How do you differentiate #f(x)=A/(B+Ce^x)#?

Question

Isaiah Allen · Accepted Answer

#f'(x)=-(ACe^x)/(B+Ce^x)^2#

If we want to differentiate with the Quotient Rule:

#f'(x)=((B+Ce^x)d/dxA-Ad/dx(B+Ce^x))/(B+Ce^x)^2#

#d/dxA=0#, the derivative of a constant is zero.

#d/dx(B+Ce^x)=Ce^x#

Thus,

#f'(x)=-(ACe^x)/(B+Ce^x)^2#

Paisley Johnson · Answer

#f'(x)=-\frac{A*C*e^x}{(B+C*e^x)^2}# Remember: #(\frac{f(x)}{g(x)})^'=\frac{f'(x)g(x)-f(x)g'(x)}{(g(x))^2}# in this case #f(x)=A# the derivative of a constant is zero and #g(x)=(B+C*e^x)#, it become: #(\frac{A}{g(x)})^'=\frac{-Ag'(x)}{(g(x))^2}# #(\frac{A}{g(x)})^'=\frac{-A(B+C*e^x)^'}{(B+C*e^x)^2}=# #=\frac{-A*C*e^x}{(B+C*e^x)^2}#

Isaiah Allen · Answer

undefined To differentiate the function $ f(x) = \frac{A}{B + Ce^x} $, you can use the quotient rule. The quotient rule states that for functions $ u(x) $ and $ v(x) $, if $ f(x) = \frac{u(x)}{v(x)} $, then $ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $. Applying this rule to the given function, we have:

$$ f(x) = \frac{A}{B + Ce^x} $$
$$ u(x) = A, \quad u'(x) = 0 $$
$$ v(x) = B + Ce^x, \quad v'(x) = Ce^x $$

Therefore, using the quotient rule, the derivative $ f'(x) $ is:

$$ f'(x) = \frac{0(B + Ce^x) - A(Ce^x)}{(B + Ce^x)^2} $$
$$ f'(x) = \frac{-ACe^x}{(B + Ce^x)^2} $$