I have an assignment
Find an equation for a tangent line to the curve $y=e^x$ which also goes through the origin.
However, in my formula it is asserted
that the slope between at a point "p" and the origin is
$m=\frac{e^p-0}{p-o}$ = $\frac{d}{dx}e^x$ (at $x=p$) = $e^p$
but surely $\frac{e^p-0}{p-o} = \frac{e^p}{p}$ how is this equal to
$\frac{d}{dx}e^x$ (at $x=p$) = $e^p$?
Am I missing something obvious?
Thank you in advance
Best Answer
If you mean the tangent line to the graph then let $(a,e^a)$ be a common point.
Thus, $$y-e^a=e^a(x-a)$$ and since our tangent line goes through the origin, we obtain $$0-e^a=e^a(0-a),$$ which gives $a=1$ and the answer: $y=ex$.