How do you differentiate # x/e^(3x)#?

Question

Luke Alvarez · Accepted Answer

Apply the quotient rule to get #(1-3x)/(e^(3x))#.

The quotient rule says that, for two functions of #x#, call them #u# and #v#, #d/dx(u/v)=(u'v-uv')/v^2# In this problem, #u=x# and #v=e^(3x)#. Step 1 is to take the derivative of #u# and #v# to make the calculations easier: #u=x# #u'=1# #v=e^(3x)# #v'=3e^(3x)# We can now make the substitutions: #d/dx(x/e^(3x))=((x)'(e^(3x))-(x)(e^(3x))')/((e^(3x))^2)# #d/dx(x/e^(3x))=(e^(3x)-3xe^(3x))/((e^(3x))^2)# And finish off with a little algebra: #d/dx(x/e^(3x))=(e^(3x)(1-3x))/((e^(3x))^2)# #d/dx(x/e^(3x))=(1-3x)/(e^(3x))#

Christopher Agresta · Answer

undefined To differentiate $ \frac{x}{e^{3x}} $, you can use the quotient rule. The quotient rule states that if you have a function in the form $ \frac{f(x)}{g(x)} $, then its derivative is given by:

$$ \frac{d}{dx} \left( \frac{f(x)}{g(x)} \right) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} $$

For $ \frac{x}{e^{3x}} $, let $ f(x) = x $ and $ g(x) = e^{3x} $. Then:

$$ f'(x) = 1 $$
$$ g'(x) = 3e^{3x} $$

Applying the quotient rule:

$$ \frac{d}{dx} \left( \frac{x}{e^{3x}} \right) = \frac{(1)(e^{3x}) - (x)(3e^{3x})}{(e^{3x})^2} $$

$$ = \frac{e^{3x} - 3xe^{3x}}{e^{6x}} $$

$$ = \frac{e^{3x}(1 - 3x)}{e^{6x}} $$

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