[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 \(\displaystyle\frac{dy}{dx}=\frac{1}{\dfrac{dx}{dy}}=\frac{1}{-\csc^2x}=-\frac{1}{1+\cot^2x}\)

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

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

\([3]\ (1)\ x\ln{x}\qquad(2)\ \dfrac{1}{x}\ln{x}\qquad(3)\ \)

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