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
\(\displaystyle\left(\frac{xf}{f-g}\right)'(3)=\frac{(xf)'(f-g)-(xf)(f-g)’}{(f-g)^2}(3)\)
\(\displaystyle=\frac{(x’f+xf’)(f-g)-(xf)(f’-g’)}{f^2-2fg+g^2}(3)\)
\(\displaystyle=\frac{(f+xf’)(f-g)-(xf)(f’-g’)}{f^2-2fg+g^2}(3)\)
\(\displaystyle=\frac{(f(3)+3f'(3))(f(3)-g(3))-(3f(3))(f'(3)-g'(3))}{\{f(3)\}^2-2f(3)g(3)+\{g(3)\}^2}\)
\(\displaystyle=\frac{(2+3(3))(2-1)-(3(2))(3-(-2))}{2^2-2(2)(1)+1^2}\)
\(\displaystyle=\frac{(2+9)(1)-(6)(3+2)}{4-4+1}=\frac{11-30}{1}=-19\)