【高校数学Ⅲ】3-1-2 導関数の性質｜問題集

\(y'=(x^2-2)'(3x^3+1)+(x^2-2)(3x^3+1)'\)
\(=2x(3x^3+1)+(x^2-2)9x^2\)
\(=6x^4+2x+9x^4-18x^2\)
\(=15x^4-18x^2+2x\)

\(y'=(x+1)'(x^3-4x)+(x+1)(x^3-4x)'\)
\(=(x^3-4x)+(x+1)(3x^2-4)\)
\(=x^3-4x+3x^3-4x+3x^2-4\)
\(=4x^3+3x^2-8x-4\)

\(y'=(3x^2-2)'(x^2+x+1)+(3x^2-2)(x^2+x+1)'\)
\(=6x(x^2+x+1)+(3x^2-2)(2x+1)\)
\(=6x^3+6x^2+6x+6x^3+3x^2-4x-2\)
\(=12x^3+9x^2+2x-2\)

\(y'=(x-1)(x-5)+(x+2)(x-5)\)
\(\ \ \ \ \ \ \ \ \ +(x+2)(x-1)\)
\(=x^2-6x+5+x^2-3x-10+x^2+x-2\)
\(=3x^2-8x-7\)

\(y'=2x(x+2)(2x-1)+(x^2-1)(2x-1)\)
\(\ \ \ \ \ \ \ \ \ +2(x^2-1)(x+2)\)
\(=8x^3+9x^2-8x-3\)

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

\(\displaystyle y'=\frac{(3x-4)'x-(3x-4)(x)'}{x^2}\)
\(\displaystyle =\frac{4}{x^2}\)

\(\displaystyle y'=-\frac{(2x-3)'}{(2x-3)^2}\)
\(\displaystyle =-\frac{2}{(2x-3)^2}\)

\(\displaystyle y'=\frac{(x^2)'(x+3)-x^2(x+3)'}{(x+3)^2}\)
\(\displaystyle =\frac{2x(x+3)-x^2}{(x+3)^2}\)
\(\displaystyle =\frac{x^2+6x}{(x+3)^2}\)

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