Find the derivative of $e^{\sqrt{x}}$
Find the derivative of $e^{\sqrt{x}}$?
My steps:
I first made it cleaner as shown below
$e^{x^{\frac{1}{2}}}$
Then I apply derivatives to this
$e^{x^{-\frac{1}{2}}}$
But anything to the e power derivative stays the same, and I apply chain rule so I get the following:
$e^{x^{\frac{1}{2}}} * \frac{1}{2}x^{-\frac{1}{2}}$
But my answer key says otherwise, why is this?
$\endgroup$ 63 Answers
$\begingroup$Your final answer is right. Define $f(x)=e^{x}$ and $g(x)=\sqrt{x}$, then $$\frac{d}{dx}e^{\sqrt{x}} = \frac{d}{dx}f(g(x))= g'(x)f'(g(x))$$ Note that $g'(x)=\frac{d}{dx}x^{1/2}=\frac{1}{2}x^{-1/2}$ and $f'(g(x))=e^{\sqrt{x}}$. It is proved.
$\endgroup$ $\begingroup$Yeah your answer is correct, an easier way to think about derivatives of the exponential function is using the chain rule. That is $$ \frac{d}{dx}\left[f(g(x))\right] = \frac{d}{d(g(x))}\left[f(g(x))\right]\frac{d(g(x))}{dx} $$ This might look a bit intimidating (especially the $\frac{d(f(g(x)))}{d(g(x))}$ part), but its really not that bad. We are essentially saying that we are looking at how $f$ changes when $g(x)$ changes. It might help to put $y=g(x)$ so then we have, $$ \frac{d}{dx}\left[e^{\sqrt{x}}\right] = \frac{d(e^y)}{dy}\frac{d(y)}{dx} = e^y(\frac{1}{2\sqrt{x}}) = \frac{e^\sqrt{x}}{2\sqrt(x)} $$
$\endgroup$ 2 $\begingroup$Let $u=\sqrt x=x^{1/2}$ and $y=e^u$.
$${\frac{dy}{du}} = \frac{d}{du}\,e^u = {e^u}=\color{blue}{e^{\sqrt x}}$$
$$\frac{du}{dx} = \frac{d}{dx}\,x^{1/2} = \frac12x^{1/2-1}=\frac12x^{-1/2}=\color{blue}{\frac{1}{2\sqrt x}}$$
$$\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}=\color{blue}{e^{\sqrt x}}\cdot\color{blue}{\frac{1}{2\sqrt x}}=\bbox[yellow,5px]{\frac{e^{\sqrt x}}{2\sqrt x}}$$
Perhaps your answer was different than the key simply because it was written in an equivalent, but still different, form. What I have put in yellow is not even considered completely rationalized.
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