미분 공식 증명

구국원·2020년 9월 6일
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미분 공식을 이해하고 내면화하기 위해 대표적인 미분 공식 증명을 정리해 보겠습니다.

1   f(x)=xnf(x)=x^{n}일때, f(x)=nxn1f'(x)=nx^{n-1}임을 증명하라

f(x)=limh0f(x+h)f(x)h=limh0(x+h)nxnh=limh0(x+hx)((x+h)n1x0+(x+h)n2x1+...+(x+h)0xn1)h=limh0((x+h)n1x0+(x+h)n2x1+...+(x+h)0xn1)=xn1x0+xn2x1+...+x0xn1=xn1+xn1+...+xn1=nxn1\begin{aligned} f'(x) &= \lim_{h \to 0} \cfrac {f(x+h) - f(x)} {h} \\ &= \lim_{h \to 0} \cfrac {(x+h)^n - x^n} {h} \\ &= \lim_{h \to 0} \cfrac {(x+h-x)((x+h)^{n-1}x^{0} + (x+h)^{n-2}x^{1} + ... + (x+h)^{0}x^{n-1})} {h} \\ &= \lim_{h \to 0} {((x+h)^{n-1}x^{0} + (x+h)^{n-2}x^{1} + ... + (x+h)^{0}x^{n-1})} \\ &= x^{n-1}x^{0} + x^{n-2}x^{1} + ...+ x^{0}x^{n-1} \\ &= x^{n-1} + x^{n-1} + ... +x^{n-1} \\ &= nx^{n-1} \end{aligned}

참고) anbn=(ab)(an1b0+an2b1+...+a0bn1)a^n - b^n = (a-b)(a^{n-1}b^{0} + a^{n-2}b^{1} + ... + a^{0}b^{n-1})

2   자연상수 ee의 정의

e=limn(1+1n)n=limx0(1+x)1/ne=\lim_{n \to\infty}\bigg(1+\cfrac{1}{n}\bigg)^n=\lim_{x\to0}\big(1+x\big)^{1/n}
참고) 마지막항은 x=1nx = \cfrac{1}{n}을 이용한 것

3   limx0ex1x=1\lim_{x\to0}\cfrac{e^x-1}{x}=1임을 증명하라.

limx0ex1x=limt0tln(1+t)ex1=t=limt01ln(1+t)t=limt011tln(1+t)=limt01ln(1+t)1t=1ln(limt0(1+t)1t)=1lnelimx0(1+x)1/n=1\begin{aligned} \lim_{x\to0}\cfrac{e^x-1}{x} &= \lim_{t\to0}\cfrac{t}{ln(1+t)} \Leftarrow e^x-1=t \\ &= \lim_{t\to0}\cfrac{1}{\cfrac{ln(1+t)}{t}} \\ &= \lim_{t\to0}\cfrac{1}{\frac{1}{t} ln(1+t)} \\ &= \lim_{t\to0}\cfrac{1}{ln(1+t)^{\frac{1}{t}}} \\ &= \cfrac{1}{ln(\lim_{t\to0}(1+t)^{\frac{1}{t}})} \\ &= \cfrac{1}{lne} \Leftarrow \lim_{x\to0}\big(1+x\big)^{1/n} \\ &= 1 \end{aligned}

4   지수함수 f(x)=exf(x) =e^x일 때, f(x)=exf'(x)=e^x임을 증명하라.

f(x)=limh0f(x+h)f(x)h=limh0ex+hexh=limh0ex(eh1)h=exlimh0eh1h=exlimx0ex1x=1\begin{aligned} f'(x) &= \lim_{h \to 0} \cfrac {f(x+h) - f(x)} {h} \\ &= \lim_{h \to 0} \cfrac {e^{x+h} - e^{x}} {h} \\ &= \lim_{h \to 0} \cfrac {e^{x}(e^{h}-1)} {h} \\ &= e^{x} \lim_{h \to 0} \cfrac {e^{h}-1} {h} \\ &= e^{x} \Leftarrow \lim_{x\to0} \cfrac{e^x-1}{x}=1 \\ \end{aligned}

4   지수함수 f(x)=axf(x) =a^x일 때, f(x)=exlnalnaf'(x)=e^{xlna}\cdot lna임을 증명하라.

다음 수식을 먼저 정의 하겠습니다.

ax=alogeex=(ex)logeaalogcb=blogca=exlogea=exlna\begin{aligned} a^x &= a^{log_{e}e^{x}} \\ &= (e^{x})^{log_{e}a} \Leftarrow a^{log_{c}b} = b^{log_{c}a} \\ &= e^{xlog_{e}a} \\ &= e^{xlna} \\ \end{aligned}

위의 수식을 이용하여 증명을 해보겠습니다.

f(x)=daxdx=ddxexlna=ddxett=xlna=detdtdtdx=etdtdx=etlna=exlnalna\begin{aligned} f'(x) &= \cfrac {da^x} {dx} \\ &= \cfrac {d} {dx} e^{xlna} \\ &= \cfrac {d} {dx} e^t \Leftarrow t=xlna \\ &= \cfrac {de^t} {dt} \cfrac {dt} {dx} \\ &= e^t \cfrac {dt} {dx} \\ &= e^t \cdot lna \\ &= e^{xlna} \cdot lna \\ \end{aligned}

5   로그함수 f(x)=lnxf(x) =lnx 일 때, f(x)=1xf'(x)=\cfrac{1}{x}임을 증명하라.

f(x)=limh0f(x+h)f(x)h=limh0ln(x+h)lnxh=limh01hlnx+hx=limh0ln(x+hx)1h=limh0ln(x+hx)xh1x=limh01xln(x+hx)xh=limh01xln(1+hx)xh=1xlnlimh0(1+hx)xh=1xlnlimh0(1+hx)xh=1xlnee=limx0(1+x)1/x=1x\begin{aligned} f'(x) &= \lim_{h \to 0} \cfrac {f(x+h) - f(x)} {h} \\ &= \lim_{h \to 0} \cfrac {ln(x+h) - lnx} {h} \\ &= \lim_{h \to 0} \cfrac {1} {h} \cdot {ln \cfrac {x+h} {x}} \\ &= \lim_{h \to 0} ln\bigg(\cfrac{x+h}{x}\bigg)^{\frac{1}{h}} \\ &= \lim_{h \to 0} ln\bigg(\cfrac{x+h}{x}\bigg)^{\frac{x}{h} \cdot \frac{1}{x}} \\ &= \lim_{h \to 0} \cfrac{1}{x} \cdot ln \bigg(\cfrac{x+h}{x}\bigg)^{\frac{x}{h}} \\ &= \lim_{h \to 0} \cfrac{1}{x} \cdot ln \bigg(1+\cfrac{h}{x}\bigg)^{\frac{x}{h}} \\ &= \cfrac{1}{x} \cdot ln \lim_{h \to 0} \bigg(1+\cfrac{h}{x}\bigg)^{\frac{x}{h}} \\ &= \cfrac{1}{x} \cdot ln \lim_{h \to 0} \bigg(1+\cfrac{h}{x}\bigg)^{\frac{x}{h}} \\ &= \cfrac{1}{x} \cdot lne \Leftarrow e=\lim_{x\to0}(1+x)^{1/x} \\ &= \cfrac {1} {x} \end{aligned}
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