高等数学（第七版）同济大学习题2-2 个人解答（后7题）

高等数学（第七版）同济大学习题2-2

8.求下列函数的导数：\begin{aligned}&8. \ 求下列函数的导数：&\end{aligned}8. 求下列函数的导数：

(1)y=(arcsinx2)2；(2)y=lntanx2；(3)y=1+ln2x；(4)y=earctanx；(5)y=sinnxcosnx；(6)y=arctanx+1x−1；(7)y=arcsinxarccosx；(8)y=lnlnlnx；(9)y=1+x−1−x1+x+1−x；(10)y=arcsin1−x1+x\begin{aligned} &\ \ (1)\ \ y=\left(arcsin\ \frac{x}{2}\right)^2；\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)\ \ y=ln\ tan\ \frac{x}{2}；\\\\ &\ \ (3)\ \ y=\sqrt{1+ln^2\ x}；\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4)\ \ y=e^{arctan\ \sqrt{x}}；\\\\ &\ \ (5)\ \ y=sin^n\ xcos\ nx；\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (6)\ \ y=arctan\ \frac{x+1}{x-1}；\\\\ &\ \ (7)\ \ y=\frac{arcsin\ x}{arccos\ x}；\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (8)\ \ y=ln\ ln\ ln\ x；\\\\ &\ \ (9)\ \ y=\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}；\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (10)\ \ y=arcsin\ \sqrt{\frac{1-x}{1+x}} & \end{aligned} (1) y=(arcsin 2x)2； (2) y=ln tan 2x； (3) y=1+ln2 x； (4) y=earctan x； (5) y=sinn xcos nx； (6) y=arctan x−1x+1； (7) y=arccos xarcsin x； (8) y=ln ln ln x； (9) y=1+x+1−x1+x−1−x； (10) y=arcsin 1+x1−x

解：

(1)设u=arcsinv，v=x2，则y′=2u⋅11−v2⋅12=arcsinx21−(x2)2=2arcsinx24−x2(2)设u=tanv，v=x2，则y′=(lnu)′⋅u′⋅v′=1u⋅sec2v⋅12=1tanx2⋅sec2x2⋅12=cosx2sinx2⋅1cos2x2⋅12=1sinx=cscx(3)设u=1+v2，v=lnx，y′=121+ln2x⋅u′⋅v′=2v⋅1x21+ln2x=lnxx1+ln2x(4)设u=arctanv，v=x，则y′=eu⋅u′⋅v′=earctanx⋅11+x⋅12x=earctanx2x(1+x)(5)y′=(sinnx)′cosnx+sinnx(cosnx)′=nsinn−1xcosxcosnx−nsinnxsinnx=nsinn−1x(cosxcosnx−sinxsinnx)=nsinn−1xcos(n+1)x(6)设u=x+1x−1，则y′=11+(x+1x−1)2⋅(x+1x−1)′=(x−1)2(x−1)2+(x+1)2⋅(x+1)′(x−1)−(x+1)(x−1)′(x−1)2=−1x2+1(7)y′=(arcsinx)′arccosx−arcsinx(arccosx)′(arccosx)2=arccosx+arcsinx1−x2(arccosx)2=π2(arccosx)21−x2(8)设u=lnv，v=lnx，则y′=1u⋅u′⋅v′=1lnlnx⋅1lnx⋅1x=1ln(x+lnx)x(9)y=1+x−1−x1+x+1−x=(1+x−1−x)(1+x−1−x)2x=1−1−x2xy′=(1−1−x2)′x−(1−1−x2)x′x2=1x21−x2−1x2(10)设u=v，v=1−x1+x，则y′=11−u2⋅u′⋅v′=11−1−x1+x⋅121−x1+x⋅(1−x)′(1+x)−(1−x)(1+x)′(1+x)2=12x1+x⋅121−x1+x⋅−2(1+x)2=−1(1+x)2x−2x2\begin{aligned} &\ \ (1)\ 设u=arcsin\ v，v=\frac{x}{2}，则y'=2u \cdot \frac{1}{\sqrt{1-v^2}} \cdot \frac{1}{2}=\frac{arcsin\ \frac{x}{2}}{\sqrt{1-\left(\frac{x}{2}\right)^2}}=\frac{2arcsin\ \frac{x}{2}}{\sqrt{4-x^2}}\\\\ &\ \ (2)\ 设u=tan\ v，v=\frac{x}{2}，则y'=(ln\ u)' \cdot u' \cdot v'=\frac{1}{u} \cdot sec^2 v \cdot \frac{1}{2}=\frac{1}{tan\ \frac{x}{2}} \cdot sec^2\ \frac{x}{2} \cdot \frac{1}{2}=\frac{cos\ \frac{x}{2}}{sin\ \frac{x}{2}} \cdot \frac{1}{cos^2\ \frac{x}{2}} \cdot \frac{1}{2}=\frac{1}{sin\ x}=csc\ x\\\\ &\ \ (3)\ 设u=1+v^2，v=ln\ x，y'=\frac{1}{2\sqrt{1+ln^2\ x}} \cdot u' \cdot v'=\frac{2v \cdot \frac{1}{x}}{2\sqrt{1+ln^2\ x}}=\frac{ln\ x}{x\sqrt{1+ln^2\ x}}\\\\ &\ \ (4)\ 设u=arctan\ v，v=\sqrt{x}，则y'=e^u \cdot u' \cdot v'=e^{arctan\ \sqrt{x}} \cdot \frac{1}{1+x} \cdot \frac{1}{2\sqrt{x}}=\frac{e^{arctan\ \sqrt{x}}}{2\sqrt{x}(1+x)}\\\\ &\ \ (5)\ y'=(sin^n\ x)'cos\ nx+sin^n\ x(cos\ nx)'=nsin^{n-1}\ xcos\ xcos\ nx-nsin^n\ xsin\ nx=\\\\ &\ \ \ \ \ \ \ \ nsin^{n-1}\ x(cos\ xcos\ nx-sin\ xsin\ nx)=nsin^{n-1}\ xcos(n+1)\ x\\\\ &\ \ (6)\ 设u=\frac{x+1}{x-1}，则y'=\frac{1}{1+\left(\frac{x+1}{x-1}\right)^2} \cdot \left(\frac{x+1}{x-1}\right)'=\frac{(x-1)^2}{(x-1)^2+(x+1)^2} \cdot \frac{(x+1)'(x-1)-(x+1)(x-1)'}{(x-1)^2}=-\frac{1}{x^2+1}\\\\ &\ \ (7)\ y'=\frac{(arcsin\ x)'arccos\ x-arcsin\ x(arccos\ x)'}{(arccos\ x)^2}=\frac{\frac{arccos\ x+arcsin\ x}{\sqrt{1-x^2}}}{(arccos\ x)^2}=\frac{\pi}{2(arccos\ x)^2\sqrt{1-x^2}}\\\\ &\ \ (8)\ 设u=ln\ v，v=ln\ x，则y'=\frac{1}{u} \cdot u' \cdot v'=\frac{1}{ln\ ln\ x} \cdot \frac{1}{ln\ x} \cdot \frac{1}{x}=\frac{1}{ln(x+ln\ x)^x}\\\\ &\ \ (9)\ y=\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}=\frac{(\sqrt{1+x}-\sqrt{1-x})(\sqrt{1+x}-\sqrt{1-x})}{2x}=\frac{1-\sqrt{1-x^2}}{x}\\\\ &\ \ \ \ \ \ \ \ y'=\frac{(1-\sqrt{1-x^2})'x-(1-\sqrt{1-x^2})x'}{x^2}=\frac{1}{x^2\sqrt{1-x^2}}-\frac{1}{x^2}\\\\ &\ \ (10)\ 设u=\sqrt{v}，v=\frac{1-x}{1+x}，则y'=\frac{1}{\sqrt{1-u^2}} \cdot u' \cdot v'=\frac{1}{\sqrt{1-\frac{1-x}{1+x}}} \cdot \frac{1}{2\sqrt{\frac{1-x}{1+x}}} \cdot \frac{(1-x)'(1+x)-(1-x)(1+x)'}{(1+x)^2}=\\\\ &\ \ \ \ \ \ \ \ \ \ \frac{1}{\sqrt{\frac{2x}{1+x}}} \cdot \frac{1}{2\sqrt{\frac{1-x}{1+x}}} \cdot \frac{-2}{(1+x)^2}=-\frac{1}{(1+x)\sqrt{2x-2x^2}} & \end{aligned} (1) 设u=arcsin v，v=2x，则y′=2u⋅1−v21⋅21=1−(2x)2arcsin 2x=4−x22arcsin 2x (2) 设u=tan v，v=2x，则y′=(ln u)′⋅u′⋅v′=u1⋅sec2v⋅21=tan 2x1⋅sec2 2x⋅21=sin 2xcos 2x⋅cos2 2x1⋅21=sin x1=csc x (3) 设u=1+v2，v=ln x，y′=21+ln2 x1⋅u′⋅v′=21+ln2 x2v⋅x1=x1+ln2 xln x (4) 设u=arctan v，v=x，则y′=eu⋅u′⋅v′=earctan x⋅1+x1⋅2x1=2x(1+x)earctan x (5) y′=(sinn x)′cos nx+sinn x(cos nx)′=nsinn−1 xcos xcos nx−nsinn xsin nx= nsinn−1 x(cos xcos nx−sin xsin nx)=nsinn−1 xcos(n+1) x (6) 设u=x−1x+1，则y′=1+(x−1x+1)21⋅(x−1x+1)′=(x−1)2+(x+1)2(x−1)2⋅(x−1)2(x+1)′(x−1)−(x+1)(x−1)′=−x2+11 (7) y′=(arccos x)2(arcsin x)′arccos x−arcsin x(arccos x)′=(arccos x)21−x2arccos x+arcsin x=2(arccos x)21−x2π (8) 设u=ln v，v=ln x，则y′=u1⋅u′⋅v′=ln ln x1⋅ln x1⋅x1=ln(x+ln x)x1 (9) y=1+x+1−x1+x−1−x=2x(1+x−1−x)(1+x−1−x)=x1−1−x2 y′=x2(1−1−x2)′x−(1−1−x2)x′=x21−x21−x21 (10) 设u=v，v=1+x1−x，则y′=1−u21⋅u′⋅v′=1−1+x1−x1⋅21+x1−x1⋅(1+x)2(1−x)′(1+x)−(1−x)(1+x)′= 1+x2x1⋅21+x1−x1⋅(1+x)2−2=−(1+x)2x−2x21

9.设函数f(x)和g(x)可导，且f2(x)+g2(x)≠0，试求函数y=f2(x)+g2(x)的导数。\begin{aligned}&9. \ 设函数f(x)和g(x)可导，且f^2(x)+g^2(x) \neq 0，试求函数y=\sqrt{f^2(x)+g^2(x)}的导数。&\end{aligned}9. 设函数f(x)和g(x)可导，且f2(x)+g2(x)=0，试求函数y=f2(x)+g2(x)的导数。

解：

y′=2f(x)f′(x)+2g(x)g′(x)2f2(x)+g2(x)=f(x)f′(x)+g(x)g′(x)f2(x)+g2(x)\begin{aligned} &\ \ y'=\frac{2f(x)f'(x)+2g(x)g'(x)}{2\sqrt{f^2(x)+g^2(x)}}=\frac{f(x)f'(x)+g(x)g'(x)}{\sqrt{f^2(x)+g^2(x)}} & \end{aligned} y′=2f2(x)+g2(x)2f(x)f′(x)+2g(x)g′(x)=f2(x)+g2(x)f(x)f′(x)+g(x)g′(x)

10.设f(x)可导，求下列函数的导数dydx:\begin{aligned}&10. \ 设f(x)可导，求下列函数的导数\frac{dy}{dx}:&\end{aligned}10. 设f(x)可导，求下列函数的导数dxdy:

(1)y=f(x2)；(2)y=f(sin2x)+f(cos2x).\begin{aligned} &\ \ (1)\ \ y=f(x^2)；\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)\ \ y=f(sin^2\ x)+f(cos^2\ x). & \end{aligned} (1) y=f(x2)； (2) y=f(sin2 x)+f(cos2 x).

解：

(1)y′=f′(x2)2x=2xf′(x2)(2)y′=f′(sin2x)2sinxcosx−f′(cos2x)2sinxcosx=2sinxcosx(f′(sin2x)−f′(cos2x))=sin2x[f′(sin2x)−f′(cos2x)]\begin{aligned} &\ \ (1)\ y'=f'(x^2)2x=2xf'(x^2)\\\\ &\ \ (2)\ y'=f'(sin^2\ x)2sin\ xcos\ x-f'(cos^2\ x)2sin\ xcos\ x=2sin\ xcos\ x(f'(sin^2\ x)-f'(cos^2\ x))=\\\\ &\ \ \ \ \ \ \ \ sin\ 2x[f'(sin^2\ x)-f'(cos^2\ x)] & \end{aligned} (1) y′=f′(x2)2x=2xf′(x2) (2) y′=f′(sin2 x)2sin xcos x−f′(cos2 x)2sin xcos x=2sin xcos x(f′(sin2 x)−f′(cos2 x))= sin 2x[f′(sin2 x)−f′(cos2 x)]

11.求下列函数的导数：\begin{aligned}&11. \ 求下列函数的导数：&\end{aligned}11. 求下列函数的导数：

(1)y=e−x(x2−2x+3)；(2)y=sin2x⋅sin(x2)；(3)y=(arctanx2)2；(4)y=lnxxn；(5)y=et−e−tet+e−t；(6)y=lncos1x；(7)y=e−sin21x；(8)y=x+x；(9)y=xarcsinx2+4−x2；(10)y=arcsin2t1+t2\begin{aligned} &\ \ (1)\ \ y=e^{-x}(x^2-2x+3)；\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)\ \ y=sin^2\ x \cdot sin(x^2)；\\\\ &\ \ (3)\ \ y=\left(arctan\ \frac{x}{2}\right)^2；\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4)\ \ y=\frac{ln\ x}{x^n}；\\\\ &\ \ (5)\ \ y=\frac{e^t-e^{-t}}{e^t+e^{-t}}；\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (6)\ \ y=ln\ cos\ \frac{1}{x}；\\\\ &\ \ (7)\ \ y=e^{-sin^2\ \frac{1}{x}}；\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (8)\ \ y=\sqrt{x+\sqrt{x}}；\\\\ &\ \ (9)\ \ y=xarcsin\ \frac{x}{2}+\sqrt{4-x^2}；\ \ \ \ \ \ \ \ (10)\ \ y=arcsin\ \frac{2t}{1+t^2} & \end{aligned} (1) y=e−x(x2−2x+3)； (2) y=sin2 x⋅sin(x2)； (3) y=(arctan 2x)2； (4) y=xnln x； (5) y=et+e−tet−e−t； (6) y=ln cos x1； (7) y=e−sin2 x1； (8) y=x+x； (9) y=xarcsin 2x+4−x2； (10) y=arcsin 1+t22t

解：

(1)y=x2−2x+3ex，则y′=(x2−2x+3)′ex−(x2−2x+3)(ex)′e2x=−x2−4x+5ex(2)y′=(sin2x)′sin(x2)+sin2x(sin(x2))=2sinxcosxsin(x2)+sin2x2xcos(x2)=sin2xsin(x2)+2xsin2xcos(x2)(3)y′=2arctanx2⋅11+x24⋅12=4arctanx2x2+4(4)y′=(lnx)′xn−lnx(xn)′x2n=xn−1−nxn−1lnxx2n=1−nlnxxn+1(5)y=e2t−1e2t+1，则y′=(e2t−1)′(e2t+1)−(e2t−1)(e2t+1)′(e2t+1)2=2e2t(e2t+1)−2e2t(e2t−1)(e2t+1)2=4e2t(e2t+1)2=4(et+e−t)2(6)y′=1cos1x⋅(−sin1x)⋅(−1x2)=tan1xx2(7)y′=e−sin21x⋅(−2sin1x)⋅(−cos1x)⋅(−1x2)=−e−sin21xsin2xx2(8)y′=12x+x⋅(1+12x)=2x+14x+xx(9)y′=(xarcsinx2)′+(4−x2)′=(x)′arcsinx2+x(arcsinx2)′+(4−x2)′=arcsinx2+x4−x2−x4−x2=arcsinx2(10)y′=11−(2t1+t2)2⋅(2t)′(1+t2)−2t(1+t2)′(1+t2)2=1+t2∣1−t2∣⋅2(1−t2)(1+t2)2={21+t2，∣t∣<1，−21+t2，∣t∣>1.\begin{aligned} &\ \ (1)\ y=\frac{x^2-2x+3}{e^x}，则y'=\frac{(x^2-2x+3)'e^x-(x^2-2x+3)(e^x)'}{e^{2x}}=-\frac{x^2-4x+5}{e^x}\\\\ &\ \ (2)\ y'=(sin^2\ x)'sin(x^2)+sin^2\ x(sin(x^2))=2sin\ xcosxsin(x^2)+sin^2\ x2xcos(x^2)=\\\\ &\ \ \ \ \ \ \ \ sin\ 2xsin(x^2)+2xsin^2\ xcos(x^2)\\\\ &\ \ (3)\ y'=2arctan\ \frac{x}{2} \cdot \frac{1}{1+\frac{x^2}{4}} \cdot \frac{1}{2}=\frac{4arctan\ \frac{x}{2}}{x^2+4}\\\\ &\ \ (4)\ y'=\frac{(ln\ x)'x^n-ln\ x(x^n)'}{x^{2n}}=\frac{x^{n-1}-nx^{n-1}ln\ x}{x^{2n}}=\frac{1-nln\ x}{x^{n+1}}\\\\ &\ \ (5)\ y=\frac{e^{2t}-1}{e^{2t}+1}，则y'=\frac{(e^{2t}-1)'(e^{2t}+1)-(e^{2t}-1)(e^{2t}+1)'}{(e^{2t}+1)^2}=\frac{2e^{2t}(e^{2t}+1)-2e^{2t}(e^{2t}-1)}{(e^{2t}+1)^2}=\\\\ &\ \ \ \ \ \ \ \ \frac{4e^{2t}}{(e^{2t}+1)^2}=\frac{4}{(e^t+e^{-t})^2}\\\\ &\ \ (6)\ y'=\frac{1}{cos\ \frac{1}{x}} \cdot \left(-sin\ \frac{1}{x}\right) \cdot \left(-\frac{1}{x^2}\right)=\frac{tan\ \frac{1}{x}}{x^2}\\\\ &\ \ (7)\ y'=e^{-sin^2\ \frac{1}{x}} \cdot \left(-2sin\ \frac{1}{x}\right) \cdot \left(-cos\ \frac{1}{x}\right) \cdot \left(-\frac{1}{x^2}\right)=-\frac{e^{-sin^2\ \frac{1}{x}}sin\ \frac{2}{x}}{x^2}\\\\ &\ \ (8)\ y'=\frac{1}{2\sqrt{x+\sqrt{x}}} \cdot \left(1+\frac{1}{2\sqrt{x}}\right)=\frac{2\sqrt{x}+1}{4\sqrt{x+x\sqrt{x}}}\\\\ &\ \ (9)\ y'=\left(xarcsin\ \frac{x}{2}\right)'+(\sqrt{4-x^2})'=(x)'arcsin\ \frac{x}{2}+x\left(arcsin\ \frac{x}{2}\right)'+(\sqrt{4-x^2})'=\\\\ &\ \ \ \ \ \ \ \ arcsin\ \frac{x}{2}+\frac{x}{\sqrt{4-x^2}}-\frac{x}{\sqrt{4-x^2}}=arcsin\ \frac{x}{2}\\\\ &\ \ (10)\ y'=\frac{1}{\sqrt{1-\left(\frac{2t}{1+t^2}\right)^2}} \cdot \frac{(2t)'(1+t^2)-2t(1+t^2)'}{(1+t^2)^2}=\frac{1+t^2}{|1-t^2|} \cdot \frac{2(1-t^2)}{(1+t^2)^2}=\begin{cases}\frac{2}{1+t^2}，\ \ \ |t| \lt 1，\\\\-\frac{2}{1+t^2}，|t| \gt 1.\end{cases} & \end{aligned} (1) y=exx2−2x+3，则y′=e2x(x2−2x+3)′ex−(x2−2x+3)(ex)′=−exx2−4x+5 (2) y′=(sin2 x)′sin(x2)+sin2 x(sin(x2))=2sin xcosxsin(x2)+sin2 x2xcos(x2)= sin 2xsin(x2)+2xsin2 xcos(x2) (3) y′=2arctan 2x⋅1+4x21⋅21=x2+44arctan 2x (4) y′=x2n(ln x)′xn−ln x(xn)′=x2nxn−1−nxn−1ln x=xn+11−nln x (5) y=e2t+1e2t−1，则y′=(e2t+1)2(e2t−1)′(e2t+1)−(e2t−1)(e2t+1)′=(e2t+1)22e2t(e2t+1)−2e2t(e2t−1)= (e2t+1)24e2t=(et+e−t)24 (6) y′=cos x11⋅(−sin x1)⋅(−x21)=x2tan x1 (7) y′=e−sin2 x1⋅(−2sin x1)⋅(−cos x1)⋅(−x21)=−x2e−sin2 x1sin x2 (8) y′=2x+x1⋅(1+2x1)=4x+xx2x+1 (9) y′=(xarcsin 2x)′+(4−x2)′=(x)′arcsin 2x+x(arcsin 2x)′+(4−x2)′= arcsin 2x+4−x2x−4−x2x=arcsin 2x (10) y′=1−(1+t22t)21⋅(1+t2)2(2t)′(1+t2)−2t(1+t2)′=∣1−t2∣1+t2⋅(1+t2)22(1−t2)=⎩⎪⎨⎪⎧1+t22， ∣t∣<1，−1+t22，∣t∣>1.

12.求下列函数的导数：\begin{aligned}&12. \ 求下列函数的导数：&\end{aligned}12. 求下列函数的导数：

(1)y=ch(shx)；(2)y=shx⋅echx；(3)y=th(lnx)；(4)y=sh3x+ch2x；(5)y=th(1−x2)；(6)y=arsh(x2+1)；(7)y=arch(e2x)；(8)y=arctan(thx)；(9)y=lnchx+12ch2x；(10)y=ch2(x−1x+1).\begin{aligned} &\ \ (1)\ \ y=ch(sh\ x)；\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)\ \ y=sh\ x \cdot e^{ch\ x}；\\\\ &\ \ (3)\ \ y=th(ln\ x)；\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4)\ \ y=sh^3\ x+ch^2\ x；\\\\ &\ \ (5)\ \ y=th(1-x^2)；\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (6)\ \ y=arsh(x^2+1)；\\\\ &\ \ (7)\ \ y=arch(e^{2x})；\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (8)\ \ y=arctan(th\ x)；\\\\ &\ \ (9)\ \ y=ln\ ch\ x+\frac{1}{2ch^2\ x}；\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (10)\ \ y=ch^2\left(\frac{x-1}{x+1}\right). & \end{aligned} (1) y=ch(sh x)； (2) y=sh x⋅ech x； (3) y=th(ln x)； (4) y=sh3 x+ch2 x； (5) y=th(1−x2)； (6) y=arsh(x2+1)； (7) y=arch(e2x)； (8) y=arctan(th x)； (9) y=ln ch x+2ch2 x1； (10) y=ch2(x+1x−1).

解：

(1)设u=shx，则y′=sh(shx)chx(2)y′=(shx)′echx+shx(echx)′=chxechx+sh2xechx=echx(chx+sh2x)(3)y′=1ch2(lnx)⋅1x=1xch2(lnx)(4)y′=(sh3x)′+(ch2x)′=3sh2xchx+2shxchx=shxchx(3shx+2)(5)y′=1ch2(1−x2)⋅(1−x2)′=−2xch2(1−x2)(6)y′=11+(x2+1)2⋅(x2+1)′=2x1+(x2+1)2(7)y′=1e4x−1⋅e2x⋅2=2e2xe4x−1(8)y′=11+th2x⋅1ch2x=1ch2x+ch2xth2x=1ch2x+sh2x=11+2sh2x(9)y′=(lnchx)′+(12ch2x)′=shxchx−shxch3x=shxch2x−shxch3x=sh3xch3x=th3x(10)y′=2ch(x−1x+1)sh(x−1x+1)2(x+1)2=2(x+1)2sh(2⋅x−1x+1)\begin{aligned} &\ \ (1)\ 设u=sh\ x，则y'=sh(sh\ x)ch\ x\\\\ &\ \ (2)\ y'=(sh\ x)'e^{ch\ x}+sh\ x(e^{ch\ x})'=ch\ xe^{ch\ x}+sh^2\ xe^{ch\ x}=e^{ch\ x}(ch\ x+sh^2\ x)\\\\ &\ \ (3)\ y'=\frac{1}{ch^2(ln\ x)} \cdot \frac{1}{x}=\frac{1}{xch^2(ln\ x)}\\\\ &\ \ (4)\ y'=(sh^3\ x)'+(ch^2\ x)'=3sh^2\ xch\ x+2sh\ xch\ x=sh\ xch\ x(3sh\ x+2)\\\\ &\ \ (5)\ y'=\frac{1}{ch^2(1-x^2)} \cdot (1-x^2)'=-\frac{2x}{ch^2(1-x^2)}\\\\ &\ \ (6)\ y'=\frac{1}{\sqrt{1+(x^2+1)^2}} \cdot (x^2+1)'=\frac{2x}{\sqrt{1+(x^2+1)^2}}\\\\ &\ \ (7)\ y'=\frac{1}{\sqrt{e^{4x}-1}} \cdot e^{2x} \cdot 2=\frac{2e^{2x}}{\sqrt{e^{4x}-1}}\\\\ &\ \ (8)\ y'=\frac{1}{1+th^2\ x} \cdot \frac{1}{ch^2\ x}=\frac{1}{ch^2\ x+ch^2\ xth^2\ x}=\frac{1}{ch^2\ x+sh^2\ x}=\frac{1}{1+2sh^2\ x}\\\\ &\ \ (9)\ y'=(ln\ ch\ x)'+\left(\frac{1}{2ch^2\ x}\right)'=\frac{sh\ x}{ch\ x}-\frac{sh\ x}{ch^3\ x}=\frac{sh\ xch^2\ x-sh\ x}{ch^3\ x}=\frac{sh^3\ x}{ch^3\ x}=th^3\ x\\\\ &\ \ (10)\ y'=2ch\ \left(\frac{x-1}{x+1}\right)sh\ \left(\frac{x-1}{x+1}\right)\frac{2}{(x+1)^2} =\frac{2}{(x+1)^2}sh\ \left(2 \cdot \frac{x-1}{x+1}\right) & \end{aligned} (1) 设u=sh x，则y′=sh(sh x)ch x (2) y′=(sh x)′ech x+sh x(ech x)′=ch xech x+sh2 xech x=ech x(ch x+sh2 x) (3) y′=ch2(ln x)1⋅x1=xch2(ln x)1 (4) y′=(sh3 x)′+(ch2 x)′=3sh2 xch x+2sh xch x=sh xch x(3sh x+2) (5) y′=ch2(1−x2)1⋅(1−x2)′=−ch2(1−x2)2x (6) y′=1+(x2+1)21⋅(x2+1)′=1+(x2+1)22x (7) y′=e4x−11⋅e2x⋅2=e4x−12e2x (8) y′=1+th2 x1⋅ch2 x1=ch2 x+ch2 xth2 x1=ch2 x+sh2 x1=1+2sh2 x1 (9) y′=(ln ch x)′+(2ch2 x1)′=ch xsh x−ch3 xsh x=ch3 xsh xch2 x−sh x=ch3 xsh3 x=th3 x (10) y′=2ch (x+1x−1)sh (x+1x−1)(x+1)22=(x+1)22sh (2⋅x+1x−1)

13.设函数f(x)和g(x)均在点x0的某一邻域内有定义，f(x)在x0处可导，f(x0)=0，g(x)在x0处连续，试讨论f(x)g(x)在x0处的可导性。\begin{aligned}&13. \ 设函数f(x)和g(x)均在点x_0的某一邻域内有定义，f(x)在x_0处可导，f(x_0)=0，g(x)在x_0处连续，\\\\&\ \ \ \ \ \ 试讨论f(x)g(x)在x_0处的可导性。&\end{aligned}13. 设函数f(x)和g(x)均在点x0的某一邻域内有定义，f(x)在x0处可导，f(x0)=0，g(x)在x0处连续，试讨论f(x)g(x)在x0处的可导性。

解：

由于f(x)在x0处可导，且f(x0)=0，则有f′(x0)=lim⁡x→x0f(x)−f(x0)x−x0=lim⁡x→x0f(x)x−x0，由于g(x)在x0处连续，则有lim⁡x→x0g(x)=g(x0)，故lim⁡x→x0f(x)g(x)−f(x0)g(x0)x−x0=lim⁡x→x0f(x)x−x0g(x)=f′(x0)g(x0)，f(x)g(x)在x0处可导，导数为f′(x0)g(x0).\begin{aligned} &\ \ 由于f(x)在x_0处可导，且f(x_0)=0，则有f'(x_0)=\lim_{x \rightarrow x_0}\frac{f(x)-f(x_0)}{x-x_0}=\lim_{x \rightarrow x_0}\frac{f(x)}{x-x_0}，\\\\ &\ \ 由于g(x)在x_0处连续，则有\lim_{x \rightarrow x_0}g(x)=g(x_0)，故\lim_{x \rightarrow x_0}\frac{f(x)g(x)-f(x_0)g(x_0)}{x-x_0}=\lim_{x \rightarrow x_0}\frac{f(x)}{x-x_0}g(x)=f'(x_0)g(x_0)，\\\\ &\ \ f(x)g(x)在x_0处可导，导数为f'(x_0)g(x_0). & \end{aligned} 由于f(x)在x0处可导，且f(x0)=0，则有f′(x0)=x→x0limx−x0f(x)−f(x0)=x→x0limx−x0f(x)，由于g(x)在x0处连续，则有x→x0limg(x)=g(x0)，故x→x0limx−x0f(x)g(x)−f(x0)g(x0)=x→x0limx−x0f(x)g(x)=f′(x0)g(x0)， f(x)g(x)在x0处可导，导数为f′(x0)g(x0).

14.设函数f(x)满足下列条件：(1)f(x+y)=f(x)⋅f(y)，对一切x，y∈R；(2)f(x)=1+xg(x)，而lim⁡x→0g(x)=1.试证明f(x)在R上处处可导，且f′(x)=f(x).\begin{aligned}&14. \ 设函数f(x)满足下列条件：\\\\&\ \ (1)\ f(x+y)=f(x) \cdot f(y)，对一切x，y \in R；\\\\&\ \ (2)\ f(x)=1+xg(x)，而\lim_{x \rightarrow 0}g(x)=1.\\\\&\ \ 试证明f(x)在R上处处可导，且f'(x)=f(x).&\end{aligned}14. 设函数f(x)满足下列条件： (1) f(x+y)=f(x)⋅f(y)，对一切x，y∈R； (2) f(x)=1+xg(x)，而x→0limg(x)=1. 试证明f(x)在R上处处可导，且f′(x)=f(x).

解：

由条件2可知f(0)=1，故f′(x)=lim⁡Δx→0f(x+Δx)−f(x)Δx=lim⁡Δx→0f(x)f(Δx)−f(x)Δx=lim⁡Δx→0[f(x)⋅f(Δx)−1Δx]=lim⁡Δx→0[f(x)⋅Δxg(Δx)Δx]=lim⁡Δx→0[f(x)g(Δx)]=f(x)⋅1=f(x).\begin{aligned} &\ \ 由条件2可知f(0)=1，故f'(x)=\lim_{\Delta x \rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}=\lim_{\Delta x \rightarrow 0}\frac{f(x)f(\Delta x)-f(x)}{\Delta x}=\lim_{\Delta x \rightarrow 0}\left[f(x) \cdot \frac{f(\Delta x)-1}{\Delta x}\right]=\\\\ &\ \ \lim_{\Delta x \rightarrow 0}\left[f(x) \cdot \frac{\Delta xg(\Delta x)}{\Delta x}\right]=\lim_{\Delta x \rightarrow 0}\left[f(x)g(\Delta x)\right]=f(x) \cdot 1=f(x). & \end{aligned} 由条件2可知f(0)=1，故f′(x)=Δx→0limΔxf(x+Δx)−f(x)=Δx→0limΔxf(x)f(Δx)−f(x)=Δx→0lim[f(x)⋅Δxf(Δx)−1]= Δx→0lim[f(x)⋅ΔxΔxg(Δx)]=Δx→0lim[f(x)g(Δx)]=f(x)⋅1=f(x).