MaxWong

Real number units = 1

Imaginary number units = i = $\sqrt{-1}$

$-2+-544343\\ = -2 - 544343\\ = -544345$

Hope that helps.

$(-\dfrac{1}{3})(21)\\ =\dfrac{-21}{3}\\ =-7$

5^1 mod 8 = 5

5^2 mod 8 = 1

5^3 mod 8 = 5

therefore 5^(2n+1) mod 8 = 5

therefore 5^137 mod 8 = 5.

$h'(1)=\dfrac{1}{\sqrt{1+1^2}}=\dfrac{\sqrt2}{2}$

$g'(1)=(e^1)((1+1^2)(1+1)+2(1)^2)=6e$

$f'(1)= 1^2-15(1)^2(\ln 1)= 1$

$\frac{d}{dx}\sqrt{1+x^2}\\\begin{array}{rl}h=\sqrt{u}&u=1+x^2\\\end{array}\\ \frac{dh}{dx}=\frac{dh}{du}\times \frac{du}{dx}=\dfrac{1}{2\sqrt{1+x^2}}\times 2x=\dfrac{x}{\sqrt{1+x^2}}$

$\frac{d}{dx} (1 + xe^x (1+x^2))$

$= \frac{d}{dx} 1 + (1+x^2)\frac{d}{dx}xe^x+(xe^x)\frac{d}{dx}(1+x^2)$

$= (1+x^2)(e^x)(1+x)+(x)(e^x)(2x)$

$=(e^x)((1+x^2)(1+x)+2x^2)$

$\frac{d}{dx}(1-x^3(5\ln x + 2))$

$=\frac{d}{dx}1 - \frac{d}{dx} 5x^3\ln x +\frac{d}{dx} 2x^3$

$=\frac{d}{dx}2x^3-(5x^3)\frac{d}{dx}\ln x - (\ln x)\frac{d}{dx} 5x^3$

$= 6x^2 - (5x^3)(\frac{1}{x})-(\ln x )(15x^2)$

$= 6x^2-5x^2-15x^2\ln x$

$= x^2-15x^2\ln x$