How i can derivative this problem

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Hi all

Queation say : x^2 - 2x + 3 / X^5

I try to solve

y1 = X^2/X^5 - 2x/x^5 + 3 X^5

If this derivative is coorect if yes how i can complet

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Use the product rule on (1/x^5) and (x^2-2x+3)

Should end up with -2/x^5+2/x^4-(5 (x^2-2 x+3))/x^6
 
r-soy said:
Hi all

Queation say : x^2 - 2x + 3 / X^5

I try to solve

y1 = X^2/X^5 - 2x/x^5 + 3 X^5
If the function is y = (x^2 - 2x + 3)/x^5, then you can rewrite the equation as y = x^2/x^5 - 2x/x^5 + 3/x^5.

Each term should be simplified. Then you can differentiate each term to get y'.
r-soy said:
If this derivative is coorect if yes how i can complet

>
 
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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