크레이지J의 탐구생활

1. $\frac{d}{dx}ax^2+bx+c$

$\begin{aligned} &\frac{d}{dx} ax^2+bx+c=2ax+b\\ \end{aligned}$

2. $\frac{d}{dx}\frac{sin(x)}{1+cos(x)}$

$\begin{aligned} &\frac{d}{dx} \frac{sin(x)}{1+cos(x)}= \frac{(sinx)'(1+cosx)-sin(x)(1+cosx)'}{(1+cosx)^2} \\ &=\frac{cosx+cos^2x+sin^2(x)}{(1+cosx)^2}==\frac{1}{1+cos(x)} \end{aligned}$

3. $\frac{d}{dx} (1+cosx)/sinx$

$\begin{aligned} &\frac{d}{dx} \frac{(1+cosx)}{sinx} = \frac{(1+cosx)'sinx-(1+cosx)(sinx)'}{sin^2x} \\ &=\frac{-sin^2x-cosx-cos^2x}{sin^2x}=-\frac{1+cosx}{1-cos^2(x)}\\ &=-\frac{1+cos x }{sin^2x}=-\frac{1}{1-cos(x)} \end{aligned}$
Alt.
$\begin{aligned} &\frac{d}{dx} \frac{(1+cosx)}{sinx} = (\frac{1}{sinx})'+(\frac{cosx}{sinx})' \\ &=(csc x)'+(cot x)' = -csc x cot x-csc^2x\\ &=-cscx(cot x+cscx) \end{aligned}$

$-cscx(cot x+cscx) = -\frac{1}{sinx}(\frac{cosx+1}{sin x})\\ =-\frac{1+cosx}{sin^2x}$

4. $\frac{d}{dx}sqrt(3x+1)$

$\begin{aligned} &\frac{d}{dx} \sqrt{3x+1}= \frac{1}{2\sqrt{3x+1}}3=\frac{3}{2\sqrt{3x+1}} \end{aligned}$

5. $\frac{d}{dx} \sin^3 {x}+sin(x^3)$

$\begin{aligned} &\frac{d}{dx} \sin^3 {x}+sin(x^3)= 3sin^2xcosx+cos(x^3)(3x^2)\\ &=3sin^2xcosx+3x^2cos(x^3) \end{aligned}$

6. $\frac{d}{dx} 1/x^4$

$\begin{aligned} &\frac{d}{dx} \frac{1}{x^4} = (x^{-4})'=-\frac{4}{x^5}\\ \end{aligned}$

7. $\frac{d}{dx}(1+cotx)^3$

$\begin{aligned} &\frac{d}{dx} (1+cotx)^3 = 3(1+cotx)^2(-csc^2x)\\ &=--3csc^2x(1+cotx)^2 \end{aligned}$

8. $\frac{d}{dx} x^2(2x^3+1)^{10}$

$\begin{aligned} &\frac{d}{dx} x^2(2x^3+1)^{10}=(x^2)'(2x^3+1)^{10}+(x^2)((2x^3+1)^{10})'\\ &=2x(2x^3+1)^{10}+x^2(10(2x^3+1)^9(6x^2))\\ &=2x(2x^3+1)^{9}( 2x^3+1+ 30x^3 )\\ &=2x(2x^3+1)^{9}(32x^3+1) \end{aligned}$

9. $\frac{d}{dx} x/(x^2+1)^2$

$\begin{aligned} &\frac{d}{dx} \frac{x}{(x^2+1)^2}=\frac{(x^2+1)^2-x2(x^2+1)2x}{(x^2+1)^4}\\ &=\frac{(x^2+1)(x^2+1-4x^2)}{(x^2+1)^4} =\frac{(x^2+1)(-3x^2+1)}{(x^2+1)^4}\\ &=\frac{(-3x^2+1)}{(x^2+1)^3} \end{aligned}$

10. $\frac{d}{dx} 20/(1+5e^{-2x})$

$\begin{aligned} &\frac{d}{dx} \frac{20}{(1+5e^{-2x})}=\frac{-20(5e^{-2x})(-2)}{(1+5e^{-2x})^2}\\ &=\frac{200e^{-2x}}{(1+5e^{-2x})^2} \end{aligned}$

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