# $\frac{\mathrm d} {\mathrm dx}\sqrt{1/e^x}=$?

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.

-

$$\dfrac{\mathrm d} {\mathrm dx}\sqrt{\frac{1}{e^x}}$$ 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.

-

Hint:

1. $\displaystyle\sqrt\frac 1 {e^x}=e^{-\frac x 2}$
2. $\displaystyle\frac d {dx} e^{f(x)}=e^{f(x)}f'(x)$
-
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.

-
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}$\$

-