크레이지J의 탐구생활

11. $\frac{d}{dx} \sqrt{e^x}+e^{\sqrt{x}}$

$\begin{aligned} &\frac{d}{dx} \sqrt{e^x}+e^{\sqrt{x}}\\ &=(e^{\frac{x}{2}})' + (e^{x^{\frac{1}{2}}})'\\ &=e^{\frac{x}{2}}(\frac{1}{2})+e^{\sqrt{x}}(\frac{1}{2\sqrt{x}})\\ &=\frac{1}{2}\sqrt{e^x}+\frac{e^{\sqrt{x}}}{2\sqrt{x}} \end{aligned}$

12. $\frac{d}{dx} sec^3(2x)$

$\begin{aligned} &\frac{d}{dx} sec^3(2x)\\ &=3sec^2(2x)(sec(2x)tan(2x))2\\ &=6sec^3(2x)tan(2x) \end{aligned}$

13. $\frac{d}{dx} \frac{1}{2} (secx)(tanx) + \frac{1}{2} ln(secx + tanx)$

$\begin{aligned} &\frac{d}{dx} \frac{1}{2} (secx)(tanx) + \frac{1}{2} ln(secx + tanx)\\ &=\frac{1}{2}(secxtanxtanx+secxsec^2x)+\frac{1}{2(secx+tanx)}(secxtanx+sec^2x)\\ &=\frac{1}{2}sec x(tan^2x+sec^2x)+\frac{secx(tanx+secx)}{2(secx+tanx)}\\ &=\frac{1}{2}secx(1+tan^2x+sec^2x)=\frac{1}{2}secx 2sec^2x\\ &=sec^3x \end{aligned}$

14. $\frac{d}{dx} (xe^x)/(1+e^x)$

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

15. $\frac{d}{dx} (e^{4x})(cos(x/2))$

$\begin{aligned} &\frac{d}{dx} (e^{4x})(cos(\frac{x}{2}))\\ &=(e^{4x})'(cos(\frac{x}{2}))+(e^{4x})(cos(\frac{x}{2}))'\\ &=(e^{4x}4)cos(\frac{x}{2})+e^{4x}(-sin(\frac{x}{2})\frac{1}{2})\\ &=4e^{4x}cos(\frac{x}{2})-\frac{1}{2}e^{4x}sin(\frac{x}{2}) \end{aligned}$

16. $\frac{d}{dx} \frac{1}{\sqrt[4]{x^3 - 2}}$

$\begin{aligned} &\frac{d}{dx} \frac{1}{\sqrt[4]{x^3 - 2}}=\frac{d}{dx} (x^3 - 2)^{-\frac{1}{4}}\\ &=-\frac{1}{4} (x^3 - 2)^{-\frac{5}{4}}(3x^2)\\ &=-\frac{3x^2}{4(x^3-2)\sqrt[4]{x^3 - 2}}\\ &=-\frac{3x^2}{4\sqrt[4]{(x^3 - 2)^5}}\\ \end{aligned}$

17. $\frac{d}{dx} arctan(sqrt(x^2-1))$

$\begin{aligned} &\frac{d}{dx} tan^{-1}(\sqrt{x^2-1})\\ & (y=arctan (x), tan y=x, sec^2ydy=dx)\\ &(R.T angle=y, a=1, o=x, h=\sqrt{1+x^2})\\ & (dy/dx = \frac{1}{sec^2y}=cos^2y=\frac{1}{1+x^2})\\ &=cos^2(tan^{-1}(\sqrt{x^2-1}))\frac{1}{2\sqrt{x^2-1}}2x\\ &=cos^2(tan^{-1}(\sqrt{x^2-1}))\frac{x}{\sqrt{x^2-1}}\\ &=\frac{1}{1+x^2-1}\frac{x}{\sqrt{x^2-1}}\\ &=\frac{1}{x\sqrt{x^2-1}} \end{aligned}$

18. $\frac{d}{dx} (lnx)/x^3$

$\begin{aligned} &\frac{d}{dx} \frac{lnx}{x^3}=\frac{\frac{1}{x}x^3-lnx(3x^2)}{x^6}\\ &=\frac{x^2-3x^2lnx}{x^6}=\frac{1-3lnx}{x^4} \end{aligned}$

19. $\frac{d}{dx} x^x$

$\begin{aligned} &\frac{d}{dx} x^x\\ &(y=x^x, log_xy=x, \frac{ln y}{ln x}=x)\\ &(lny = xlnx, \frac{1}{y}dy=(lnx+1)dx)\\ &\frac{dy}{dx}=(1+lnx)y=(1+lnx)x^x \end{aligned}$
Alt.
$\frac{d}{dx} x^x, (x=e^{lnx})\\ =\frac{d}{dx} (e^{lnx})^x=\frac{d}{dx} (e^{xlnx})\\ =e^{xlnx}(lnx+1)\\ =x^x(lnx+1)$

20. $\frac{d}{dx}(x^3+y^3=6xy)$

$\begin{aligned} &\frac{d}{dx}(x^3+y^3=6xy) \\ &3x^2dx+3y^2dy=6ydx+6xdy\\ &(3x^2-6y)dx=(6x-3y^2)dy\\ &\frac{dy}{dx}=\frac{3x^2-6y}{6x-3y^2}\\ &=\frac{x^2-2y}{2x-y^2} \end{aligned}$
Alt.
$\begin{aligned} &x^3+y^3=6xy :(Dx)\\ &3x^2+3y^2y'=6y+6xy'\\ &(3x^2-6y)=(6x-3y^2)y'\\ &y'=\frac{3x^2-6y}{6x-3y^2}\\ &=\frac{x^2-2y}{2x-y^2} \end{aligned}$

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