【微分積分】3-3-1 導関数の計算｜問題集

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

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

\(y'=(-2xe^{-x^2+1})\sin2x+e^{-x^2+1}・2\cos2x\)
\(=2e^{-x^2+1}(\cos2x-x\sin2x)\)

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

\(y'=2x・2^x+x^2・2^x\log2\)
\(=2^x(2x+x^2\log2)\)

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

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

両辺の対数をとると、
\(\log y=\log x^{\frac{1}{x}}\)
\(\displaystyle \log y=\frac{1}{x}\log x\)
両辺を\(x\)で微分すると、
\(\displaystyle \frac{y'}{y}=\frac{\frac{1}{x}・x-\log x・1}{x^2}\)
\(\displaystyle y'=y\frac{1-\log x}{x^2}\)
\(\displaystyle =\frac{x^{\frac{1}{x}}(1-\log x)}{x^2}\)

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

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

\(\displaystyle y'=\frac{\frac{1}{\sqrt{1-x^2}}・\cos^{-1}x-\sin^{-1}x・\left(-\frac{1}{\sqrt{1-x^2}}\right)}{(\cos^{-1}x)^2}\)
\(\displaystyle =\frac{\cos^{-1}x+\sin^{-1}x}{\sqrt{1-x^2}(\cos^{-1}x)^2}\)
\(\displaystyle =\frac{\pi}{2\sqrt{1-x^2}(\cos^{-1}x)^2}\)

両辺の対数をとると、
\(\log y=\log x・\log x\)
\(\displaystyle \log y=(\log x)^2\)
両辺を\(x\)で微分すると、
\(\displaystyle \frac{y'}{y}=2\log x・\frac{1}{x}\)
\(\displaystyle y'=y\frac{2\log x}{x}\)
\(\displaystyle =2x^{\log x-1}\log x\)

両辺の対数をとると、
\(\log y=\cos x・\log(\tan x)\)
両辺を\(x\)で微分すると、
\(\displaystyle \frac{y'}{y}=-\sin x・\log(\tan x)+\cos x・\frac{\sec^2 x}{\tan x}\)
\(\displaystyle =-\sin x・\log(\tan x)+\cos x・\frac{1}{\sin x\cos x}\)
\(\displaystyle y'=y\left\{-\sin x\log(\tan x)+\frac{1}{\sin x}\right\}\)
\(\displaystyle =(\tan x)^{\cos x}\left\{\frac{1}{\sin x}-\sin x\log(\tan x)\right\}\)

両辺の対数をとると、
\(\log y=x\log(\log x)\)
両辺を\(x\)で微分すると、
\(\displaystyle \frac{y'}{y}=1・\log(\log x)+x・\frac{1}{\log x}・\frac{1}{x}\)
\(\displaystyle y'=y\left\{\log(\log x)+\frac{1}{\log x}\right\}\)
\(\displaystyle =(\log x)^x\left\{\log(\log x)+\frac{1}{\log x}\right\}\)

両辺の対数をとると、
\(\log y=x\log\left(1+\frac{1}{x}\right)\)
両辺を\(x\)で微分すると、
\(\displaystyle \frac{y'}{y}=\log\left(1+\frac{1}{x}\right)+x・\frac{1}{1+\frac{1}{x}}・\left(-\frac{1}{x^2}\right)\)
\(\displaystyle y'=y\left\{\log\left(1+\frac{1}{x}\right)-\frac{1}{x+1}\right\}\)
\(\displaystyle =\left(1+\frac{1}{x}\right)^x\left\{\log\left(1+\frac{1}{x}\right)-\frac{1}{x+1}\right\}\)

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

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

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

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

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

\(y'=3x^2\tan2x+x^3・2\sec^22x\)
\(=2x^3\sec^22x+3x^2\tan2x\)

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

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

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

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

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

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

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

\(y'=2e^{2x}\cos x-e^{2x}\sin x\)
\(=e^{2x}(2\cos x-\sin x)\)

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

\(y'=2x\cos(x^2+1)\)

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

\(\displaystyle y'=e^{\sin x}\cos x\)