Double Differentiation of (e^-x)(1-x)^1/2: Correct Solutions

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I have to differentiate (e^-x) ((1-x)^1/2) twice:

The first time i differentiate i get

(-e^-x) ((1-x)^1/2) + (e^-x) ((1-x)^-1/2) (-1/2)

then differentiating this i get

[(e^-x) ((1-x)^1/2)] - [(e^-x) ((1-x)^-1/2) (-1/2)] + [(-1/2e^-x) (1/2(1-x)^3/2) + ((1-x)^-1/2) (1/2e^-x)]

are these correct?

thankyou for checking
 
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If you want someone to check something for details, you should make sure you've shown what you want checked clearly and unambiguously. What does this mean '(1/2(1-x)^3/2)'? Is it (1/2)*(1-x)^(3/2)), or ((1/(2(1-x))^3)/2) etc etc. That said there should be terms with (1-x) to the power of 1/2, -1/2 and -3/2 (not 3/2). Other than that ok, I think.
 
First derivative is OK. Before differentiating for the second time please write the first derivative nicer by taking (1-x) out of the first term and do some adding. I am sure that simplification will result.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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