同濟高等數(shù)學第七版2.2習題精講（續(xù)二 ）

6、求下列函數(shù)的導數(shù)：

\begin{array}{ll}{\text { (1) } y=(2 x+5)^{4} ;} & {\text { (2) } y=\cos (4-3 x)} \\ {\text { (3) } y=e^{-3 x^{2}} ;} & {\text { (4) } y=\ln \left(1+x^{2}\right)}\\{\text { (5) } y=\sin ^{2} x ;} & {\text { (6) } y=\sqrt{a^{2}-x^{2}}} \\ {\text { (7) } y=\tan x^{2} ;} & {\text { (8) } y=\arctan \left(\mathrm{e}^{x}\right)} \\ {\text { (9) } y=(\arcsin x)^{2} ;} & {\text { (10) } y=\ln \cos x}\end{array}

根據(jù)求導的運算法則等直接進行求導。

解： (1) y^{\prime}=4(2 x+5)^{3} \cdot 2=8(2 x+5)^{3}
(2) y^{\prime}=-\sin (4-3 x)(-3)=3 \sin (4-3 x)
(3) y^{\prime}=\mathrm{e}^{-3 x^{2}} \cdot(-6 x)=-6 x \mathrm{e}^{-3 x^{2}}
(4) y^{\prime}=\frac{1}{1+x^{2}} \cdot 2 x=\frac{2 x}{1+x^{2}}
(5) y^{\prime}=2 \sin x \cos x=\sin 2 x
(6) y^{\prime}=\frac{1}{2 \sqrt{a^{2}-x^{2}}}(-2 x)=-\frac{x}{\sqrt{a^{2}-x^{2}}}
(7) y^{\prime}=\sec ^{2} x^{2} \cdot 2 x=2 x \sec ^{2} x^{2}
(8) y^{\prime}=\frac{1}{1+\left(e^{x}\right)^{2}} \cdot \mathrm{e}^{x}=\frac{\mathrm{e}^{x}}{1+\mathrm{e}^{2 x}}
(9) y^{\prime}=2 \arcsin x \cdot \frac{1}{\sqrt{1-x^{2}}}=\frac{2}{\sqrt{1-x^{2}}} \arcsin x
(10) y^{\prime}=\frac{1}{\cos x}(-\sin x)=-\tan x

7、求下列函數(shù)的導數(shù)：

\begin{array}{ll}{\text { (1) } y=\arcsin (1-2 x) ;} & {(2) y=\frac{1}{\sqrt{1-x^{2}}}} \\ {\text { (3) } y=\mathrm{e}^{-\frac{x}{2}} \cos 3 x ;} & {\text { (4) } y=\arccos \frac{1}{x}} \\ {\text { (5) } y=\frac{1-\ln x}{1+\ln x} ;} & {\text { (6) } y=\frac{\sin 2 x}{x}} \\ {\text { (7) } y=\arcsin \sqrt{x} ;} & {\text { (8) } y=\ln (x+\sqrt{a^{2}+x^{2}})} \\ {\text { (9) } y=\ln (\sec x+\tan x) ;} & {(10) y=\ln (\csc x-\cot x)}\end{array}

解：

\text { (1) } y^{\prime}=\frac{1}{\sqrt{1-(1-2 x)^{2}}} \cdot(-2)=-\frac{1}{\sqrt{x-x^{2}}} \\\text { (2) } y^{\prime}=\frac{\frac{(-2 x)}{(\sqrt{1-x^{2}})^{2}}}{(\sqrt{1-x^2})^2}=\frac{x}{\sqrt{\left(1-x^{2}\right)^{3}}} \\ \text { (3) } y^{\prime}=-\frac{1}{2} \mathrm{e}^{-\frac{x}{2}} \cos 3 x-3 \mathrm{e}^{-\frac{x}{2}} \sin 3 x \\=-\frac{1}{2} \mathrm{e}^{-\frac{x}{2}}(\cos 3 x+6 \sin 3 x)\\\text { (4) } y^{\prime}=-\frac{1}{\sqrt{1-\left(\frac{1}{x}\right)^{2}}} \cdot\left(-\frac{1}{x^{2}}\right)=\frac{|x|}{x^{2} \sqrt{x^{2}-1}}\\{\text { (5) } y^{\prime}=\frac{-\frac{1}{x}(1+\ln x)-(1-\ln x) \cdot \frac{1}{x}}{(1+\ln x)^{2}}=-\frac{2}{x(1+\ln x)^{2}}} \\ {\text { (6) } y^{\prime}=\frac{2 x \cos 2 x-\sin 2 x}{x^{2}}} \\ {\text { (7) } y^{\prime}=\frac{1}{\sqrt{1-(\sqrt{x})^{2}}} \cdot \frac{1}{2 \sqrt{x}}=\frac{1}{2 \sqrt{x-x^{2}}}}

(8) y^{\prime} =\frac{1}{x+\sqrt{a^{2}+x^{2}}}\left(1+\frac{2 x}{2 \sqrt{a^{2}+x^{2}}}\right)=\frac{1}{x+\sqrt{a^{2}+x^{2}}} \cdot \frac{x+\sqrt{a^{2}+x^{2}}}{\sqrt{a^{2}+x^{2}}} \\ =\frac{1}{\sqrt{a^{2}+x^{2}}} \\(9) y^{\prime} =\frac{1}{\sec x+\tan x}\left(\sec x \tan x+\sec ^{2} x\right)=\sec x \\(10) y^{\prime} =\frac{1}{\csc x-\cot x}\left(-\csc x \cot x+\csc ^{2} x\right)=\csc x

8、求下列函數(shù)的導數(shù)：

\begin{array}{ll}{\text { (1) } y=\left(\arcsin \frac{x}{2}\right)^{2}} & {\text { (2) } y=\ln \tan \frac{x}{2}} \\ {\text { (3) } y=\sqrt{1+\ln ^{2} x} ;} & {\text { (4) } y=\mathrm{e}^{\text {arctan } \sqrt{x}}} \\ {\text { (5) } y=\sin ^{n} x \cos n x ;} & {\text { (6) } y=\mathrm{e}^{\text {arctan } \sqrt{x}} ;} \\ {\text { (7) } y=\frac{\arcsin x}{\text { arccos } x} ;} & {\text { (10) } y=\arcsin \sqrt{\frac{1-x}{1+x}}} \\ {\text { (9) } y=\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}} & {\text { (10) } y=\arcsin \sqrt{\frac{1-x}{1+x}}}\end{array}

解：(1) y^{\prime}=2 \arcsin \frac{x}{2} \cdot \frac{1}{\sqrt{1-\left(\frac{x}{2}\right)^{2}}} \cdot \frac{1}{2}=\frac{2 \arcsin \frac{x}{2}}{\sqrt{4-x^{2}}}
(2) y^{\prime}=\frac{1}{\tan \frac{x}{2}} \cdot \sec ^{2} \frac{x}{2} \cdot \frac{1}{2}=\frac{1}{2 \sin \frac{x}{2} \cos \frac{x}{2}}=\frac{1}{\sin x}=\csc x
(3) y^{\prime}=\frac{1}{2 \sqrt{1+\ln ^{2} x}} \cdot 2 \ln x \cdot \frac{1}{x}=\frac{\ln x}{x \sqrt{1+\ln ^{2} x}}

(4) y^{\prime}=\mathrm{e}^{\arctan \sqrt{x}} \cdot \frac{1}{1+(\sqrt{x})^{2}} \cdot \frac{1}{2 \sqrt{x}}=\frac{1}{2 \sqrt{x}(1+x)} \mathrm{e}^{\arctan \sqrt{x}}
(5) y^{\prime}=n \sin ^{n-1} x \cos x \cos n x+\sin ^{n} x(-\sin n x) \cdot n
=n \sin ^{n-1} x(\cos x \cos n x-\sin x \sin n x) \cdot n
=n \sin ^{n-1} x(\cos x \cos n x-\sin x \sin n x)
=n \sin ^{n-1} x \cos (n+1) x
(6)y'=\frac{1} {1+\left(\frac{x+1}{x-1}\right)^{2} }\cdot \frac{(x-1)-(x+1)}{(x-1)^2}=\frac{-2}{(x-1)^{2}+(x+1)^{2}}
=-\frac{1}{1+x^{2}}

\begin{aligned}(7) y^{\prime} &=\frac{\frac{1}{\sqrt{1-x^{2}}} \arccos x-\arcsin x\left(-\frac{1}{\sqrt{1-x^{2}}}\right)}{(\arccos x)^{2}} \\ &=\frac{\arccos x+\arcsin x}{\sqrt{1-x^{2}}(\arccos x)^{2}}=\frac{\pi}{2 \sqrt{1-x^{2}}(\arccos x)^{2}} \\(8) y^{\prime} &=\frac{1}{\ln \ln x} \cdot \frac{1}{\ln x} \cdot \frac{1}{x}=\frac{1}{x \ln x \ln \ln x} \end{aligned}

(9) y^{\prime}=
\frac{\left(\frac{1}{2 \sqrt{1+x}}+\frac{1}{2 \sqrt{1-x}}\right)(\sqrt{1+x}+\sqrt{1-x})-(\sqrt{1+x}-\sqrt{1-x})\left(\frac{1}{2 \sqrt{1+x}}-\frac{1}{2 \sqrt{1-x}}\right)}{(\sqrt{1+x}+\sqrt{1-x})^2}
\frac{1}{2} \cdot \frac{\frac{1}{\sqrt{1+x} \sqrt{1-x}}(\sqrt{1+x}+\sqrt{1-x})^{2}+\frac{1}{\sqrt{1+x}\sqrt{1-x}}(\sqrt{1+x}-\sqrt{1-x})^{2}}{(2+2\sqrt{1-x^2})}
=\frac{1}{4} \frac{2+2}{(1+\sqrt{1-x^{2}}) \sqrt{1-x^{2}}}=\frac{1-\sqrt{1-x^{2}}}{x^{2} \sqrt{1-x^{2}}}

(10)y^{\prime}=\frac{1}{\sqrt{1-(\sqrt{\frac{1-x}{1+x}})^{2}}} \cdot \frac{1}{2 \sqrt{\frac{1-x}{1+x}}} \cdot \frac{-(1+x)-(1-x)}{(1+x)^{2}}

=-\frac{1}{\sqrt{1-\frac{1-x}{1+x}}} \cdot \frac{1}{\sqrt{\frac{1-x}{1+x}}} \cdot \frac{1}{(1+x)^{2}}
=-\frac{1}{\sqrt{2 x}(1+x) \sqrt{1-x}}=-\frac{1}{(1+x) \sqrt{2 x(1-x)}}

導數(shù)這部分實在是太多公式了，估計以后的也差不多都是這樣。還不知道能堅持多久………………

最后編輯于 ：
?著作權歸作者所有,轉載或內容合作請聯(lián)系作者
【社區(qū)內容提示】社區(qū)部分內容疑似由AI輔助生成，瀏覽時請結合常識與多方信息審慎甄別。
平臺聲明：文章內容（如有圖片或視頻亦包括在內）由作者上傳并發(fā)布，文章內容僅代表作者本人觀點，簡書系信息發(fā)布平臺，僅提供信息存儲服務。

相關閱讀更多精彩內容

友情鏈接更多精彩內容