[연습문제] 미분법의 공식

1. 다음 함수의 도함수를 구하여라.

\([1]\ (1)\ x^n(x^n+1)\qquad(2)\ (x+a)(x+b)(x+c)\qquad(3)\ (x^n+1)^3\qquad(4)\ x^n\sin{x}\)

\(\quad\ (5)\ \sin{x}\cos{x}\qquad\,(6)\ \sin^3{x}\qquad(7)\ \dfrac{ax+b}{cx+d}\qquad(8)\ \dfrac{ax^2+b}{cx^2+d}\qquad(9)\ \dfrac{a\sin{x}+b}{c\sin{x}+d}\)

\(\quad\ (10)\ \dfrac{\sin{x}}{x}\qquad(11)\ \dfrac{1}{\sin{x}\cos{x}}\qquad(12)\ \tan{x}+\dfrac{1}{3}\tan^3x\)

<풀이>

\((1)\ \{x^n(x^n+1)\}'=nx^{n-1}(x^n+1)+x^nnx^{n-1}=nx^{n-1}(2x^n+1)\)

\(\begin{align}(2)\ \{(x+a)(x+b)(x+c)\}'&=(x+b)(x+c)+(x+a)(x+c)+(x+a)(x+b)\\&=3x^2+2(a+b+c)x+ab+bc+ca\end{align}\)

\((3)\ \{(x^n+1)^3\}'=3(x^n+1)^2nx^{n-1}=3nx^{n-1}(x^n+1)^2\)

\((4)\ (x^n\sin{x})'=nx^{n-1}\sin{x}+x^n\cos{x}=x^{n-1}(n\sin{x}+x\cos{x})\)

\((5)\ (\sin{x}\cos{x})'=\left(\dfrac{\sin2x}{2}\right)'=\cos2x\)

\((6)\ (\sin^3x)'=2\sin^2x\cos{x}\)

\((7)\ \left(\dfrac{ax+b}{cx+d}\right)'=\dfrac{a(cx+d)-(ax+b)c}{(cx+d)^2}=\dfrac{ad-bc}{(cx+d)^2}\)

\((8)\ \left(\dfrac{ax^2+b}{cx^2+d}\right)'=\dfrac{2ax(cx^2+d)-(ax^2+b)2cx}{(cx^2+d)^2}=\dfrac{2x(ad-bc)}{(cx^2+d)^2}\)

\((9)\ \left(\dfrac{a\sin{x}+b}{c\sin{x}+d}\right)'=\dfrac{a\cos{x}(c\sin{x}+d)-(a\sin{x}+b)c\cos{x}}{(c\sin{x}+d)^2}=\dfrac{\cos{x}(ad-bc)}{(c\sin{x}+d)^2}\)

\((10)\ \left(\dfrac{\sin{x}}{x}\right)'=\dfrac{x\cos{x}-\sin{x}}{x^2}\)

\((11)\ \left(\dfrac{1}{\sin{x}\cos{x}}\right)'=-\dfrac{\cos2x}{\sin^2x\cos^2x}\)

\((12)\ \left(\tan{x}+\dfrac{1}{3}\tan^3x\right)'=\sec^2x+\tan^2x\sec^2x=\sec^4x\)

\(\begin{align}[2]\ &(1)\ x{\rm Sin}^{-1}x\qquad(2)\ x{\rm Tan}^{-1}x\qquad(3)\ ({\rm Sin}^{-1}x)^2\qquad(4)\ ({\rm Tan}^{-1}x)^2\qquad(5)\ {\rm Cot}^{-1}x\\&(6)\ {\rm Csc}^{-1}x\end{align}\)

<풀이>

\((1)\ (x{\rm Sin}^{-1}x)'={\rm Sin}^{-1}x+\dfrac{x}{\sqrt{1-x^2}}\)

\((2)\ (x{\rm Tan}^{-1}x)'={\rm Tan}^{-1}x+\dfrac{x}{1+x^2}\)

\((3)\ \{({\rm Sin}^{-1}x)^2\}'=\dfrac{2{\rm Sin}^{-1}x}{\sqrt{1-x^2}}\)

\((4)\ \{({\rm Tan}^{-1}x)^2\}'=\dfrac{2{\rm Tan}^{-1}x}{1+x^2}\)

\((5)\ {\rm Cot}^{-1}x=y\)로 두면 \(\cot{y}=x\). 따라서 \(\displaystyle\frac{dy}{dx}=\frac{1}{\dfrac{dx}{dy}}=\frac{1}{-\csc^2x}=-\frac{1}{1+\cot^2x}\)

\(=-\dfrac{1}{1+x^2}\)

\((6)\ ({\rm Csc}^{-1}x)'=\left({\rm Sin}^{-1}\dfrac{1}{x}\right)'=\dfrac{1}{\sqrt{1-\left(\dfrac{1}{x}\right)^2}}\left(-\dfrac{1}{x^2}\right)=-\dfrac{1}{|x|\sqrt{x^2-1}}\)

\([3]\ (1)\ x\ln{x}\qquad(2)\ \dfrac{1}{x}\ln{x}\qquad(3)\ e^{-x}\qquad(4)\ x^ne^x\qquad(5)\ e^x\sin{x}\qquad(6)\ e^x\cos{x}\)

\(\quad\ (7)\ a^xb^{1-x}(a>0,\,b>0)\qquad(8)\ \sinh{x}\qquad(9)\ \cosh{x}\qquad(10)\ \dfrac{e^{ax}-e^{-ax}}{e^{ax}+e^{-ax}}\)

\(\quad(11)\ x^{\sin{x}}(x>0)\qquad(12)\ \sqrt[x]{x}(x>0)\qquad(13)\ e^{x^x}(x>0)\qquad(14)\ (ax^2+2bx+x)^x\)

<풀이>

\((1)\ (x\ln{x})'=\ln{x}+1\)

\((2)\ \left(\dfrac{1}{x}\ln{x}\right)'=-\dfrac{1}{x^2}\ln{x}+\dfrac{1}{x^2}=\dfrac{1-\ln{x}}{x^2}\)

\((3)\ (e^{-x})'=-e^{-x}\)

\((4)\ (x^ne^x)'=nx^{n-1}e^x+x^ne^x=x^{n-1}e^x(n+x)\)

\((5)\ (e^x\sin{x})'=e^x\sin{x}+e^x\cos{x}=e^x(\sin{x}+\cos{x})\)

\((6)\ (e^x\cos{x})'=e^x\cos{x}+e^x(-\sin{x})=e^x(\cos{x}-\sin{x})\)

\((7)\ (a^xb^{1-x})'=(a^x\ln{a})b^{1-x}+a^xb^{1-x}\ln{b}(-1)=a^xb^{1-x}\ln\frac{a}{b}\)

\((8)\ (\sinh{x})'=\left(\dfrac{e^x-e^{-x}}{2}\right)'=\dfrac{e^x+e^{-x}}{2}=\cosh{x}\)

\((9)\ (\cosh{x})'=\left(\dfrac{e^x+e^{-x}}{2}\right)'=\dfrac{e^x-e^{-x}}{2}=\sinh{x}\)

\((10)\ \left(\dfrac{e^{ax}-e^{-ax}}{e^{ax}+e^{-ax}}\right)'=\left(\dfrac{\sinh{ax}}{\cosh{ax}}\right)'=\dfrac{a(\cosh^2ax-\sinh^2ax)}{\cosh^2ax}=\dfrac{4a}{(e^{ax}+e^{-ax})^2}\)

\((11)\ y=x^{\sin{x}}\)로 두고 양변에 대수를 취하면 \(\ln{y}=\sin{x}\ln{x}\).

양변을 \(x\)로 미분하면 \(\dfrac{y'}{y}=\cos{x}\ln{x}+\dfrac{\sin{x}}{x},\,y'=x^{\sin{x}}\left(\cos{x}\ln{x}+\dfrac{\sin{x}}{x}\right)\)

\((12)\ y=\sqrt[x]{x}=x^{1\over x}\)로 두고 양변에 대수를 취하면 \(\ln{y}=\dfrac{1}{x}\ln{x}\).

양변을 \(x\)로 미분하면 \(\dfrac{y'}{y}=-\dfrac{1}{x^2}\ln{x}+\dfrac{1}{x^2},\,y'=\sqrt[x]{x}\left(\dfrac{1-\ln{x}}{x^2}\right)\)

\((13)\ (e^{x^x})'=e^{x^x}(x^x)'=e^{x^x}x^x(\ln{x}+1)\)

\((14)\ y=(ax^2+2bx+c)^x\)로 두고 양변에 대수를 취하면 \(\ln{y}=x\ln(ax^2+2bx+c)\).

양변을 \(x\)로 미분하면 \(\dfrac{y'}{y}=\ln(ax^2+2bx+c)+\dfrac{2x(ax+b)}{ax^2+2bx+c}\),

\(y'=(ax^2+2bx+c)^x\left\{\ln(ax^2+2bx+c)+\dfrac{2x(ax+b)}{ax^2+2bx+c}\right\}\)

\([4]\ (1)\ \left(a+\dfrac{b}{x}\right)^3\qquad(2)\ \dfrac{1}{(x+1)^n}\qquad(3)\ \sin(ax+b)\qquad(4)\ \sin^nx\qquad(5)\ e^{-hx^2}\)

\(\quad\ (6)\ \ln(\sin{x})\qquad(7)\ \ln(x+\sqrt{x^2+a^2})\qquad(8)\ x^n(\ln{x})^m\qquad(9)\ e^{ax}(a\sin{bx}-b\cos{bx})\)

\(\quad\ (10)\ x({\rm Sin}^{-1}x)^2+2\sqrt{1-x^2}{\rm Sin}^{-1}x\qquad(11)\ f(x)^\mu g(x)^\nu\qquad(12)\ \dfrac{f(x)^\mu}{g(x)^\nu}\)

<풀이>

\(\displaystyle(1)\ \left\{\left(a+{b\over x}\right)^3\right\}'=3\left(a+{b\over x}\right)^2\left(-{b\over x^2}\right)=-{3b\over x^2}\left(a+{b\over x}\right)^2\)

\((2)\ \left\{\dfrac{1}{(x+1)^n}\right\}'=\{(x+1)^{-n}\}'=-n(x+1)^{-n-1}=-\dfrac{n}{(x+1)^{n+1}}\)

\((3)\ \{\sin(ax+b)\}'=a\cos(ax+b)\qquad(4)\ (\sin^nx)'=n\sin^{n-1}x\cos{x}\)

\((5)\ (e^{-hx^2})'=-2hxe^{-hx^2}\qquad\qquad\qquad(6)\ \{\ln(\sin{x})\}'=\dfrac{\cos{x}}{\sin{x}}=\cot{x}\)

\((7)\ \{\ln(x+\sqrt{x^2+a^2})\}'=\dfrac{1+\dfrac{x}{\sqrt{x^2+a^2}}}{x+\sqrt{x^+a^2}}=\dfrac{1}{\sqrt{x^2+a^2}}\)

\((8)\ \{x^n(\ln{x})^m\}'=nx^{n-1}(\ln{x})^m+mx^{n-1}(\ln{x})^{m-1}=x^{n-1}(\ln{x})^{m-1}(n\ln{x}+m)\)

\((9)\ \{e^{ax}(a\sin{bx}-b\cos{bx})\}'=ae^{ax}(a\sin{bx}-b\cos{bx})+e^{ax}(ab\cos{bx}+b^2\sin{bx})\)

\(\qquad\qquad\qquad\qquad\qquad\qquad\,\,=e^{ax}(a^2+b^2)\sin{bx}\)

\((10)\ \{x({\rm Sin}^{-1}x)^2+2\sqrt{1-x^2}{\rm Sin}^{-1}x\}'=({\rm Sin}^{-1}x)^2+\dfrac{2x}{\sqrt{1-x^2}}{\rm Sin}^{-1}x\)

\(\qquad-\dfrac{2x}{\sqrt{1-x^2}}{\rm Sin}^{-1}x+2=({\rm Sin}^{-1})^2+2\)

\((11)\ \{f(x)^\mu g(x)^\nu\}'=\mu f(x)^{\mu-1}f'(x)g(x)^\nu+\nu f(x)^\mu g(x)^{\nu-1}g'(x)\)

\(\qquad\qquad\qquad\qquad=f(x)^{\mu-1}g(x)^{\nu-1}\{\mu f'(x)g(x)+\nu f(x)g'(x)\}\)

\((12)\ \left\{\dfrac{f(x)^\mu}{g(x)^\nu}\right\}'=\dfrac{\mu f(x)^{\mu-1}f'(x)g(x)^\nu-\nu f(x)^\mu g(x)^{\nu-1}g'(x)}{g(x)^{2\nu}}\)

\(\qquad\qquad\qquad\ \ =\dfrac{f(x)^{\mu-1}\{\mu f'(x)g(x)-\nu f(x)g'(x)\}}{g(x)^{\nu+1}}\)

2. 다음 함수에 대한 \(dy/dx\)를 구하여라.

\((1)\ x=a(\theta-\sin\theta),\,y=a(1-\cos\theta)\ (a>0)\)

\((2)\ x=3at/(1+t^3),\,y=3at^2/(1+t^3)\)

\((3)\ x=a\theta\cos\theta,\,y=a\theta\sin\theta\ (a>1)\)

<풀이>

\((1)\ \dfrac{dx}{d\theta}=a(1-\cos\theta),\,\dfrac{dy}{d\theta}=a\sin\theta,\,\dfrac{dy}{dx}=\dfrac{dy/d\theta}{dx/d\theta}=\dfrac{\sin\theta}{1-\cos\theta}\)

\((2)\ \dfrac{dx}{dt}=\dfrac{3a(1-2t^3)}{(1+t^3)^2},\,\dfrac{dy}{dt}=\dfrac{3at(2-t^3)}{(1+t^3)^2},\,\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}=\dfrac{t(2-t^3)}{1-2t^3}\)

\((3)\ \dfrac{dx}{d\theta}=a^\theta(\cos\theta\ln{a}-\sin\theta),\,\dfrac{dy}{d\theta}(\sin\theta\ln{a}+\cos\theta),\)

\(\quad\ \,\dfrac{dy}{dx}=\dfrac{dy/d\theta}{dx/d\theta}=\dfrac{\sin\theta\ln{a}+\cos\theta}{\cos\theta\ln{a}-\sin\theta}=\dfrac{\tan\theta\ln{a}+1}{\ln{a}-\tan\theta}\)

3. 음함수의 미분법에 의해 다음 함수에 대한 \(dy/dx\)를 구하여라.

\((1)\ x^2+xy+y^2=0\qquad(2)\ x^3y+x^2+y^2=10\qquad(3)\ y\sin{x}+x\sin{y}=\pi\)

\((4)\ y^2=\dfrac{x-y}{x+y}\qquad(5)\ \sqrt{2x}+\sqrt{3y}-5=0\ (x>0,\,y>0)\)

\((6)\ x^{2\over3}+y^{2\over3}=a^{2\over3}\ (|x|\le a,\,|y|\le a)\)

<풀이>

\((1)\ 2x+y+xy'+2yy'=0,\,y'=-\dfrac{2x+y}{x+2y}\)

\((2)\ 3x^2y+x^3y'+2x+2yy'=0,\,y'=-\dfrac{x(2+3xy)}{x^3+2y}\)

\((3)\ y'\sin{x}+y\cos{x}+\sin{y}+xy'\cos{y}=0,\,y'=-\dfrac{y\cos{x}+\sin{y}}{\sin{x}+x\cos{y}}\)

\((4)\ xy^2+y^3-x+y=0,\,y^2+2xyy'+3y^2y'-1+y'=0,\,y'=\dfrac{1-y^2}{3y^2+2xy+1}\)

\((5)\ \dfrac{1}{\sqrt{2x}}+\dfrac{3y'}{2\sqrt{3y}}=0,\,y'=-\sqrt{\dfrac{2y}{3x}}\)

\((6)\ \dfrac{2}{3x^{1/3}}+\dfrac{2y'}{3y^{1/3}}=0,\,y'=-\left(\dfrac{y}{x}\right)^{1/3}\)