How to find the $n$-th derivative of $y=e^{ax+b}$
Please provide an explanation of the steps. Thanks.
$\endgroup$ 02 Answers
$\begingroup$Recall that for $y = e^{f(x)},\; y' = f'(x)e^{f(x)}$, where $f(x)$ is a function of $x$. This is making use of the chain rule, as it relates to the exponential function.
$\;y' = a \,e^{ax + b}$
$\,y'' = a^2 e^{ax + b}$
$y''' = a^3 e^{ax + b}$
$\vdots$
$y^{(n)} = a^n e^{ax + b}$.
$\endgroup$ 7 $\begingroup$If $f(x)=e^{ax+b}$, then
$$ \frac{df}{dx} = \frac{d(ax+b)}{dx}e^{ax+b} = ae^{ax+b} $$
$$ \frac{df^2}{dx^2} = \frac{d(ax+b)}{dx}ae^{ax+b} = a^2e^{ax+b} $$
$$ ... $$
$$ \frac{df^n}{dx^n} = \frac{d(ax+b)}{dx}a^{n-1}e^{ax+b} = a^ne^{ax+b} $$
$\endgroup$
