 Tangent through the origin to an exponential curve

For a nonzero real constant $a$, the tangent line to $y=e^{ax}$ at $x_{\ast }$ passes through the origin precisely when $e^{a{x}_{\ast }}-x_{\ast }a{e}^{a{x}_{\ast }}=0$. Thus $x_{\ast }=1/a$, $y_{\ast }=e$, and its slope is $k=ae$. This gives the unique such tangent line. If $a=0$, the graph is the horizontal line $y=1$ and no tangent line through the origin exists. The equality between the derivative and the slope from the origin, $f^{\prime }(x_{\ast })=f(x_{\ast })/x_{\ast }$, is useful for other differentiable functions too.