What is the derivative of $f(x) = e^-xcos(x^2)+e^xsin(x)$ ?

Question

1678 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-09T13:01:55

$=-e^(-x)cos(x^2) - 2xe^(-x)sin(x^2) + e^xsin(x) + e^xcos(x)$

Product rule is our friend here.

$d/(dx) (e^(-x)cos(x^2)) + d/(dx)(e^xsin(x))$

$=d/(dx)(e^(-x))cos(x^2) + e^(-x)d/(dx)(cos(x^2)) + d/(dx)(e^x)sin(x) + e^xd/(dx)(sin(x))$

$=-e^(-x)cos(x^2) - 2xe^(-x)sin(x^2) + e^xsin(x) + e^xcos(x)$

NB: for $d/(dx)(cos(x^2))$ I have used the chain rule because $x^2$ is also a function of x

What is the derivative of f(x) = e^-xcos(x^2)+e^xsin(x)f(x) = e^-xcos(x^2)+e^xsin(x)?