【高校数学Ⅲ】3-1-3 合成関数と逆関数の微分法｜問題集

\(y'=4(3x+1)^3・(3x+1)'\)
\(=4(3x+1)^3・3\)
\(=12(3x+1)^3\)

\(y'=4(2x^2+5)^3・(2x^2+5)'\)
\(=4(2x^2+5)^3・4x\)
\(=16x(2x^2+5)^3\)

\(y'=3(1-2x^2)^2・(1-2x^2)'\)
\(=3(1-2x^2)^2・-4x\)
\(=-12x(1-2x^2)^2\)

\(y'=-2(4x+3)^{-3}・(4x+3)'\)
\(=-2(4x+3)^{-3}・4\)
\(\displaystyle =\frac{-8}{(4x+3)^3}\)

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

\(y'=-3(x^2+1)^{-4}・(x^2+1)'\)
\(=-3(x^2+1)^{-4}・2x\)
\(\displaystyle =\frac{-6x}{(x^2+1)^4}\)

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

\(y'=8x^{-3}\)
\(\displaystyle =\frac{8}{x^3}\)

\(y'=-x^{-4}\)
\(\displaystyle =-\frac{1}{x^4}\)

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

\(\displaystyle y'=\frac{1}{6}x^{-\frac{5}{6}}\)
\(\displaystyle =\frac{1}{6\sqrt[6]{x^5}}\)

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

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

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

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