How do you find the derivative of #sinx/e^x#?

Question

Charles Anctil · Accepted Answer

#(cosx-sinx)/e^x#

differentiate using the #color(blue)"quotient rule"#

#"Given "f(x)=(g(x))/(h(x))" then"#

#color(red)(bar(ul(|color(white)(2/2)color(black)(f'(x)=(h(x)g'(x)-g(x)h'(x))/(h(x))^2)color(white)(2/2)|)))to(A)#

#"here " f(x)=(sinx)/e^x#

#g(x)=sinxrArrg'(x)=cosx#

#"and "h(x)=e^xrArrh'(x)=e^x#

Substituting these values into (A)

#f'(x)=(e^x(cosx)-sinx(e^x))/e^(2x)#

#rArrf'(x)=(e^x(cosx-sinx))/e^(2x)=(cosx-sinx)/e^x#

Emily Ammerman · Answer

undefined To find the derivative of $ \sin(x)/e^x $, we use the quotient rule. The quotient rule states that if we have a function of the form $ f(x) = \frac{g(x)}{h(x)} $, then the derivative of $ f(x) $ is given by $ f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2} $. Applying this rule to our function, with $ g(x) = \sin(x) $ and $ h(x) = e^x $, we have $ g'(x) = \cos(x) $ and $ h'(x) = e^x $. Plugging these values into the formula, we get:

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

So, the derivative of $ \sin(x)/e^x $ is $ (\cos(x) - \sin(x))/e^x $.