[월:] 2025년 04월

다음 함수의 이계도함수를 구하라.

$$y=\frac{x}{x^2-1}$$

풀이

$\displaystyle\frac{1}{(x+1)^2}+\frac{1}{(x-1)^2}=\frac{(x-1)^2+(x+1)^2}{(x+1)^2(x-1)^2}$

$\displaystyle=\frac{x^2-2x+1+x^2+2x+1}{\{(x+1)(x-1)\}^2}=\frac{2(x^2+1)}{\{(x+1)(x-1)\}^2}$

$\displaystyle y’=\frac{x'(x^2-1)-x(x^2-1)’}{(x^2-1)^2}=\frac{x^2-1-x(2x)}{(x^2-1)^2}$

$\displaystyle=\frac{x^2-1-2x^2}{(x^2-1)^2}=\frac{-(x^2+1)}{\{(x+1)(x-1)\}^2}$

$\displaystyle=-\frac{1}{2}\left\{\frac{1}{(x+1)^2}+\frac{1}{(x-1)^2}\right\}$

$\displaystyle y'{}’=-\frac{1}{2}\left\{\left\{\frac{1}{(x+1)^2}\right\}’+\left\{\frac{1}{(x-1)^2}\right\}’\right\}$

$\displaystyle=-\frac{1}{2}\left\{\frac{-((x+1)^2)’}{(x+1)^4}+\frac{-((x-1)^2)’}{(x-1)^4}\right\}$

$\displaystyle=-\frac{1}{2}\left\{\frac{-(x^2+2x+1)’}{(x+1)^4}+\frac{-(x^2-2x+1)’}{(x-1)^4}\right\}$

$\displaystyle=-\frac{1}{2}\left\{\frac{-(2x+2)}{(x+1)^4}+\frac{-(2x-2)}{(x-1)^4}\right\}$

$\displaystyle=-\frac{1}{2}\left\{\frac{-2(x+1)}{(x+1)^4}+\frac{-2(x-1)}{(x-1)^4}\right\}$

$\displaystyle=-\frac{1}{2}\left\{\frac{-2}{(x+1)^3}+\frac{-2}{(x-1)^3}\right\}$

$\displaystyle=\frac{1}{(x+1)^3}+\frac{1}{(x-1)^3}$

다음 함수의 이계도함수를 구하라.

$$y=\sqrt{x}$$

풀이

$\displaystyle y’=\{x^{\frac{1}{2}}\}’=\frac{1}{2}x^{-\frac{1}{2}}$

$\displaystyle y'{}’=\frac{1}{2}\left(-\frac{1}{2}\right)x^{-\frac{3}{2}}=-\frac{1}{4}\left(\frac{1}{\sqrt{x^3}}\right)=-\frac{1}{4x\sqrt{x}}$

다음 함수의 이계도함수를 구하라.

$$y=x^3+3x^2-x-1$$

풀이

$\displaystyle y’=3x^2+6x-1$

$\displaystyle y'{}’=6x+6$

도함수의 공식을 이용하여 다음 함수의 도함수를 구하라.

$$f(x)=(x^2+1)(x^2+x+1)(3x-1)$$

풀이

$\displaystyle f'(x)=((x^2+1)(x^2+x+1))'(3x-1)$

$\displaystyle+(x^2+1)(x^2+x+1)(3x-1)’$

$\displaystyle=((x^2+1)'(x^2+x+1)+(x^2+1)(x^2+x+1)’)(3x-1)$

$\displaystyle+3(x^2+1)(x^2+x+1)$

$\displaystyle=2x(x^2+x+1)(3x-1)+(x^2+1)(2x+1)(3x-1)$

$\displaystyle+3(x^2+1)(x^2+x+1)$

도함수의 공식을 이용하여 다음 함수의 도함수를 구하라.

$$f(x)=\sqrt{x}+\frac{2}{\sqrt{x}}$$

풀이

$\displaystyle f'(x)=\left\{\sqrt{x}\right\}’+\left\{\frac{2}{\sqrt{x}}\right\}’=\left\{x^{\frac{1}{2}}\right\}’+\left\{2x^{-\frac{1}{2}}\right\}’$

$\displaystyle=\frac{1}{2}x^{-\frac{1}{2}}+2(-\frac{1}{2})x^{-\frac{3}{2}}=\frac{1}{2}x^{-\frac{1}{2}}-x^{-\frac{3}{2}}$

$\displaystyle=\frac{1}{2\sqrt{x}}-\frac{1}{\sqrt{x^3}}=\frac{1}{2\sqrt{x}}-\frac{1}{x\sqrt{x}}$

도함수의 공식을 이용하여 다음 함수의 도함수를 구하라.

$$f(x)=\frac{x^2-1}{x-2}$$

풀이

$\displaystyle f'(x)=\frac{(x^2-1)'(x-2)-(x^2-1)(x-2)’}{(x-2)^2}$

$\displaystyle=\frac{2x(x-2)-(x^2-1)}{(x-2)^2}=\frac{2x^2-4x-x^2+1}{(x-2)^2}$

$\displaystyle=\frac{x^2-4x+1}{(x-2)^2}$

도함수의 공식을 이용하여 다음 함수의 도함수를 구하라.

$$f(x)=(x^2+x+1)(x^3+1)$$

풀이

$\displaystyle f'(x)=(x^2+x+1)'(x^3+1)+(x^2+x+1)(x^3+1)’$

$\displaystyle=(2x+1)(x^3+1)+(x^2+x+1)3x^2$

$\displaystyle=2x^4+2x+x^3+1+3x^4+3x^3+3x^2$

$\displaystyle=5x^4+4x^3+3x^2+2x+1$

두 함수 $f(x)$, $g(x)$에 대해

$$f(3)=2,\quad g(3)=1,\quad f'(3)=3,\quad g'(3)=-2$$

일 때 다음을 구하라.

$$\left(\frac{xf}{f-g}\right)'(3)$$

풀이

$\displaystyle\left(\frac{xf}{f-g}\right)'(3)=\frac{(xf)'(f-g)-(xf)(f-g)’}{(f-g)^2}(3)$

$\displaystyle=\frac{(x’f+xf’)(f-g)-(xf)(f’-g’)}{f^2-2fg+g^2}(3)$

$\displaystyle=\frac{(f+xf’)(f-g)-(xf)(f’-g’)}{f^2-2fg+g^2}(3)$

$\displaystyle=\frac{(f(3)+3f'(3))(f(3)-g(3))-(3f(3))(f'(3)-g'(3))}{\{f(3)\}^2-2f(3)g(3)+\{g(3)\}^2}$

$\displaystyle=\frac{(2+3(3))(2-1)-(3(2))(3-(-2))}{2^2-2(2)(1)+1^2}$

$\displaystyle=\frac{(2+9)(1)-(6)(3+2)}{4-4+1}=\frac{11-30}{1}=-19$