I know how to take the derivative of this function(It would be $e^{-5x} *( -5)$ right?) but I'm stuck on trying to take the anti derivative of it. This is a small part of a larger integration problem I'm working on.
$\endgroup$ 13 Answers
$\begingroup$Derivative of $e^{ax}$ is $ae^{ax}$, this can be easily seen by applying the chain rule and using the property of exponential function that derivative of $e^x$ is again $e^x$.
Since derivative is a linear operator, derivative of $be^{ax}$ is $b$ times the derivative of $e^{ax}$, i.e. it is $abe^{ax}$. By choosing $b=a^{-1}$, one gets that derivative of $a^{-1}e^{ax}$ is $e^{ax}$. Derivative of a function plus constant is equal to the derivative of the function because derivative of the constant is zero. This means that derivative of $a^{-1}e^{ax}+C$ is also $e^{ax}$, where $C$ is any constant.
Anti-derivative works another way, so anti-derivative of $e^{ax}$ is $a^{-1}e^{ax}+C$. Therefore, for special case of $a=-5$, one gets that anti-derivative of $e^{-5x}$ is:
\begin{equation} \frac{e^{-5x}}{-5}+C \end{equation}
$\endgroup$ $\begingroup$hint: considering that $e^x$ has an anti-derivative of $e^x$, try to find a function that has as derivative $e^{-5x}$. Can you go on from here?
$\endgroup$ $\begingroup$You are correct about the derivative. Since the derivative is the original function multiplied by $-5$, the antiderivative is the original function divided by $-5$ plus a constant. Try taking the derivative to confirm it.
$\endgroup$ 2
