MaxWong

Oh I see!!! Thank you for pointing it out :D

Thank you for telling us the mistake :D

Pie is a kind of food.

I think you mean 'pi'

'pi' is the ratio of the circumference of any circle to the diameter of the same circle.

~The smartest cookie in the world

Ta(7) is still not found yet. Maybe we will wait for the super mathematicians for few more years and they get the value of it.

So far, the ever-found biggest Taxicab number is 

\(\text{Ta}(6)=\;24153319581254312065344\\ =582162^3 + 28906206^3\\ =3064173^3+ 28894803^3\\ = 8519281^3 + 28657487^3\\ =16218068^3 + 27093208^3\\ = 17492496^3 + 26590452^3\\ =18289922^3 + 26224366^3\)

Enough mind-blowing isn't it?

~The smartest cookie in the world

Maybe.

Let's think about \((i^i)^i\) which is \(e^{-i\pi/2}=e^{1/2\cdot\ln(-1)}=\left(e^{\ln(-1)}\right)^{1/2}=\sqrt{-1} = i\)

Then think about \(\left(\left(i^i\right)^i\right)^i\\ \) which is

\(i^i = e^{-i\pi/2}\)

We just go in imaginary -> real -> imaginary -> real -> ......... circles.

We can conclude that:

\(\left(\left(i^i\right)^{.......}\right)^i\text{ with odd number of }i = i\\ \left(\left(i^i\right)^{.......}\right)^i\text{ with even number of }i = e^{-\pi/2}\)

Thank you!! That is really awesome!! 

Thank you for answering :D

\(\infty \times \infty\\ = \dfrac{1}{0}\times \dfrac{1}{0}\\ =\dfrac{1\times 1}{0\times 0}\\ =\dfrac{1}{0}\\ =\infty\)

\(1.\\ f(x) = 5\pi^2\\ f'(x) = 0\text{ as it is a constant}\\ ......................................................\\ 2.\\ f(x) = 5\sqrt x\\ f'(x) = \dfrac{5}{2\sqrt x}\\......................................................\\ 3.\\ f(x) = \dfrac{1}{x^4}=x^{-4}\\ f'(x) = -4x^{-5}=-\dfrac{4}{x^5}\\......................................................\\ 4.\\ f(x) = 2x^7 - 5x^{-7}\\ f'(x) = 14x^6 +35x^{-6}=14x^6 + \dfrac{35}{x^6}\\......................................................\\ 5.\\ f(x) = \dfrac{2x-1}{x^3}=\left(x^{-3}\right)(2x-1)=2x^{-2}-x^{-3}\\ f'(x) = -4x^{-3} +3x^{-4}=-\dfrac{4}{x^3}+\dfrac{3}{x^4}\\ f'(1) = -\dfrac{4}{1^3}+\dfrac{3}{1^4}=-4 + 3 = -1 \\\text{P.S.: Need not use quotient rule nor product rule}\\ ......................................................\\ 6.\\ f(x) = 3x^4 \ln x\\ f'(x) = (12x^3)(\ln x)+(3x^4)(\dfrac{1}{x})=12x^3\ln x + 3x^3\\ f'(1) = 12(1^3)\ln 1 + 3(1^3)=12(0) + 3(1) = 3 \\......................................................\\ 7.\\ f(x)=8\sqrt x + 6x^{3/4}=8x^{1/2}+6x^{3/4}\\ f'(x) = \dfrac{4}{\sqrt x}+\dfrac{9}{2x^{1/4}}\\......................................................\\8.\\ y=\dfrac{5}{x^2}-\dfrac{3}{\sqrt x}+5x^4 = 5x^{-2} - 3x^{-1/2} + 5x^4\\ \dfrac{dy}{dx} = -10x^{-3} +\dfrac{3}{2x^{3/2}}+20x^3 = -\dfrac{10}{x^3}+\dfrac{3}{2x^{3/2}}+20x^3\\​......................................................\\ 9.\\ y = (6x^3 + 2)(5x - 3)\\ \dfrac{dy}{dx} = (18x^2)(5x - 3) + (6x^3 + 2)(5) = 120 x^3 -54x^2+10\\​......................................................\\ 10.\\ f(t) = \dfrac{(8t-3)(2t+5)}{t-7}= \dfrac{16t^2 +34t-15}{t-7} = 16t + 146 +\dfrac{1007}{t-7}\\ f'(t) = 16 - 1007(t-7)^{-2}=16-\dfrac{1007}{(t-7)^2}\\ f'(1) = 16 - \dfrac{1007}{(1-7)^2}=16 - 27\dfrac{35}{36} = -\dfrac{431}{36}\)

Hey :)

How ya doing?

~The smartest cookie in the world