第12类RD Sharma解决方案–第11章求差–练习11.3 |套装3

问题33.区分$y=tan^{-1}\left(\frac{x^{\frac{1}{3}}+a^\frac{1}{3}}{1-(ax)^{\frac{1}{3}}}\right)$关于x。

解决方案：

We have,$y=tan^{-1}\left(\frac{x^{\frac{1}{3}}+a^\frac{1}{3}}{1-(ax)^{\frac{1}{3}}}\right)$

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

Differentiating with respect to x, we get,

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

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

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

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

问题34.区分$y=sin^{-1}\left(\frac{2^{x+1}}{1+4^x}\right)$关于x。

解决方案：

We have,$y=sin^{-1}\left(\frac{2^{x+1}}{1+4^x}\right)$

On putting 2^x = tan θ, we get,

$y=sin^{-1}\left(\frac{2^{x+1}}{1+4^x}\right)$

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

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

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

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

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

=$sin^{-1}\left(2sinθcosθ\right)$

=$sin^{-1}\left(sin2θ\right)$

= 2θ

= 2 tan⁻¹ (2^x)

Differentiating with respect to x, we get,

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

=$\frac{2×2^xlog2}{1+(2^x)^2}$

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

问题35.如果$y=sin^{-1}\left(\frac{2x}{1+x^2}\right)+sec^{-1}\left(\frac{1+x^2}{1-x^2}\right)$ ，0 $\frac{dy}{dx}=\frac{4}{1+x^2}$ 。

解决方案：

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

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

On putting x = tan θ, we get,

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

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

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

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

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

=$sin^{-1}\left(sin2θ\right)+cos^{-1}\left(cos2θ\right)$

Now, 0 < x < 1

=> 0 < tan θ < 1

=> 0 < θ < π/4

=> 0 < 2θ < π/2

So, y = 2θ + 2θ

= 4θ

= 4 tan⁻¹ x

Now, L.H.S. =$\frac{dy}{dx} = \frac{d}{dx}\left(4tan^{−1}x\right)$

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

= R.H.S.

Hence proved.

问题36.如果$y=sin^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right)+cos^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right)$ ，0 $\frac{dy}{dx}=\frac{2}{1+x^2}$ 。

解决方案：

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

On putting x = tan θ, we get,

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

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

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

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

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

Now, 0 < x < ∞

=> 0 < tan θ < ∞

=> 0 < θ < π/2

So, y = θ + θ

= 2θ

= 2 tan⁻¹ x

Now, L.H.S. =$\frac{dy}{dx} = \frac{d}{dx}\left(2tan^{−1}x\right)$

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

= R.H.S.

Hence proved.

问题37关于x区分以下几点：

(i)cos ^-1 (sin x)

解决方案：

We have, y = cos⁻¹ (sin x)

=$cos^{−1}(cos(\frac{π}{2}-x))$

=$\frac{π}{2}-x$

Differentiating with respect to x, we get,

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

= 0 − 1

= −1

(ii) $cot^{−1}\left(\frac{1-x}{1+x}\right)$

解决方案：

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

On putting x = tan θ, we get,

$y=cot^{−1}\left(\frac{1-tanθ}{1+tanθ}\right)$

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

=$cot^{−1}\left(tan(\frac{π}{4}-θ)\right)$

=$cot^{−1}\left(cot(\frac{π}{2}-\frac{π}{4}+θ)\right)$

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

=$\frac{π}{4}+θ$

=$\frac{π}{4}+tan^{-1}x$

Differentiating with respect to x, we get,

$\frac{dy}{dx} = \frac{d}{dx}(\frac{π}{4}+tan^{-1}x)$

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

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

问题38.区分$y=cot^{-1}\left[\frac{\sqrt{1+sinx}+\sqrt{1-sinx}}{\sqrt{1+sinx}-\sqrt{1-sinx}}\right]$ ，相对于x ，0 π/ 2。

解决方案：

We have,$y=cot^{-1}\left[\frac{\sqrt{1+sinx}+\sqrt{1-sinx}}{\sqrt{1+sinx}-\sqrt{1-sinx}}\right]$

=$cot^{-1}\left[\frac{(\sqrt{1+sinx}+\sqrt{1-sinx})^2}{(\sqrt{1+sinx}-\sqrt{1-sinx})(\sqrt{1+sinx}+\sqrt{1-sinx})}\right]$

=$cot^{-1}\left[\frac{1+sinx+1-sinx+2(\sqrt{1+sinx})(\sqrt{1-sinx})}{1+sinx-1+sinx}\right]$

=$cot^{-1}\left[\frac{2+2(\sqrt{1-sin^2x})}{2sinx}\right]$

=$cot^{-1}\left[\frac{2+2cosx}{2sinx}\right]$

=$cot^{-1}\left[\frac{1+cosx}{sinx}\right]$

=$cot^{-1}\left[\frac{2cos^2\frac{x}{2}}{2sin\frac{x}{2}cos\frac{x}{2}}\right]$

=$cot^{-1}\left[cot\frac{x}{2}\right]$

=$\frac{x}{2}$

Differentiating with respect to x, we get,

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

=$\frac{1}{2}$

问题39.如果$y=tan^{-1}\left(\frac{2x}{1-x^2}\right)+sec^{-1}\left(\frac{1+x^2}{1-x^2}\right)$ ，x> 0，证明$\frac{dy}{dx}=\frac{4}{1+x^2}$ 。

解决方案：

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

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

On putting x = tan θ, we get,

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

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

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

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

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

=$tan^{-1}\left(\frac{\frac{2sinθ}{cosθ}}{\frac{cos2θ}{cos^2θ}}\right)+cos^{-1}\left(\frac{\frac{cos2θ}{cos^2θ}}{\frac{1}{cos^2θ}}\right)$

=$tan^{-1}\left(\frac{2sinθcosθ}{cos2θ}\right)+cos^{-1}\left(cos2θ\right)$

=$tan^{-1}\left(\frac{sin2θ}{cos2θ}\right)+cos^{-1}\left(cos2θ\right)$

=$tan^{-1}\left(tan2θ\right)+cos^{-1}\left(cos2θ\right)$

Here, 0 < x < ∞

=> 0 < tan θ < ∞

=> 0 < θ < π/2

=> 0 < 2θ < π

So, y = 2θ + 2θ

= 4θ

= 4 tan⁻¹ x

Now, L.H.S. =$\frac{dy}{dx} = \frac{d}{dx}\left(4tan^{−1}x\right)$

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

= R.H.S.

Hence proved.

问题40.如果$y=sec^{-1}\left(\frac{x+1}{x-1}\right)+sin^{-1}\left(\frac{x-1}{x+1}\right)$ ，x> 0，找到$\frac{dy}{dx}$ 。

解决方案：

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

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

=$\frac{π}{2}$

Differentiating with respect to x, we get,

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

= 0

问题41.如果$y=sin\left[2tan^{-1}\sqrt{(\frac{1-x}{1+x})}\right]$ ， 找$\frac{dy}{dx}$ 。

解决方案：

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

On putting x = cos 2θ, we get,

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

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

=$sin\left[2tan^{-1}(\sqrt{tan^2θ})\right]$

=$sin\left[2tan^{-1}(tanθ)\right]$

=$sin\left(2θ\right)$

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

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

=$sin\left(sin^{-1}(\sqrt{1-x^2})\right)$

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

Differentiating with respect to x, we get,

$\frac{dy}{dx} = \frac{d}{dx}(\sqrt{1-x^2})$

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

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

问题42.如果$y=cos^{-1}(2x)+2cos^{-1}(\sqrt{1-4x^2})$ ，0 $\frac{dy}{dx}$ 。

解决方案：

We have,$y=cos^{-1}(2x)+2cos^{-1}(\sqrt{1-4x^2})$

On putting 2x = cos θ, we get,

$y=cos^{-1}(cosθ)+2cos^{-1}(\sqrt{1-cos^2θ})$

=$cos^{-1}(cosθ)+2cos^{-1}(sinθ)$

=$cos^{-1}(cosθ)+2cos^{-1}(cos(\frac{π}{2}-θ))$

Now, 0 < x < 1/2

=> 0 < 2x < 1

=> 0 < cos θ < 1

=> 0 < θ < π/2

and 0 > −θ > −π/2

=> π/2 > (π/2 −θ) > 0

So, y =$θ+2(\frac{π}{2}-θ)$

= π − θ

= π − cos⁻¹ (2x)

Differentiating with respect to x, we get,

$\frac{dy}{dx} = \frac{d}{dx}(π−cos^{−1}(2x))$

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

=$\frac{2}{\sqrt{1-4x^2}}$

问题43.如果tan ^-1 (a + bx)的导数在x = 0时取值为1，则证明1 + a ² = b。

解决方案：

We have, y = tan⁻¹ (a + bx)

Differentiating with respect to x, we get,

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

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

At x = 0, we have,

=>$\frac{b}{1+(a+b(0))^2}$= 1

=>$\frac{b}{1+a^2}$= 1

=> 1 + a² = b

Hence proved.

问题44.如果$y=cos^{-1}(2x)+2cos^{-1}(\sqrt{1-4x^2})$ ，−1/2 $\frac{dy}{dx}$ 。

解决方案：

We have,$y=cos^{-1}(2x)+2cos^{-1}(\sqrt{1-4x^2})$

On putting 2x = cos θ, we get,

$y=cos^{-1}(cosθ)+2cos^{-1}(\sqrt{1-cos^2θ})$

=$cos^{-1}(cosθ)+2cos^{-1}(sinθ)$

=$cos^{-1}(cosθ)+2cos^{-1}(cos(\frac{π}{2}-θ))$

Now, −1/2 < x < 0

=> −1 < 2x < 0

=> −1 < cos θ < 0

=> π/2 < θ < π

and −π/2 > −θ > −π

=> 0 > (π/2 −θ) > −π/2

So, y =$θ+2(-\frac{π}{2}+θ)$

= −π + 3θ

= −π + 3 cos⁻¹ (2x)

Differentiating with respect to x, we get,

$\frac{dy}{dx} = \frac{d}{dx}(−π+3cos^{−1}(2x))$

= 0 +$\frac{-6}{\sqrt{1-(2x)^2}}$

=$\frac{-6}{\sqrt{1-4x^2}}$

问题45.如果$y=tan^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)$ ， 找$\frac{dy}{dx}$ 。

解决方案：

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

On putting x = cos 2θ, we get,

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

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

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

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

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

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

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

=$\frac{π}{4}-θ$

=$\frac{π}{4}-\frac{1}{2}cos^{-1}x$

Differentiating with respect to x, we get,

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

= 0 −$\left(\frac{-1}{2\sqrt{1-x^2}}\right)$

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

问题46.如果$y=cos^{-1}\left(\frac{2x-3\sqrt{1-x^2}}{\sqrt{13}}\right)$ ， 找$\frac{dy}{dx}$ 。

解决方案：

We have,$y=cos^{-1}\left(\frac{2x-3\sqrt{1-x^2}}{\sqrt{13}}\right)$

On putting x = cos θ, we get,

$y=cos^{-1}\left(\frac{2cosθ-3\sqrt{1-cos^2θ}}{\sqrt{13}}\right)$

=$cos^{-1}\left(\frac{2cosθ-3sinθ}{\sqrt{13}}\right)$

=$cos^{-1}\left(\frac{2}{\sqrt{13}}cosθ-\frac{3}{\sqrt{13}}sinθ\right)$

Let$cosØ=\frac{2}{\sqrt{13}}$

=> sin Ø =$\sqrt{1-cos^2Ø}$

=> sin Ø =$\sqrt{1-\left(\frac{2}{\sqrt{13}}\right)^2}$

=> sin Ø =$\sqrt{1-\frac{4}{13}}$

=> sin Ø =$\sqrt{\frac{9}{13}}$

=> sin Ø =$\frac{3}{\sqrt{13}}$

So, y =$cos^{-1}\left(cosØcosθ-sinØsinθ\right)$

=$cos^{-1}\left(cos(Ø+θ)\right)$

= Ø + θ

=$cos^{-1}\left(\frac{2}{\sqrt{13}}\right)+cos^{-1}x$

Differentiating with respect to x, we get,

$\frac{dy}{dx} = \frac{d}{dx}(cos^{-1}\left(\frac{2}{\sqrt{13}}\right)+cos^{-1}x)$

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

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

问题47.区分$y=sin^{-1}\left[\frac{2^{x+1}×3^x}{1+(36)^x}\right]$关于x。

解决方案：

We have,$y=sin^{-1}\left[\frac{2^{x+1}×3^x}{1+(36)^x}\right]$

=$sin^{-1}\left[\frac{2×2^x×3^x}{1+(36)^x}\right]$

=$sin^{-1}\left[\frac{2×6^x}{1+(6)^{2x}}\right]$

On putting 6^x = tan θ, we get,

=$sin^{-1}\left[\frac{2tanθ}{1+tan^2θ}\right]$

=$sin^{-1}\left[\frac{\frac{2sinθ}{cosθ}}{1+\frac{sin^2θ}{cos^2θ}}\right]$

=$sin^{-1}\left[\frac{\frac{2sinθ}{cosθ}}{\frac{cos^2θ+sin^2θ}{cos^2θ}}\right]$

=$sin^{-1}\left[\frac{\frac{2sinθ}{cosθ}}{\frac{1}{cos^2θ}}\right]$

=$sin^{-1}\left[{\frac{2sinθcos^2θ}{cosθ}}\right]$

=$sin^{-1}\left(2sinθcosθ\right)$

=$sin^{-1}\left(sin2θ\right)$

= 2θ

= 2 tan⁻¹ (6^x)

Differentiating with respect to x, we get,

$\frac{dy}{dx} = \frac{d}{dx}(2tan^{−1}(6x))$

=$\frac{2×6^xlog6}{1+(6^x)^2}$

=$\frac{2×6^xlog6}{1+36^x}$