How do you differentiate # x/(e^(2x))#?

Question

Evelyn Agard · Accepted Answer

#(1-2x)/e^(2x)#

differentiate using the #color(blue)"quotient and chain rules"#

Quotient rule :

f(x) = g(x).h(x) then #f'(x)=(h(x).g'(x)-g(x).h'(x))/(h(x)^2#

Chain rule :

#d/dx(f(g(x))=f'(g(x)).g'(x)# #"-------------------------------------------------------------"# #g(x)=xrArrg'(x)=1#

#h(x)=e^(2x)rArrh'(x)=e^(2x).2=2e^(2x)# #"------------------------------------------------------------"# Substitute these values into f'(x) in the quotient rule

#f'(x)=(e^(2x).1-x.2e^(2x))/(e^(2x))^2=(e^(2x)-2xe^(2x))/e^(4x)#

#=(e^(2x)(1-2x))/e^(4x)=(1-2x)/e^(2x)#

Charles Anctil · Answer

undefined To differentiate the function $ \frac{x}{e^{2x}} $, you can use the quotient rule.

The quotient rule states that if you have a function of the form $ \frac{u}{v} $, where $ u $ and $ v $ are differentiable functions of $ x $, then the derivative is given by:
$$ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} $$

Using this rule, differentiate $ \frac{x}{e^{2x}} $:

$$ u = x $$
$$ v = e^{2x} $$

$$ u' = 1 $$
$$ v' = 2e^{2x} $$

So applying the quotient rule:

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

$$ = \frac{e^{2x} - 2xe^{2x}}{e^{4x}} $$

$$ = \frac{e^{2x}(1 - 2x)}{e^{4x}} $$

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