)^2#

Finding both of the internal derivatives, we see that

#d/dx(1)=0#

To find the second derivative, use the chain rule:

#d/dx(e^x)=e^x" "=>" "d/dx(e^u)=e^u*(du)/dx#

Since #u=3-x#,

#d/dx(e^(3-x)+1)=e^(3-x)d/dx(3-x)=e^(3-x)(-1)=-e^(3-x)#

The #1# will be neglected since it is a constant.

Plug these both back in:

#f'(x)=((e^(3-x)+1)(0)-1(-e^(3-x)))/(e^(3-x)+1)^2#

#f'(x)=e^(3-x)/(e^(3-x)+1)^2#

We can try to simplify further:

#f'(x)=(e^(3-x))/(e^3/e^x+e^x/e^x)^2#

#f'(x)=(e^(3-x))/(((e^3+e^x)^2)/e^(2x))#

#f'(x)=(e^(3-x)e^(2x))/(e^x+e^3)^2#

#f'(x)=(e^(x+3))/(e^x+e^3)^2#

Isaiah Allen · Answer

undefined To differentiate $ f(x) = \frac{1}{e^{3-x} + 1} $ using the quotient rule, follow these steps:

1. Identify $ u(x) $ as the numerator and $ v(x) $ as the denominator.
2. Compute $ u'(x) $ and $ v'(x) $.
3. Apply the quotient rule formula: $$ \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} $$
4. Substitute the values obtained into the formula.

So, the differentiation of $ f(x) $ with respect to $ x $ using the quotient rule is: $$ f'(x) = \frac{(0)(e^{3-x} + 1) - (1)(-e^{3-x})}{(e^{3-x} + 1)^2} $$
$$ f'(x) = \frac{e^{3-x}}{(e^{3-x} + 1)^2} $$