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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?

amWhy
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John Rawls
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    What does the answer key say? It's correct. – Mathematician 42 Sep 04 '17 at 05:29
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    @Mathematician42 it says $\frac{e^{\sqrt{x}}}{2*\sqrt{x}}$ – John Rawls Sep 04 '17 at 05:32
  • Your second step has a negative sign that is redundant....But, your final answer is right! – Susan_Math123 Sep 04 '17 at 05:36
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    Did you notice that $x^{-1/2}=1/\sqrt{x}$? – Susan_Math123 Sep 04 '17 at 05:38
  • This is not a calculus problem, but rather an arithmetical problem in which you have failed to acknowledge that $$\frac 12 x^{\frac {-1}{2} } = \frac 1{2x^{\frac 12}} = \frac 1{2\sqrt x}$$ Now just multiply by $e^{\large x^{\large \frac 12}}$ to get $$\frac {e^{\sqrt x}}{2\sqrt x}$$ – amWhy Sep 04 '17 at 21:31
  • @JohnRawls Your differentiation was fine. But you need to work on the basics of arithmetic if you want to be successful in math. $x^{1/2} = \sqrt x$, and $x^{-b} = \frac 1{x^b},$ and all the other fun (and more difficult) stuff you need to know to succeed in calculus. In other words, while your differentiation is indeed correct, you failed to recognize your answer is equivalent to the given answer you state above. So you need to review exponents, fraction exponents (roots), and negative exponents. – amWhy Sep 04 '17 at 21:42

3 Answers3

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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.

Susan_Math123
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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, $$ \frac{d}{dx}\left[e^{\sqrt{x}}\right] = \frac{d(e^y)}{dy}\frac{d(y)}{dx} = e^y(\frac{1}{2\sqrt{x}}) = \frac{e^\sqrt{x}}{2\sqrt(x)} $$

paulw
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  • I say this not as a competitive answerer but as a student of calculus and as a careful notator: I find this rampant parenthesizing fairly jarring. It is totally your business, but I thought I would put it out there. – gen-ℤ ready to perish Sep 04 '17 at 06:36
  • @ChaseRyanTaylor Chill. It is quite clear what paulw is saying, and in writing so, makes it unambiguous. So please, mind your own post, and improve your post prior to criticizing another's post. And, I beg to differ that you are a "careful notator" when you feel the need to scream out (highlight) the final answer, which everyone else obtained as well. – amWhy Sep 04 '17 at 21:48
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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.

gen-ℤ ready to perish
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