What is the tangent line to $y=e^{^{\frac{x}{2}}}$ through (0,0)?

Question

I'm trying to solve

$y=e^{^{\frac{x}{2}}}$

The derivative:

$\frac{\sqrt{e^{^x}}}{2}$

So, I need to find the "slope" to the linear function $y\:-\:y_1\:=m\left(x-x_1\right)$, solving the derivative by replacing $x$ by $0$ is $m=\frac{1}{2}$, so the answer is:

$y-y_1=m\left(x-x_1\right),\:y-0=\frac{1}{2}\left(x-0\right),\:y=\frac{1}{2}x$

But the answer that Wolfram Alpha gives me is:

$\frac{ex}{2}$

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So, does this problem requires another formula or process to be solved? Or did I just fail in the process?

Greetings!

Feel free to edit the post if there are any English issues in the post, I appreciate it so much!

Answer 1

Firs of all note that $(0,0)$ is not a point of the graph of the function. So fix $x_0\in\mathbb{R}.$ The tangent line at $(x_0,y_0)$ is given by $$y-e^{x_0/2}=\frac 12 e^{x_0/2}(x-x_0).$$ (Note that $f'(x)=\frac12 e^{x/2}$.) If this line contains the point $(0,0)$ then we have $$e^{x_0/2}=\frac 12 e^{x_0/2}x_0.$$ Since $e^{x_0/2}\ne 0$ we get that $x_0=2.$ Thus the tangent line is

$$y-e=\frac{e}{2}(x-2)$$ That is $$y=\frac{e}{2}x.$$

Answer 2

How did you take the derivative here? Remember that $\frac{d}{dx}(e^x) = e^x$ and try using the chain rule.