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
$\endgroup$ 53 Answers
$\begingroup$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: $y=ex$.
$\endgroup$ $\begingroup$The troublesome equal sign isn't a conclusion, it's a condition. The slope of the line is expressed in two different ways. $m = \frac{e^p}{p}$ by algebra methods and $m=e^p$ by calculus methods. They are set equal to each other so that you can solve for $p$:
$$\frac{e^p}{p} = e^p$$
$$\frac{1}{p} = 1$$
$$p=1$$
So the slope of the tangent line you're seeking is $e^1$. Its $y$-intercept is $0$, so the equation is $y=ex$.
$\endgroup$ 3 $\begingroup$Problem:
Find an equation for a tangent line to the curve $y=e^{x}$ which also goes through the origin.
Here's how I would solve it:
For a line to be tangent to a given curve, the line and the curve must share a common point. Let's call it $(x_{0},e^{x_{0}})$. Then, the slope of the line tangent to the curve $y=e^{x}$ at $x_{0}$ must be: $\frac{d}{dx}e^{x}|_{x=x_{0}}=e^{x_{0}}$. From that we can fine the line equation of the sought tangent line in terms of the point $x_{0}$:
$$ y-e^{x_{0}}=e^{x_{0}}(x-x_{0})\implies y=e^{x_{0}}x-e^{x_{0}}(x_{0}-1) $$
According to the problem statement, the line should also go through the origin:
$$ 0=e^{x_{0}}\cdot 0-e^{x_{0}}(x_{0}-1)\implies x_{0}=1\\ $$
Now, just plug in $x_{0}=1$ into our original equation:
$$ y=e^{1}x-e^{1}(1-1)\implies y=ex $$
Answer: $y=ex$.
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