91. $\frac{d}{dx}x^3, definition$

$\begin{aligned} &\frac{d}{dx}x^3\\ &=\lim_{h\to0}\frac{(x+h)^3-x^3}{h}=\frac{3hx^2+3h^2x+h^3}{h}\\ &=\lim_{h\to0}3x^2+3hx+h^2=3x^2 \end{aligned}$

92. $\frac{d}{dx} \sqrt{3x+1}, def.$

$\begin{aligned} &\frac{d}{dx} \sqrt{3x+1}\\ &=\lim_{h\to0} \frac{\sqrt{3(x+h)+1}-\sqrt{3x+1}}{h}\\ &=\lim_{h\to0} \frac{(3x+3h+1)-(3x+1)}{h(\sqrt{3x+3h+1}+\sqrt{3x+1})}\\ &=\lim_{h\to0}\frac{3}{(\sqrt{3x+3h+1}+\sqrt{3x+1})}\\ &=\frac{3}{2\sqrt{3x+1}} \end{aligned}$

93. $\frac{d}{dx} \frac{1}{2x+5}, def.$

$\begin{aligned} &\frac{d}{dx} \frac{1}{2x+5}\\ &=\lim_{h\to0} \frac{\frac{1}{2(x+h)+5}-\frac{1}{2x+5}}{h}=\frac{ \frac{-2h}{(2x+2h+5)(2x+5)}}{h}\\ &=\lim_{h\to0} -\frac{2}{(2x+2h+5)(2x+5)}\\ &=-\frac{2}{(2x+5)^2} \end{aligned}$

94. $\frac{d}{dx}\frac{1}{x^2}, def.$

$\begin{aligned} &\frac{d}{dx}\frac{1}{x^2}\\ &=\lim_{h\to0} \frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}= \frac{\frac{-2xh-h^2}{x^2(x+h)^2}}{h}=\frac{-2x-h}{x^2(x+h)^2}\\ &=\frac{-2x}{x^4}=-\frac{2}{x^3} \end{aligned}$

95. $\frac{d}{dx}sinx, def.$

$\begin{aligned} &\frac{d}{dx} sinx=\lim_{h\to0} \frac{sin(x+h)-sin(x)}{h}\\ &=\lim_{h\to0} \frac{ sin(x)cos(h)+cos(x)sin(h)-sin(x)}{h}\\ &=\lim_{h\to0} \frac{sinx(cos(h)-1)}{h}+cos(x)\frac{sin(h)}{h}\\ & cos(h)=1-\frac{h^2}{2!}+\frac{h^4}{4!}-...\\ & \lim_{h\to0} \frac{cos(h)-1}{h}=hc+h^3c+..=O(h)=0 \\ & sin(h)=h-\frac{h^3}{3!}+...\\ & \lim_{h\to0} \frac{sin(h)}{h}=1-O(h^2)=1\\ & \therefore =sin(x)0+cos(x)1=cos(x) \end{aligned}$

96. $\frac{d}{dx}secx, def.$

$\begin{aligned} &\frac{d}{dx}sec(x)=\lim_{h\to0}\frac{\sec(x+h)-\sec(x)}{h}\\ &=\lim_{h\to0}\frac{1/\cos(x+h)-1/\cos(x)}{h}=\frac{cos(x)-cos(x+h)}{cos(x+h)cos(x)h}\\ &=\frac{cos(x)-cos(x)cos(h)+sin(x)sin(h)}{(cos(x)cos(h)-sin(x)sin(h))cos(x)h}\\ &=\frac{1-cos(h)+tan(x)sin(h)}{h(cos(x)cos(h)-sin(x)sin(h))}\\ &=\lim \frac{1}{cos(x)cos(h)-sin(x)sin(h)} (\lim \frac{1-cos(h)}{h}+\lim \frac{tan(x)sin(h)}{h}) \\ &=\frac{1}{cos(x)}(0+tan(x))=sec(x)tan(x) \end{aligned}$

97. $\frac{d}{dx}arcsinx, def.$

$\begin{aligned} &\frac{d}{dx}arcsinx=\lim_{h\to0} \frac{arcsin(x+h)-arcsin(x)}{h}\\ \end{aligned}$

fail

$\begin{aligned} &\frac{d}{dx}arcsinx=\lim_{h\to0} \frac{arcsin(x+h)-arcsin(x)}{h}\\ & sin(a-b) = sinacosb-sinbcosa\\ & \arcsin {sin(a-b)} = \arcsin {(sinacosb-sinbcosa)}\\ & a-b = \arcsin {(sinacosb-sinbcosa)}\\ & a=arcsin(x+h), b=arcsin(x)\\ &arcsin(x+h)-arcsin(x) = arcsin(sin(arcsin(x+h))cos(arcsin(x))-sin(arcsin(x))cos(arcsin(x+h)))\\ &=arcsin((x+h)cos(arcsin(x))-xcos(arcsin(x+h)))\\ & (cos(x) = \sqrt {1-sin^2(x)}) \\ & \text{We know,} \lim_{x\to0} \frac{sin x}{x}=1, \lim _{x\to0} sin(x) = \lim_{x\to0} x\\ &So, \lim_{x\to0} sin^{-1}(x)=\lim_{x\to0}x &=arcsin( (x+h) \sqrt{1-x^2}-x\sqrt{1-(x+h)^2} )\\ &\frac{d}{dx}arcsinx=\lim_{h\to0} \frac{arcsin(x+h)-arcsin(x)}{h}\\ &=\lim_{h\to0} \frac{arcsin( (x+h) \sqrt{1-x^2}-x\sqrt{1-(x+h)^2} )}{h}\\ &=\lim_{h\to0} \frac{ (x+h) \sqrt{1-x^2}-x\sqrt{1-(x+h)^2} }{h} \\ &=\lim_{h\to0} \frac{ (x+h)^2(1-x^2)-x^2(1-(x+h)^2)} {h ((x+h) \sqrt{1-x^2}+x\sqrt{1-(x+h)^2} )} \\ &Numerator=(x+h)^2-x^2(x+h)^2-x^2+x^2(x+h)^2\\ &Numerator=(x+h)^2-x^2=2xh+h^2\\ &=\lim_{h\to0} \frac{ 2x+h} {(x+h) \sqrt{1-x^2}+x\sqrt{1-(x+h)^2} } \\ &=\frac{2x}{x\sqrt{1-x^2}+x\sqrt{1-x^2}}=\frac{2x}{2x\sqrt{1-x^2}}\\ &=\frac{1}{\sqrt{1-x^2}} \end{aligned}$

98. $\frac{d}{dx}arctanx, def.$

$\begin{aligned} &\frac{d}{dx}arctan(x)=\lim_{h\to0} \frac{arctan(x+h)-arctan(x)}{h}\\ & tan(a-b)=\frac{tan(a)-tan(b)}{1+tan(a)tan(b)}\\ & a-b = arctan(\frac{tan(a)-tan(b)}{1+tan(a)tan(b)})\\ &Numerator=arctan(x+h)-arctan(x)\\ &N=arctan( \frac{tan(arctan(x+h))-tan(arctan(x))}{1+tan(arctan(x+h))tan(arctan(x))} )\\ &=arctan( \frac{x+h-x}{1+(x+h)x} )=artan(\frac{h}{1+x^2+hx})\\ &So, \\ &\frac{d}{dx}arctan(x)=\lim_{h\to0} \frac{arctan(x+h)-arctan(x)}{h}\\ &=\lim_{h\to0} \frac{artan(\frac{h}{1+x^2+hx})}{h}\\ &=\lim_{h\to0} \frac{1}{1+x^2+hx}=\frac{1}{1+x^2}\\ \end{aligned}$

99. $\frac{d}{dx}f(x)g(x), def.$

$\begin{aligned} &\frac{d}{dx}f(x)g(x)=\lim_{h\to0}\frac{f(x+h)g(x+h)-f(x)g(x)}{h}\\ &=\lim_{h\to0}\frac{f(x+h)g(x+h)-f(x)g(x)-g(x+h)f(x)+g(x+h)f(x)}{h}\\ &=\lim_{h\to0} \frac{g(x+h)(f(x+h)-f(x))+f(x)(g(x+h)-g(x))}{h}\\ &=\lim_{h\to0} g(x+h)\frac{f(x+h)-f(x)}{h}+f(x)\frac{g(x+h)-g(x)}{h}\\ &=g(x)f'(x)+f(x)g'(x)=f'(x)g(x)+f(x)g'(x) \end{aligned}$

100. $\frac{d}{dx} \frac{f(x)}{g(x)}, def.$

$\begin{aligned} &\frac{d}{dx} \frac{f(x)}{g(x)}=\lim_{h\to0}\frac{ \frac{f(x+h)}{g(x+h)} - \frac{f(x)}{g(x)}}{h} =\frac{ \frac{f(x+h)g(x)-f(x)g(x+h)}{g(x)g(x+h)} }{h}\\ &=\frac{ f(x+h)g(x)-f(x)g(x+h)-g(x)f(x)+g(x)f(x)}{hg(x)g(x+h)}\\ &=\frac{g(x)(f(x+h)-f(x))-f(x)(g(x+h)-g(x))}{hg(x)g(x+h)}\\ &=\frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2} \end{aligned}$

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

First,$\frac{d}{dx}x^x\\ y=x^x, lny=xlnx, (1/y)y'=lnx+x(1/x)\\ y'=ylnx+y=x^xlnx+x^x$
Second,
$\begin{aligned} &\frac{d}{dx} x^{x^{x}}\\ &y=x^{x^{x}} \\ &ln y = x^x ln(x) \\ &\frac{1}{y} y' = (x^xlnx+x^x)ln(x)+x^x\frac{1}{x}\\ &=x^x((lnx)^2+ln(x)+\frac{1}{x})\\ &y'=x^{x^x}x^x((lnx)^2+ln(x)+\frac{1}{x})\\ \end{aligned}$

The END

Author: crazyj7@gmail.com

81. $\frac{d}{dx}e^x sinhx$

$\begin{aligned} &\frac{d}{dx}e^x sinhx\\ &=e^xsinhx+e^xcoshx=e^x(sinhx+coshx)\\ &=e^x(\frac{2e^x}{2})=e^{2x} \end{aligned}$

82. $\frac{d}{dx}sech(\frac{1}{x})$

$\begin{aligned} &\frac{d}{dx}sech(\frac{1}{x})\\ & sech (x)=1/cosh(x), D(sech(x))=-sinh(x)/cosh^2(x)\\ &=-sech(x)tanh(x)\\ &\frac{d}{dx}sech(\frac{1}{x})=-sinh(1/x)/cosh^2(1/x)(-1/x^2)\\ &=\frac{sinh(1/x)}{x^2cosh^2(1/x)}=\frac{sech(\frac{1}{x})tanh(\frac{1}{x})}{x^2}\\ \end{aligned}$

83. $\frac{d}{dx}cosh(lnx))$

$\begin{aligned} &\frac{d}{dx}cosh(lnx)=\frac{sinh(lnx)}{x}=\frac{e^{lnx}-e^{-lnx}}{2x}\\ &=\frac{x-(1/x)}{2x}=\frac{x^2-1}{2x^2}\\ \end{aligned}$

84. $\frac{d}{dx}ln(coshx)$

$\begin{aligned} &\frac{d}{dx}ln(coshx)=\frac{sinhx}{coshx}=tanh(x)\\ \end{aligned}$

85. $\frac{d}{dx}\frac{sinhx}{1+coshx}$

$\begin{aligned} &\frac{d}{dx}\frac{sinhx}{1+coshx}=\frac{coshx(1+coshx)-sinhxsinhx}{1+cosh^2x+2cosh(x)}\\ &=\frac{coshx+cosh^2x-sinh^2x}{(1+coshx)^2}=\frac{1+coshx}{(1+coshx)^2}\\ &=\frac{1}{1+coshx} \end{aligned}$

86. $\frac{d}{dx}arctanh(cosx)$

$\begin{aligned} &\frac{d}{dx}arctanh(cosx)\\ &D(arctanh(x)) = \frac{1}{1-x^2}\\ &y=arctanh(x), x=tanh(y), dx=sech^2(y)dy\\ &dy/dx = 1/sech^2(y)=cosh^2(y)=\frac{1}{(1/cosh^2(y))}\\ &=\frac{1}{( cosh^2(y)-sinh^2(y))/cosh^2(y)}=\frac{1}{1-tanh^2(y)}\\ &=\frac{1}{1-x^2}\\ &\frac{d}{dx}arctanh(cosx)=-\frac{sinx}{1-cos^2x}\\ &=-\frac{sin(x)}{sin^2(x)}=-csc(x)\\ \end{aligned}$

87. $\frac{d}{dx}(x)(arctanhx)+ln(\sqrt{(1-x^2}))$

$\begin{aligned} &\frac{d}{dx}(x)(arctanhx)+ln(\sqrt{1-x^2})\\ &=arctanh(x)+x\frac{1}{1-x^2}+\frac{1}{\sqrt{1-x^2}}\frac{-2x}{2\sqrt{1-x^2}}\\ &=arctanh(x)+\frac{x}{1-x^2}-\frac{x}{1-x^2}\\ &=arctanh(x) \end{aligned}$

88. $\frac{d}{dx}arcsinh(tanx)$

$\begin{aligned} &\frac{d}{dx}arcsinh(tanx)\\ &=\frac{1}{\sqrt{1+tan^2x}}sec^2x=\frac{sec^2x}{sec x}\\ &=sec(x) & \end{aligned}$

89. $\frac{d}{dx}arcsin(tanhx)$

$\begin{aligned} &\frac{d}{dx}arcsin(tanhx)\\ &=\frac{1}{\sqrt{1-tanh^2x}} sech^2x\\ &1-tanh^2x = \frac{cosh^2x-sinh^2x}{cosh^2x}=sech^2x\\ &=\frac{sech^2x}{\sqrt {sech^2x} }=sech(x) \end{aligned}$

90. $\frac{d}{dx} \frac{arctanhx}{1-x^2}$

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

Author: crazyj7@gmail.com

Remove Dict Key

딕셔너리에서 키를 안전하게 삭제하는 방법.
샘플로 다음과 같은 dictionary가 있다고 하자.

dic = dict()
dic['apple']=100
dic['banana']=200
dic['orange']=300
# 간단하게 하면 
dic = {'apple':100, 'banana':200, 'orange':300 }

위에서 apple를 지워보자.

del 로 삭제할때…

만약 키가 없으면 어떻게 될까?
del dic['mango']
KeyError!
안전하게 삭제하려면

# 1안
if 'mango' in dic:
	del dic['mango']

위와 같이 존재하는지 검사한다음, 있으면 삭제하는 것으로 할 수 있다.
매번 이렇게 if를 해 주는게 귀찮긴 한데…
일단 삭제를 시도하고, 에러나면 에러 처리하도록 할 수 있다.

# 2안
try:
	del dic['mango']
except KeyError:
	pass

pop으로 삭제할때…

dict()의 pop()을 사용하면 값을 꺼내올 수 도 있고, 해당 키가 삭제되는 효과도 있다. 이것도 키가 없다면? 에러가 발생할 것 같은데…
del과 마찬가지로 KeyError 예외가 발생한다.
즉, 해결방법은 1안과 2안처럼, 존재체크를 하거나 예외 처리를 할 수 있다.

dic.pop('mango')  # mango not found -> KeyError!!!

다른 방법은???
pop()의 다른 형태로는 파라미터에 값을 추가하면 키가 없을 때 디폴트 값을 리턴하도록 하는 기능이 있다. 이 방식을 사용하면 쉽게 안전한 삭제가 가능하다.

# 3안
dic.pop('mango', None) # if not found -> None. (no error!)

이제 있을지 모르는 키를 삭제하려면 pop( [key], None)을 사용하자.

dic.pop('apple', None)

위와 같이 하면 해당 키가 존재하는지 체크를 하거나 except 처리를 할 필요가 없다.

Author: crazyj7@gmail.com