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

Question

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

William Abdalla · Accepted Answer

#d/(dx) (e^x/(x-3)) = (e^x(x-4))/(x-3)^2#

As per the quotient rule,

#d/(dx)( f(x)/g(x) )= (f'(x)g(x) -f(x)g'(x))/(g(x))^2#

For #f(x) = e^x# and #g(x) = x-3# we have:

#d/(dx) (e^x/(x-3)) = (e^x(x-3) - e^x)/(x-3)^2 = (e^x(x-4))/(x-3)^2#

Cameron Alexander · Answer

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

1. Identify $ u(x) $ as the numerator $ e^x $ and $ v(x) $ as the denominator $ x - 3 $.
2. Compute $ u'(x) $ and $ v'(x) $, which are the derivatives of $ u(x) $ and $ v(x) $ respectively.
3. Apply the quotient rule formula: 
$$ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$
4. Substitute the values of $ u(x) $, $ v(x) $, $ u'(x) $, and $ v'(x) $ into the formula.
5. Simplify the expression if necessary.

Applying the quotient rule to the given function, we get:

$$ f'(x) = \frac{(e^x)(x - 3) - (e^x)(1)}{(x - 3)^2} $$

$$ = \frac{e^x(x - 3) - e^x}{(x - 3)^2} $$

$$ = \frac{e^x(x - 3 - 1)}{(x - 3)^2} $$

$$ = \frac{e^x(x - 4)}{(x - 3)^2} $$

So, the derivative of $ f(x) = \frac{e^x}{x - 3} $ using the quotient rule is $ f'(x) = \frac{e^x(x - 4)}{(x - 3)^2} $.