次の4つの関数を微分する問題です。 (i) $cos(sin^{-1}x)$ (ii) $e^{tan^{-1}(2x)}$ (iii) $log(log(logx))$ (iv) $(cosx)^{cosx}$

1. 問題の内容

次の4つの関数を微分する問題です。

(i) $cos(sin^{-1}x)$

(ii) $e^{tan^{-1}(2x)}$

(iii) $log(log(logx))$

(iv) $(cosx)^{cosx}$

2. 解き方の手順

(i) $y = cos(sin^{-1}x)$

$sin^{-1}x = t$と置くと、$x = sint$

$y = cost = \sqrt{1-sin^2t} = \sqrt{1-x^2}$

$\frac{dy}{dx} = \frac{1}{2\sqrt{1-x^2}} \times (-2x) = \frac{-x}{\sqrt{1-x^2}}$

(ii) $y = e^{tan^{-1}(2x)}$

$\frac{dy}{dx} = e^{tan^{-1}(2x)} \times \frac{d}{dx} (tan^{-1}(2x))$

$\frac{d}{dx} (tan^{-1}(2x)) = \frac{1}{1+(2x)^2} \times 2 = \frac{2}{1+4x^2}$

$\frac{dy}{dx} = e^{tan^{-1}(2x)} \times \frac{2}{1+4x^2} = \frac{2e^{tan^{-1}(2x)}}{1+4x^2}$

(iii) $y = log(log(logx))$

$\frac{dy}{dx} = \frac{1}{log(logx)} \times \frac{d}{dx} (log(logx))$

$\frac{d}{dx} (log(logx)) = \frac{1}{logx} \times \frac{1}{x} = \frac{1}{xlogx}$

$\frac{dy}{dx} = \frac{1}{log(logx)} \times \frac{1}{xlogx} = \frac{1}{xlogxlog(logx)}$

(iv) $y = (cosx)^{cosx}$

両辺の対数をとると、

$logy = cosxlog(cosx)$

両辺をxで微分すると、

$\frac{1}{y} \frac{dy}{dx} = -sinxlog(cosx) + cosx \times \frac{-sinx}{cosx} = -sinxlog(cosx) - sinx = -sinx(log(cosx)+1)$

$\frac{dy}{dx} = y \times (-sinx(log(cosx)+1)) = (cosx)^{cosx} (-sinx(log(cosx)+1)) = -(cosx)^{cosx} sinx(log(cosx)+1)$

3. 最終的な答え

(i) $\frac{dy}{dx} = \frac{-x}{\sqrt{1-x^2}}$

(ii) $\frac{dy}{dx} = \frac{2e^{tan^{-1}(2x)}}{1+4x^2}$

(iii) $\frac{dy}{dx} = \frac{1}{xlogxlog(logx)}$

(iv) $\frac{dy}{dx} = -(cosx)^{cosx} sinx(log(cosx)+1)$