【高校数学Ⅲ】3-1-4 三角関数の導関数｜問題集

\(y'=-\sin2x・(2x)'\)
\(=-2\sin2x\)

\(\displaystyle y'=\sqrt{2}\cos\left(3x+\frac{\pi}{4}\right)・\left(3x+\frac{\pi}{4}\right)'\)
\(\displaystyle =3\sqrt{2}\cos\left(3x+\frac{\pi}{4}\right)\)

\(y'=2\sin x・(\sin x)'\)
\(=2\sin x\cos x\)

\(y'=2\tan x・(2x)'\)
\(\displaystyle =\frac{2\tan x}{\cos^2x}\)

\(y'=-(\sin x)^{-2}・(\sin x)'\)
\(\displaystyle =-\frac{\cos x}{\sin^2 x}\)

\(y'=2\cos3x・(\cos3x)'\)
\(=2\cos3x・-3\sin3x\)
\(=-6\sin3x\cos3x\)

\(y'=\cos x-\cos x+x\sin x\)
\(=x\sin x\)

\(y'=\sin x+x\cos x-\sin x\)
\(=x\cos x\)

\(y'=\cos x-x\sin x-\cos x\)
\(=-x\sin x\)

\(y'=2\tan x・(\tan x)'\)
\(\displaystyle =2\frac{\sin x}{\cos x}・\frac{1}{\cos^2x}\)
\(\displaystyle =\frac{2\sin x}{\cos^3x}\)

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

\(y'=2\sin x\cos x\cos2x-2\sin^2x\sin2x\)
\(=2\sin2x\cos2x-2\sin^2x\sin2x\)
\(=\sin2x(\cos2x-2\sin^2x)\)
\(=\sin2x(1-2\sin^2x-2\sin^2x)\)
\(=\sin2x(1-4\sin^2x)\)