第12类RD Sharma解决方案–第11章求差–练习11.3 |套装2

问题17.区分$y=tan^{-1}\left(\frac{2^{x+1}}{1-4^x}\right)$ ，相对于x，-∞＜x ＜0。

解决方案：

We have,$y=tan^{-1}\left(\frac{2^{x+1}}{1-4^x}\right)$, −∞ < x < 0

On putting 2^x = tan θ, we get,

$y=tan^{-1}\left(\frac{2tanθ}{1-tan^2θ}\right)$

=$tan^{-1}\left(tan2θ\right)$

Now, −∞ < x < 0

=> 0 < 2^x < 1

=> 0 < θ < π/4

=> 0 < 2θ < π/2

So, y = 2θ

= 2 tan⁻¹ (2^x)

Differentiating with respect to x, we get,

$\frac{dy}{dx}=\frac{d}{dx}\left(2 tan^{−1} (2^x)\right)$

=$\frac{2.2^xlog2}{1+(2^x)^2}$

=$\frac{2^{x+1}log2}{1+4^x}$

问题18.区分$y=tan^{-1}\left(\frac{2a^{x}}{1-a^{2x}}\right)$ ，相对于x，a> 1，-∞

解决方案：

We have,$y=tan^{-1}\left(\frac{2a^{x}}{1-a^{2x}}\right)$, −∞ < x < 0

On putting a^x = tan θ, we get,

$y=tan^{-1}\left(\frac{2tanθ}{1-tan^2θ}\right)$

=$tan^{-1}\left(tan2θ\right)$

Now, −∞ < x < 0

=> 0 < a^x < 1

=> 0 < θ < π/4

=> 0 < 2θ < π/2

So, y = 2θ

= 2 tan⁻¹ (a^x)

Differentiating with respect to x, we get,

$\frac{dy}{dx}=\frac{d}{dx}\left(2 tan^{−1} (a^x)\right)$

=$\frac{2a^xloga}{1+(a^x)^2}$

=$\frac{2a^{x}loga}{1+a^{2x}}$

问题19.区分$y=sin^{-1}\left(\frac{\sqrt{1+x}+\sqrt{1-x}}{2}\right)$ ，相对于x，0

解决方案：

We have,$y=sin^{-1}\left(\frac{\sqrt{1+x}+\sqrt{1-x}}{2}\right)$, 0 < x < 1

On putting x = cos 2θ, we get,

$y=sin^{-1}\left(\frac{\sqrt{1+cos2θ}+\sqrt{1-cos2θ}}{2}\right)$

=$sin^{-1}\left(\frac{\sqrt{2cos^2θ}+\sqrt{2sin^2θ}}{2}\right)$

=$sin^{-1}\left(\frac{\sqrt{2}cosθ+\sqrt{2}sinθ}{2}\right)$

=$sin^{-1}\left(cosθ\frac{1}{\sqrt{2}}+sinθ\frac{1}{\sqrt{2}}\right)$

=$sin^{-1}\left(sin(θ+\frac{π}{4})\right)$

Now, 0 < x < 1

=> 0 < cos 2θ < 1

=> 0 < 2θ < π/2

=> 0 < θ < π/4

=> π/4 < (θ+π/4) < π/2

So, y =$θ+\frac{π}{4}$

=$\frac{1}{2}cos^{-1}x+\frac{π}{4}$

Differentiating with respect to x, we get,

$\frac{dy}{dx}=\frac{d}{dx}\left(\frac{1}{2}cos^{-1}x+\frac{π}{4}\right)$

=$\frac{1}{2}(\frac{-1}{\sqrt{1-x^2}})+0$

=$\frac{-1}{2\sqrt{1-x^2}}$

问题20.区分$y=tan^{-1}\left(\frac{\sqrt{1+a^2x^2}-1}{ax}\right)$ ，相对于x ，x ≠0。

解决方案：

We have,$y=tan^{-1}\left(\frac{\sqrt{1+a^2x^2}-1}{ax}\right)$

On putting ax = tan θ, we get,

$y=tan^{-1}\left(\frac{\sqrt{1+tan^2θ}-1}{tanθ}\right)$

=$tan^{-1}\left(\frac{secθ-1}{tanθ}\right)$

=$tan^{-1}\left(\frac{1-cosθ}{sinθ}\right)$

=$tan^{-1}\left(\frac{2sin^2\frac{θ}{2}}{2sin\frac{θ}{2}cos\frac{θ}{2}}\right)$

=$tan^{-1}\left(tan\frac{θ}{2}\right)$

=$\frac{θ}{2}$

=$\frac{1}{2}tan^{-1}\left(ax\right)$

Differentiating with respect to x, we get,

$\frac{dy}{dx}=\frac{d}{dx}\left(\frac{1}{2}tan^{-1}\left(ax\right)\right)$

=$\frac{a}{2(1+a^2x^2)}$

问题21.区分$y=tan^{-1}\left(\frac{sinx}{1+cosx}\right)$ ，相对于x，-π＜x ＜π。

解决方案：

We have,$y=tan^{-1}\left(\frac{sinx}{1+cosx}\right)$, −π < x < π

=$tan^{-1}\left(\frac{2sin\frac{x}{2}cos\frac{x}{2}}{2cos^2\frac{x}{2}}\right)$

=$tan^{-1}\left(tan\frac{x}{2}\right)$

=$\frac{x}{2}$

Differentiating with respect to x, we get,

$\frac{dy}{dx}=\frac{d}{dx}\left(\frac{x}{2}\right)$

=$\frac{1}{2}$

问题22.区分$y=sin^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right)$关于x。

解决方案：

We have,$y=sin^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right)$

On putting x = cot θ, we get,

$y=sin^{-1}\left(\frac{1}{\sqrt{1+cot^2θ}}\right)$

=$sin^{-1}\left(\frac{1}{cosecθ}\right)$

=$sin^{-1}\left(sinθ\right)$

= θ

= cot⁻¹ x

Differentiating with respect to x, we get,

$\frac{dy}{dx}=\frac{d}{dx}\left(cot^{-1}x\right)$

=$\frac{-1}{1+x^2}$

问题23.区分$y=cos^{-1}\left(\frac{1-x^{2n}}{1+x^{2n}}\right)$ ，相对于x，0

解决方案：

We have,$y=cos^{-1}\left(\frac{1-x^{2n}}{1+x^{2n}}\right)$,0 < x < ∞

On putting xⁿ = tan θ, we get,

$y=cos^{-1}\left(\frac{1-tan^{2}θ}{1+tan^{2}θ}\right)$

=$cos^{-1}\left(cos2θ\right)$

Now, 0 < x < ∞

=> 0 < xⁿ < ∞

=> 0 < θ < π/2

=> 0 < 2θ < π

So, y = 2θ

= 2 tan^–1 (xⁿ)

Differentiating with respect to x, we get,

$\frac{dy}{dx}=\frac{d}{dx}\left(2 tan^{-1}(x^n)\right)$

=$\frac{2}{1+x^{2n}}×(nx^{n-1})$

=$\frac{2nx^{n-1}}{1+x^{2n}}$

问题24.区分$y=sin^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)+sec^{-1}\left(\frac{1+x^{2}}{1-x^{2}}\right)$ ，相对于x，x∈R。

解决方案：

We have,$y=sin^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)+sec^{-1}\left(\frac{1+x^{2}}{1-x^{2}}\right)$

=$sin^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)+cos^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)$

=$\frac{π}{2}$

Differentiating with respect to x, we get,

$\frac{dy}{dx}=\frac{d}{dx}(\frac{π}{2})$

= 0

问题25.区分$y=tan^{-1}\left(\frac{a+x}{1-ax}\right)$关于x。

解决方案：

We have,$y=tan^{-1}\left(\frac{a+x}{1-ax}\right)$

=$tan^{-1}a+tan^{-1}x$

Differentiating with respect to x, we get,

$\frac{dy}{dx}=\frac{d}{dx}(tan^{-1}a+tan^{-1}x)$

= 0 +$\frac{1}{1+x^2}$

=$\frac{1}{1+x^2}$

问题26.区分$y=tan^{-1}\left(\frac{\sqrt{x}+\sqrt{a}}{1-\sqrt{xa}}\right)$关于x。

解决方案：

We have,$y=tan^{-1}\left(\frac{\sqrt{x}+\sqrt{a}}{1-\sqrt{xa}}\right)$

=$tan^{-1}\sqrt{x}+tan^{-1}\sqrt{a}$

Differentiating with respect to x, we get,

$\frac{dy}{dx}=\frac{d}{dx}\left(tan^{-1}\sqrt{x}+tan^{-1}\sqrt{a}\right)$

=$\frac{1}{1+(\sqrt{x})^2}×\frac{1}{2\sqrt{x}} + 0$

=$\frac{1}{2\sqrt{x(}1+x)}$

问题27.区分$y=tan^{-1}\left(\frac{a+btanx}{b-atanx}\right)$关于x。

解决方案：

We have,$y=tan^{-1}\left(\frac{a+btanx}{b-atanx}\right)$

=$tan^{-1}\left(\frac{\frac{a+btanx}{b}}{\frac{b-atanx}{b}}\right)$

=$tan^{-1}\left(\frac{\frac{a}{b}+tanx}{1-\frac{atanx}{b}}\right)$

=$tan^{-1}(\frac{a}{b})+tan^{-1}(tanx)$

=$tan^{-1}(\frac{a}{b})+x$

Differentiating with respect to x, we get,

$\frac{dy}{dx}=\frac{d}{dx}\left(tan^{-1}(\frac{a}{b})+x\right)$

= 0 + 1

= 1

问题28.区分$y=tan^{-1}\left(\frac{a+bx}{b-ax}\right)$关于x。

解决方案：

We have,$y=tan^{-1}\left(\frac{a+bx}{b-ax}\right)$

=$tan^{-1}\left(\frac{\frac{a+bx}{b}}{\frac{b-ax}{b}}\right)$

=$tan^{-1}\left(\frac{\frac{a}{b}+x}{1-\frac{ax}{b}}\right)$

=$tan^{-1}(\frac{a}{b})+tan^{-1}x$

Differentiating with respect to x, we get,

$\frac{dy}{dx}=\frac{d}{dx}\left(tan^{-1}(\frac{a}{b})+tan^{-1}x\right)$

= 0 +$\frac{1}{1+x^2}$

=$\frac{1}{1+x^2}$

问题29.区分$y=tan^{-1}\left(\frac{x-a}{x+a}\right)$关于x。

解决方案：

We have,$y=tan^{-1}\left(\frac{x-a}{x+a}\right)$

=$tan^{-1}\left(\frac{\frac{x-a}{a}}{\frac{x+a}{a}}\right)$

=$tan^{-1}\left(\frac{\frac{x}{a}-1}{1+\frac{x}{a}}\right)$

=$tan^{-1}(\frac{x}{a})-tan^{-1}1$

Differentiating with respect to x, we get,

$\frac{dy}{dx}=\frac{d}{dx}\left(tan^{-1}(\frac{x}{a})-tan^{-1}1\right)$

=$\frac{1}{a(1+\frac{x^2}{a^2})}$

=$\frac{a^2}{a(a^2+x^2)}$

=$\frac{a}{a^2+x^2}$

问题30.区分$y=tan^{-1}\left(\frac{x}{1+6x^2}\right)$关于x。

解决方案：

We have,$y=tan^{-1}\left(\frac{x}{1+6x^2}\right)$

=$tan^{-1}\left(\frac{3x-2x}{1+(3x)(2x)}\right)$

=$tan^{-1}(3x)-tan^{-1}(2x)$

Differentiating with respect to x, we get,

$\frac{dy}{dx}=\frac{d}{dx}\left((tan^{-1}(3x)-tan^{-1}(2x)\right)$

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

=$\frac{3}{1+9x^2}-\frac{2}{1+4x^2}$

问题31.区分$y=tan^{-1}\left(\frac{5x}{1-6x^2}\right)$关于x。

解决方案：

We have,$y=tan^{-1}\left(\frac{5x}{1-6x^2}\right)$

=$tan^{-1}\left(\frac{3x+2x}{1-(3x)(2x)}\right)$

=$tan^{-1}(3x)+tan^{-1}(2x)$

Differentiating with respect to x, we get,

$\frac{dy}{dx}=\frac{d}{dx}\left((tan^{-1}(3x)+tan^{-1}(2x)\right)$

=$\frac{3}{1+(3x)^2}+\frac{2}{1+(2x)^2}$

=$\frac{3}{1+9x^2}+\frac{2}{1+4x^2}$

问题32.区分$y=tan^{-1}\left(\frac{cosx+sinx}{cosx-sinx}\right)$相对于x为-π/ 4 ＜x ＜π/ 4。

解决方案：

We have,$y=tan^{-1}\left(\frac{cosx+sinx}{cosx-sinx}\right)$, −π/4 < x < π/4

=$tan^{-1}\left(\frac{\frac{cosx+sinx}{cosx}}{\frac{cosx-sinx}{cosx}}\right)$

=$tan^{-1}\left(\frac{1+tanx}{1-tanx}\right)$

=$tan^{-1}\left(\frac{tan\frac{π}{4}+tanx}{1-tan\frac{π}{4}tanx}\right)$

=$tan^{-1}\left(tan(\frac{π}{4}+x)\right)$

=$\frac{π}{4}+x$

Differentiating with respect to x, we get,

$\frac{dy}{dx}=\frac{d}{dx}\left(\frac{π}{4}+x\right)$

= 0 + 1

= 1