第12类RD Sharma解决方案–第11章区分–练习11.1

问题1.将以下功能与第一原理e ^-x区别开

解决方案：

We have,

Let,

f(x)=e^-x

f(x+h)=e^-(x+h)

$\frac{d}{dx}f(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$

$=\lim_{h\to0}\frac{e^{-(x+h)}-e^{-x}}{h}$

$=\lim_{h\to0}\frac{e^{-x}.e^{-h}-e^{-x}}{h}$

$=\lim_{h\to0}[e^{-x}(\frac{e^{-h}-1}{-h})](-1)$

$=-e^{-x}[\lim_{h\to0}(\frac{e^{-h}-1}{-h})]$

=-e^-x

问题2.将以下功能与第一性原理区^分开3x

解决方案：

We have,

Let,

f(x)=e^3x

f(x+h)=e^3(x+h)

$\frac{d}{dx}f(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$

$=\lim_{h\to0}\frac{e^{3(x+h)}-e^{3x}}{h}$

$=\lim_{h\to0}\frac{e^{3x}.e^{3h}-e^{3x}}{h}$

$=\lim_{h\to0}[e^{3x}(\frac{e^{3h}-1}{3h})](3)$

$=3e^{3x}[\lim_{h\to0}(\frac{e^{3h}-1}{3h})]$

=3e^3x

问题3.将以下功能与第一原理e ^{ax + b区别开}

解决方案：

We have,

Let,

f(x)=e^ax+b

f(x+h)=e^a(x+h)+b

$\frac{d}{dx}f(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$

$=\lim_{h\to0}\frac{e^{a(x+h)+b}-e^{ax+b}}{h}$

$=\lim_{h\to0}\frac{e^{ax+b}.e^{ah}-e^{ax+b}}{h}$

$=\lim_{h\to0}[e^{ax+b}(\frac{e^{ah}-1}{ah})](a)$

$=ae^{ax+b}[\lim_{h\to0}(\frac{e^{ah}-1}{ah})]$

=ae^ax+b

问题4.将以下功能与第一原理和^{cosx区别开}

解决方案：

We have,

Let,

f(x)=e^cosx

f(x+h)=e^cos(x+h)

$\frac{d}{dx}f(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$

$=\lim_{h\to0}\frac{e^{cos(x+h)}-e^{cosx}}{h}$

$=\lim_{h\to0}[e^{cosx}(\frac{e^{cos(x+h)-cosx}-1}{h})]$

$=e^{cosx}\lim_{h\to0}(\frac{e^{cos(x+h)-cosx}-1}{cos(x+h)-cosx})(\frac{cos(x+h)-cosx}{h})$

$=e^{cosx}\lim_{h\to0}(\frac{cos(x+h)-cosx}{h})$

$=e^{cosx}\lim_{h\to0}(\frac{-2sin(\frac{x+h+x}{2})sin(\frac{x+h-x}{2})}{h})$

$=e^{cosx}\lim_{h\to0}\frac{-2sin(\frac{2x+h}{2})}{2}\frac{sin\frac{h}{2}}{\frac{h}{2}}$

$=e^{cosx}\lim_{h\to0}\frac{-2sin(\frac{2x+h}{2})}{2}\frac{1}{2}$

=e^cosx(-sinx)

=-sinx.e^cosx

问题5.将以下功能与第一原理^{e√2x区分开来}

解决方案：

We have,

Let,

f(x)=e^√2x

f(x+h)=e^√2(x+h)

$\frac{d}{dx}f(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$

$=\lim_{h\to0}\frac{e^{\sqrt{2(x+h)}}-e^{\sqrt{2x}}}{h}$

$=\lim_{h\to0}[e^{\sqrt{2x}}(\frac{e^{\sqrt{2(x+h)}-\sqrt{2x}}-1}{h})]$

$=e^{\sqrt{2x}}\lim_{h\to0}(\frac{e^{\sqrt{2(x+h)}-\sqrt{2x}}-1}{\sqrt{2(x+h)}-\sqrt{2x}})(\frac{\sqrt{2(x+h)}-\sqrt{2x}}{h})$

$=e^{\sqrt{2x}}\lim_{h\to0}(\frac{\sqrt{2(x+h)}-\sqrt{2x}}{h})$

$=e^{\sqrt{2x}}\lim_{h\to0}(\frac{2(x+h)-2x}{h(\sqrt{2(x+h)}-\sqrt{2x})})$   (After rationalising the numerator)

$=e^{\sqrt{2x}}\lim_{h\to0}(\frac{2x+2h-2x}{h(\sqrt{2(x+h)}-\sqrt{2x})})$

$=e^{\sqrt{2x}}\lim_{h\to0}(\frac{2h}{h(\sqrt{2(x+h)}-\sqrt{2x})})$

$=e^{\sqrt{2x}}\lim_{h\to0}(\frac{2}{(\sqrt{2(x+h)}-\sqrt{2x})})$

$=\frac{e^{\sqrt{2x}}}{\sqrt{2x}}$

问题6.将以下功能与第一原理log(cosx)区别开

解决方案：

We have,

Let,

f(x)=log(cosx)

f(x+h)=log(cos(x+h))

$\frac{d}{dx}f(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$

$=\lim_{h\to0}\frac{log(cos(x+h))-log(cosx)}{h}$

$=\lim_{h\to0}\frac{log\frac{cos(x+h)}{cosx}}{h}$

$=\lim_{h\to0}\frac{log[1+\frac{cos(x+h)}{cosx}-1]}{h}$

$=\lim_{h\to0}\frac{log[1+\frac{cos(x+h)}{cosx}]}{\frac{cos(x+h)-cosx}{cosx}}×\lim_{h\to0}\frac{cos(x+h)-cosx}{cosx}$

$=1×\lim_{h\to0}\frac{cos(x+h)-cosx}{h×cosx}$

$=\lim_{h\to0}\frac{-2sin(\frac{x+h+x}{2})sin(\frac{x+h-x}{2})}{h×cosx}$

$=-2\lim_{h\to0}\frac{sin(\frac{2x+h}{2})sin(\frac{h}{2})}{2(\frac{h}{2})×cosx}$

Since, $\lim_{h\to0}\frac{sinx}{x}=1$

=-(2sinx)/(2cosx)

=-tanx

问题7.将以下功能与第一原则^{e√cotx区别开}

解决方案：

We have,

Let,

f(x)=e^√cotx

f(x+h)=e^√cot(x+h)

$\frac{d}{dx}f(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$

$=\lim_{h\to0}\frac{e^{\sqrt{cot(x+h)}}-e^{\sqrt{cotx}}}{h}$

$=\lim_{h\to0}[e^{\sqrt{cotx}}(\frac{e^{\sqrt{cot(x+h)}-\sqrt{cotx}}-1}{h})]$

$=e^{\sqrt{cotx}}\lim_{h\to0}(\frac{e^{\sqrt{cot(x+h)}-\sqrt{cotx}}-1}{\sqrt{cot(x+h)}-\sqrt{cotx}})(\frac{\sqrt{cot(x+h)}-\sqrt{cotx}}{h})$

$=e^{\sqrt{cotx}}\lim_{h\to0}(\frac{\sqrt{cot(x+h)}-\sqrt{cotx}}{h})$

since, $\lim_{h\to0}\frac{e^x-1}{x}=1$

$=e^{\sqrt{cotx}}\lim_{h\to0}{\frac{cot(x+h)-cotx}{h(\sqrt{cot(x+h)}-\sqrt{cotx})}}$  (After rationalising the numerator)

$=e^{\sqrt{cotx}}\lim_{h\to0}(\frac{\frac{cot(x+h)cotx+1}{cot(x-x-h)}}{h(\sqrt{cot(x+h)}-\sqrt{cotx})})$

$=e^{\sqrt{cotx}}\lim_{h\to0}(\frac{cot(x+h)cotx+1}{hcot(-h)(\sqrt{cot(x+h)}-\sqrt{cotx})})$

$=-e^{\sqrt{cotx}}\lim_{h\to0}(\frac{cot(x+h)cotx+1}{\frac{h}{tanh}(\sqrt{cot(x+h)}-\sqrt{cotx})})$

Since, $\lim_{h\to0}\frac{tanx}{x}=1$

$=\frac{e^{\sqrt{cotx}}×(cot^2x+1)}{2\sqrt{cotx}}$

$=\frac{e^{\sqrt{cotx}}×cosec^2x}{2\sqrt{cotx}}$

问题8.将以下功能与第一原理x ² e ^x区别开

解决方案：

We have,

Let,

f(x)=x²e^x

f(x+h)=(x+h)²e^(x+h)

$\frac{d}{dx}f(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$

$=\lim_{h\to0}\frac{(x+h)^2e^{(x+h)}-x^2e^x}{h}$

$=\lim_{h\to0}(\frac{x^2e^{x+h}-x^2e^x}{h}+\frac{2hxe^{x+h}}{h}+\frac{h^2e^{x+h}}{h})$

$=\lim_{h\to0}(\frac{x^2e^x(e^h-1)}{h}+2xe^{x+h}+{he^{x+h}})$

Since, $\frac{e^h-1}{h}=1$

=x²e^x+2xe^x+0

=e^x(x²+2x)

问题9.将以下功能与第一原理日志(cosecx)区别开

解决方案：

We have,

Let,

f(x)=log(cosecx)

f(x+h)=log(cosec(x+h))

$\frac{d}{dx}f(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$

$=\lim_{h\to0}\frac{log(cosec(x+h))-log(cosecx)}{h}$

$=\lim_{h\to0}\frac{log\frac{cosec(x+h)}{cosecx}}{h}$

$=\lim_{h\to0}\frac{log[1+\frac{cosec(x+h)}{cosecx}-1]}{h}$

$=\lim_{h\to0}\frac{log[1+\frac{sinx}{sin(x+h)}-1]}{h}$

$=\lim_{h\to0}\frac{log[1+\frac{sinx-sin(x+h)}{sin(x+h)}]}{\frac{sinx-sin(x+h)}{sin(x+h)}}×\lim_{h\to0}\frac{\frac{sinx-sin(x+h)}{sin(x+h)}}{h}$

$=\lim_{h\to0}\frac{sinx-sin(x+h)}{sin(x+h)}$

$=\lim_{h\to0}\frac{2cos(\frac{x+x+h}{2})sin(\frac{x-x-h}{2})}{hsin(x+h)}$

$=\lim_{h\to0}\frac{2cos(\frac{2x+h}{2})sin(\frac{x-x-h}{2})}{sin(x+h)}\frac{sin(-\frac{h}{2})}{-\frac{h}{2}.(-2)}$

=-cotx

问题10.将以下功能与第一原理sin ^-1 (2x + 3)区别开

解决方案：

We have,

Let,

f(x)=sin^-1(2x+3)

f(x+h)=sin^-1[2(x+h)+3]

f(x+h)=sin^-1(2x+2h+3)

$\frac{d}{dx}f(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$

$=\lim_{h\to0}\frac{sin^{-1}(2x+2h+3)-sin^{-1}(2x+3)}{h}$

$=\lim_{h\to0}\frac{sin^{-1}[(2x+2h+3)\sqrt{1-(2x+3)^2}-(2x+3)\sqrt{1-(2x+2h+3)^2}]}{h}$

$=\lim_{h\to0}\frac{sin^{-1}t}{t}\frac{t}{h}$

Where $t=[(2x+2h+3)\sqrt{1-(2x+3)^2}-(2x+3)\sqrt{1-(2x+2h+3)^2}]$

$=\lim_{h\to0}\frac{t}{h}$

$=\lim_{h\to0}\frac{[(2x+2h+3)\sqrt{1-(2x+3)^2}-(2x+3)\sqrt{1-(2x+2h+3)^2}]}{h}$

$=\lim_{h\to0}\frac{[(2x+2h+3)^2[1-(2x+3)^2]-(2x+3)^2[1-(2x+2h+3)^2]}{h[(2x+2h+3)\sqrt{1-(2x+3)^2}+(2x+3)\sqrt{1-(2x+2h+3)^2]}}$      (After rationalising the numerator)

Solving above equation

$=\lim_{h\to0}\frac{4h[h+(2x+3)]}{h[(2x+2h+3)\sqrt{1-(2x+3)^2}+(2x+3)\sqrt{1-(2x+2h+3)^2]}}$

$=\frac{4(2x+3)}{[(2x+3)\sqrt{1-(2x+3)^2}+(2x+3)\sqrt{1-(2x+3)^2]}}$

$=\frac{4(2x+3)}{2[(2x+3)\sqrt{1-(2x+3)^2}]}$

$=\frac{2}{2[\sqrt{1-(2x+3)^2}]}$