How do you differentiate #f(x)= 1/(e^(3x) -5x)# using the quotient rule?

Question

How do you differentiate #f(x)=  1/(e^(3x)  -5x)# using the quotient rule?

Emma Aldridge · Accepted Answer

#(5-3e^(3x))/( e^(3x) - 5x)^2# The quotient Rule states that the derivative of a division of two functions #f(x)/g(x)# Is equal to #(f'(x)g(x) - f(x)g'(x))/ (g(x))^2# therefore in your Question, let #f(x) = 1# and #g(x) = e^(3x) - 5x# their respective derivatives are #f'(x) = 0# #g'(x) = 3e^(3x) - 5# therefore, the derivative of the entire equation using the Quotient rule is #((0*e^(3x) - 5x) - (1*(3e^(3x) - 5)))/ ( e^(3x) - 5x)^2# # = - (3e^(3x) - 5)/( e^(3x) - 5x)^2# which is equal to #(5-3e^(3x))/( e^(3x) - 5x)^2# = and you can simplify it more if you want but that's basically it

Aurora Alvarado · Answer

undefined To differentiate $ f(x) = \frac{1}{e^{3x} - 5x} $ using the quotient rule, let $ u(x) = 1 $ and $ v(x) = e^{3x} - 5x $.

Then, applying the quotient rule $ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} $, where $ u' $ and $ v' $ denote the derivatives of $ u $ and $ v $ respectively:

$$ u'(x) = 0 $$
$$ v'(x) = 3e^{3x} - 5 $$

$$ f'(x) = \frac{0 \cdot (e^{3x} - 5x) - 1 \cdot (3e^{3x} - 5)}{(e^{3x} - 5x)^2} $$
$$ f'(x) = \frac{-3e^{3x} + 5}{(e^{3x} - 5x)^2} $$

Isaiah Allen · Answer

undefined To differentiate $ f(x) = \frac{1}{{e^{3x} - 5x}} $ using the quotient rule:

1. Identify the numerator and denominator of the function $ f(x) $:
   - Numerator: 1
   - Denominator: $ e^{3x} - 5x $

2. Apply the quotient rule, which states that for a function $ \frac{u(x)}{v(x)} $, the derivative is given by:
   $$ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

3. Differentiate the numerator and denominator separately:
   - $ u(x) = 1 $, so $ u'(x) = 0 $
   - $ v(x) = e^{3x} - 5x $, so $ v'(x) = 3e^{3x} - 5 $

4. Apply the quotient rule formula to find $ \frac{d}{dx} \left( \frac{1}{{e^{3x} - 5x}} \right) $:
   $$ \frac{d}{dx} \left( \frac{1}{{e^{3x} - 5x}} \right) = \frac{0 \cdot (e^{3x} - 5x) - 1 \cdot (3e^{3x} - 5)}{(e^{3x} - 5x)^2} $$

5. Simplify the expression:
   $$ \frac{-3e^{3x} + 5}{{(e^{3x} - 5x)^2}} $$

Therefore, the derivative of $ f(x) = \frac{1}{{e^{3x} - 5x}} $ using the quotient rule is $ \frac{-3e^{3x} + 5}{{(e^{3x} - 5x)^2}} $.