User:Michael Hardy/Abelian


 * $$ \begin{align}

\frac {dy}{dx} = {} & \frac d {dx} \Big( (2x+1)^3 (3x-1)^5 \Big) = (2x+1)^3 \cdot \frac d {dx} (3x-1)^5 + (3x-1)^5 \frac d {dx} (2x+1)^3 \\[12pt] = {} & (2x+1)^3\cdot 5(3x-1)^4 \cdot 3 + (3x-1)^3\cdot 3(2x+1)^2 \cdot 2 \\[12pt] = {} & 15(2x+1)^3(3x-1)^4 + 6(3x-1)^3 (2x+1)^2 \\[12pt] = {} & (3x-1)^3(2x+1)^2 \Big( 15(2x+2)(3x-1) + 6 \Big) \\[12pt] = {} & 6(3x-1)^3(2x+1)^2 \Big( 15x^2 + 10x - 4\Big) \\[12pt] = {} & 6(3x-1)^3(2x+1)^2 \cdot 15\left( x - \tfrac{5+\sqrt{85}} {15} \right) \left( x - \tfrac{5-\sqrt{85}} {15} \right) \\[12pt] = {} & 90 \cdot 3^3 \cdot 2^2 \cdot \left(x-\frac 1 3 \right)^3 \left(x+\frac 1 2\right)^2 \left( x - \tfrac{5+\sqrt{85}} {15} \right) \left( x - \tfrac{5-\sqrt{85}} {15} \right) \end{align} $$


 * $$ \begin{align} {} \\[8pt]

\frac {dy}{dx} = {} & \frac d {dx}\, \frac{\tan x}{1+ \cos x} = \frac{(1+\cos x)\frac d {dx} \tan x - (\tan x)\frac d {dx} (1+\cos x)}{(1+\cos x)^2} \\[12pt] = {} & \frac{(1+\cos x) \sec^2 x + (\tan x) \sin x}{(1+\cos x)^2} \\ [12pt] {} \end{align} $$

Problem #3 starts here:
 * $$ \begin{align}

& x e^y = y \sin x \\[12pt] & x \frac d {dx} e^y + e^y \frac d {dx} x = y \left( \frac d {dx} \sin x\right) + (\sin x)\frac {dy}{dx} \\[12pt] & x e^y \frac{dy}{dx} + e^y = y\cos x + (\sin x) \frac{dy}{dx} \\[12pt] & xe^y \frac{dy}{dx} - (\sin x)\frac{dy}{dx} = y\cos x - e^y \\[12pt] & (xe^y - \sin x) \frac {dy}{dx} = y\cos x - e^y \\[12pt] & \frac{dy}{dx} = \frac{y\cos x - e^y}{xe^y - \sin x} \\[12pt] {} \end{align} $$

Problem #4 starts here:
 * $$ \begin{align}

y = {} & (\arcsin(2x))^2 + \cot(\csc x) \\[12pt] \frac{dy}{dx} = {} & 2(\arcsin(2x))\cdot \frac d{dx} \arcsin(2x) - \csc^2(\csc x) \cdot (-\csc x \cot x) \\[12pt] = {} & 2(\arcsin(2x)) \cdot \frac 1 {\sqrt{1 - (2x)^2}} \cdot \frac d {dx} (2x) + \csc^2(\csc x) (\csc x \cot x) \\[12pt] = {} & 2(\arcsin(2x)) \cdot \frac 1 {\sqrt{1 - (2x)^2}} \cdot 2 + \csc^2(\csc x) (\csc x \cot x) \\[12pt] = {} & 2(\arcsin(2x)) \cdot \frac 1 {\sqrt{1 - (2x)^2}} \cdot 2 + \big( \csc(\csc x) \big)^2 (\csc x \cot x) \\[12pt] = {} & \frac {4\arcsin(2x)} {\sqrt{1 - (2x)^2}} + \big( \csc(\csc x) \big)^2 (\csc x \cot x) \\[12pt] {} \end{align} $$

Problem #5 starts here:
 * $$ \begin{align}

y = {} & \tan(10^x) + 10^{\tan(\pi x)} \\[12pt] \text{Recall that } & \frac d {dx} 10^x = 10^x \cdot\log_e 10 \\[12pt] \frac {dy}{dx} = {} & \sec^2(10^x) \cdot \frac d {dx} 10^x + 10^{\tan(\pi x)} \cdot \frac d {dx} \tan(\pi x) \\[12pt] = {} & \sec^2(10^x)\cdot 10^x \cdot \ln 10 + 10^{\tan(\pi x)} \cdot \sec^2(\pi x) \cdot \frac d {dx} (\pi x) \\[12pt] = {} & \sec^2(10^x)\cdot 10^x \cdot \ln 10 + 10^{\tan(\pi x)} \cdot \sec^2(\pi x) \cdot \pi \end{align} $$