Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The equation I have is $$\sqrt{1/e^x}.$$ I have been asked to do this with the chain rule. Couldn't find a clue.

share|cite|improve this question
up vote 2 down vote accepted

$$\dfrac{\mathrm d} {\mathrm dx}\sqrt{\frac{1}{e^x}}$$ Let $$u=\frac{1}{e^x}$$ then $$=\dfrac{\mathrm d \sqrt{u}}{\mathrm du} \dfrac{\mathrm du}{\mathrm dx}=\dfrac{1}{2\sqrt u} \dfrac{\mathrm du}{\mathrm dx}$$

This is how to do it with purely chain rule, but if instead one uses $$\sqrt{\frac{1}{e^x}}=e^{-\dfrac{x}{2}}$$ then it becomes trivial.

share|cite|improve this answer


  1. $\displaystyle\sqrt\frac 1 {e^x}=e^{-\frac x 2}$
  2. $\displaystyle\frac d {dx} e^{f(x)}=e^{f(x)}f'(x)$
share|cite|improve this answer
I got to that point but how do I put it in the format $$\frac{dy}{du}\frac{du}{dx}$$? – Mohaimenul Haque Adnan Mar 10 '13 at 2:42

If you are having trouble, its always best to get the hang of it by explicitly stating the functions which are composed. Let the function given be $f$. I will write this as a composition of the functions $g(x) = \sqrt{x}, h(x) = \frac{1}{x}, k(x) = e^x$. Then we have $$ f(x) = g(h(k(x))) $$ $$ f'(x) = g'(h(k(x))) \cdot h'(k(x)) \cdot k'(x) \tag{1} \\ f'(x) = \frac{1}{2\sqrt{\frac{1}{e^x}}} \cdot \left(-\frac{1}{(e^x)^2}\right) \cdot e^x. $$ Simplify the answer to get $$ f'(x) = -\frac{e^{-\frac{x}{2}}}{2}. $$

If you prefer the $\frac{dy}{dx}$ notation rewrite in the following way. Let $y = f(x), p = h(k(x)) = \frac{1}{e^x}, q = k(x) = e^x$. Then we have $$ \frac{dy}{dx} = \frac{dy}{dp} \cdot \frac{dp}{dq} \cdot \frac{dq}{dx}. $$ This line and $(1)$ above is equivalent. The steps after this is exactly the same, just different notation to show derivatives.

share|cite|improve this answer
Thats cool. Will you please explain this part for me: $$f(g(h(k(x))))= g'(h(k(x))).h'(k(x)).k'(x)$$ – Mohaimenul Haque Adnan Mar 10 '13 at 3:00
@MohaimenulHaqueAdnan Its actually $f'(x) = g'(h(k(x))) \cdot h'(k(x)) \cdot k'(x)$. To see it even more clearly, let $m(x) = h(k(x))$. Then $f(x) = g(m(x))$. Using the chain rule, $f'(x) = g'(m(x)) \cdot m'(x)$ (this is exactly what the chain rule states). Now, since $m(x) = h(k(x))$, we can use chain rule on this to get $m'(x) = h'(k(x)) \cdot k'(x)$. So, $f'(x) = g'(m(x)) \cdot m'(x) = g'(h(k(x))) \cdot h'(k(x)) \cdot k'(x)$ – Pratyush Sarkar Mar 10 '13 at 3:06
@MohaimenulHaqueAdnan Or just stick to the $dy/dx$ notation since its much easier to follow. I explained this in my answer. – Pratyush Sarkar Mar 10 '13 at 3:07
I got it. Thanks. – Mohaimenul Haque Adnan Mar 10 '13 at 3:09
@MohaimenulHaqueAdnan You're welcome. You might want to look at the other answers as well as they pointed out that you can rewrite the function in a much simpler form (which I missed as I was answering quickly) which is very easy to differentiate. – Pratyush Sarkar Mar 10 '13 at 3:15

just go through this like same as your problem

$For\ the\ exponential\ function, $$\frac{d}{dx}\left(e^{f(x)}\right) = e^{f(x)}f'(x).$

Here, $e^{f(x)} = e^{-\frac{x}{2}}$, so $f(x) = -\frac{x}{2}$. So then we must have $$\frac{d}{dx}\left(e^{-\frac{x}{2}}\right) = e^{-\frac{x}{2}}\left(-\frac{1}{2}\right) = -\frac{e^{-\frac{x}{2}}}{2}$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.