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

1 Answer
Jim G.
Feb 8, 2017

#(cosx-sinx)/e^x#

Explanation:

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#

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