次の関数を微分せよ。 (a) $y = (1-3x)^2$ (b) $y = \frac{1}{(5x-2)^5}$解析学微分合成関数導関数2025/6/131. 問題の内容次の関数を微分せよ。(a) y=(1−3x)2y = (1-3x)^2y=(1−3x)2(b) y=1(5x−2)5y = \frac{1}{(5x-2)^5}y=(5x−2)512. 解き方の手順(a)合成関数の微分を使う。y=u2y = u^2y=u2 とおくと、u=1−3xu = 1 - 3xu=1−3x。dydx=dydu⋅dudx\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}dxdy=dudy⋅dxdudydu=2u=2(1−3x)\frac{dy}{du} = 2u = 2(1 - 3x)dudy=2u=2(1−3x)dudx=−3\frac{du}{dx} = -3dxdu=−3よって、dydx=2(1−3x)⋅(−3)=−6(1−3x)=−6+18x=18x−6\frac{dy}{dx} = 2(1 - 3x) \cdot (-3) = -6(1 - 3x) = -6 + 18x = 18x - 6dxdy=2(1−3x)⋅(−3)=−6(1−3x)=−6+18x=18x−6(b)y=1(5x−2)5=(5x−2)−5y = \frac{1}{(5x-2)^5} = (5x - 2)^{-5}y=(5x−2)51=(5x−2)−5y=u−5y = u^{-5}y=u−5 とおくと、u=5x−2u = 5x - 2u=5x−2。dydx=dydu⋅dudx\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}dxdy=dudy⋅dxdudydu=−5u−6=−5(5x−2)−6\frac{dy}{du} = -5u^{-6} = -5(5x - 2)^{-6}dudy=−5u−6=−5(5x−2)−6dudx=5\frac{du}{dx} = 5dxdu=5よって、dydx=−5(5x−2)−6⋅5=−25(5x−2)−6=−25(5x−2)6\frac{dy}{dx} = -5(5x - 2)^{-6} \cdot 5 = -25(5x - 2)^{-6} = \frac{-25}{(5x - 2)^6}dxdy=−5(5x−2)−6⋅5=−25(5x−2)−6=(5x−2)6−253. 最終的な答え(a) 18x−618x - 618x−6(b) −25(5x−2)6\frac{-25}{(5x - 2)^6}(5x−2)6−25