【微分積分】3-1-2 導関数の性質｜問題集

\(y'=4x^3\)

\(y'=55x^4-18x^2\)

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

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

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

\(\displaystyle y'=\frac{(x^2-1)'(2x+3)-(x^2-1)(2x+3)'}{(2x+3)^2}\)
\(\displaystyle =\frac{2x(2x+3)-2(x^2-1)}{(2x+3)^2}\)
\(\displaystyle =\frac{2x^2+6x+2}{(2x+3)^2}\)

\(\displaystyle y'=\frac{(6-\frac{1}{x})'(x-2)-(6-\frac{1}{x})(x-2)'}{(x-2)^2}\)
\(\displaystyle =\frac{\frac{1}{x^2}(x-2)-(6-\frac{1}{x})}{(x-2)^2}\)
\(\displaystyle =\frac{-6x^2+2x-2}{x^2(x-2)^2}\)

\(\displaystyle y'=\frac{(1+x^4)'x^2-(1+x^4)(x^2)'}{(x^2)^2}\)
\(\displaystyle =\frac{4x^3x^2-(1+x^4)・2x}{x^4}\)
\(\displaystyle =\frac{2x^5-2x}{x^4}\)
\(\displaystyle =\frac{2x^4-2}{x^3}\)

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

\(y'=2004(x^2+1)^{2003}・(x^2+1)'\)
\(=2004(x^2+1)^{2003}・2x\)
\(=4008x(x^2+1)^{2003}\)

\(\displaystyle y'=3\left(x^2+\frac{1}{x^2}\right)^2・\left(x^2+\frac{1}{x^2}\right)'\)
\(\displaystyle =3\left(x^2+\frac{1}{x^2}\right)^2・\left(2x-\frac{2}{x^3}\right)\)
\(\displaystyle =6\left(x-\frac{1}{x^3}\right)\left(x^2+\frac{1}{x^2}\right)^2\)

\(\{(2x+1)^2+(x+1)^2\}'\)
\(=2(2x+1)・2+2(x+1)\)
\(=10x+6\)
\(y'=3\{(2x+1)^2+(x+1)^2\}^2・(10x+6)\)
\(=6(5x+3)(5x^2+6x+2)^2\)