第12类RD Sharma解决方案–第11章求差–练习11.3 |套装1

问题1.区分$y=cos^{-1}(2x\sqrt{1-x^2})$ ，相对于x，1 /√2

解决方案：

We have,

$y=cos^{-1}(2x\sqrt{1-x^2})$, 1/√2 < x < 1.

On putting x = cos θ, we get,

y =$cos^{-1}(2cosθ\sqrt{1-cos^2θ})$

=$cos^{-1}(2cosθ\sqrt{sin^2θ})$

= cos⁻¹(2cos θ sin θ)

= cos⁻¹(sin 2θ)

=$cos^{-1}(cos(\frac{π}{2}-2θ))$

Now, 1/√2 < x < 1

=> 1/√2 < cos θ < 1

=> 0 < θ < π/4

=> 0 < 2θ < π/2

=> 0 > −2θ > −π/2

=> π/2 > (π/2−2θ) > 0

So, y =$\frac{π}{2}-2cos^{-1}x$

Differentiating with respect to x, we get,

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

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

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

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

解决方案：

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

On putting x = cos 2θ, we get,

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

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

=$cos^{-1}\left(\sqrt{cos^2θ}\right)$

=$cos^{-1}\left(cosθ\right)$

Now, −1 < x < 1

=> −1 < cos 2θ < 1

=> 0 < 2θ < π

=> 0 < θ < π/2

So, y =$\frac{1}{2}cos^{-1}x$

Differentiating with respect to x, we get,

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

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

问题3.区分$y=sin^{-1}\left(\sqrt{\frac{1-x}{2}}\right)$ ，相对于x，0

解决方案：

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

On putting x = cos 2θ, we get,

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

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

=$sin^{-1}\left(sinθ\right)$

Now, 0 < x < 1

=> 0 < cos 2θ < 1

=> 0 < 2θ < π/2

=> 0 < θ < π/4

So,$y = \frac{1}{2}cos^{-1}x$

Differentiating with respect to x, we get,

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

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

问题4 。区分$y=sin^{-1}(\sqrt{1-x^2})$ ，相对于x，0

解决方案：

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

On putting x = cos θ, we get,

y =$sin^{-1}(\sqrt{1-cos^2θ})$

=$sin^{-1}(\sqrt{sin^2θ})$

=$sin^{-1}\left(sinθ\right)$

Now, 0 < x < 1

=> 0 < cos θ < 1

=> 0 < θ < π/2

So, y = cos⁻¹x

Differentiating with respect to x, we get,

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

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

问题5.区分$y=tan^{-1}\left(\frac{x}{\sqrt{a^2-x^2}}\right)$ ，相对于x，-a ＜x ＜a。

解决方案：

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

On putting x = a sin θ, we get,

y =$tan^{-1}\left(\frac{asinθ}{\sqrt{a^2-a^2sin^2θ}}\right)$

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

=$tan^{-1}\left(\frac{asinθ}{acosθ}\right)$

=$tan^{-1}\left(tanθ\right)$

Now, −a < x < a

=> −1 < x/a < 1

=> −π/2 < θ < π/2

So,$y=sin^{-1}(\frac{x}{a})$

Differentiating with respect to x, we get,

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

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

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

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

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

解决方案：

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

On putting x = a tan θ, we get,

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

=$sin^{-1}\left(\frac{atanθ}{\sqrt{a^2sec^2θ}}\right)$

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

=$sin^{-1}(sinθ)$

= θ

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

Differentiating with respect to x, we get,

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

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

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

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

问题7.区分$y=sin^{-1}\left(2x^2-1\right)$ ，相对于x，0

解决方案：

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

On putting x = cos θ, we get,

y =$sin^{-1}\left(2cos^2θ-1\right)$

=$sin^{-1}\left(cos2θ\right)$

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

Now, 0 < x < 1

=> 0 < cos θ < 1

=> 0 < θ < π/2

=> 0 < 2θ < π

=> π/2 > (π/2−2θ) > −π/2

So, y =$\frac{π}{2}-2cos^{-1}x$

Differentiating with respect to x, we get,

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

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

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

问题8.区分$y=sin^{-1}\left(1-2x^2\right)$ ，相对于x，0

解决方案：

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

On putting x = sin θ, we get,

y =$sin^{-1}\left(1-2sin^2θ\right)$

=$sin^{-1}\left(cos2θ\right)$

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

Now, 0 < x < 1

=> 0 < sin θ < 1

=> 0 < θ < π/2

=> 0 < 2θ < π

=> π/2 > (π/2−2θ) > −π/2

So, y =$\frac{π}{2}-2sin^{-1}x$

Differentiating with respect to x, we get,

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

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

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

问题9.区分$y=cos^{-1}\left(\frac{x}{\sqrt{a^2+x^2}}\right)$关于x。

解决方案：

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

Putting x = cot θ, we get,

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

=$cos^{-1}\left(\frac{acotθ}{\sqrt{a^2cosec^2θ}}\right)$

=$cos^{-1}\left(\frac{acotθ}{acosecθ}\right)$

=$cos^{-1}(cosθ)$

= θ

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

Differentiating with respect to x, we get,

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

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

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

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

问题10.区分$y=sin^{-1}\left(\frac{sinx+cosx}{\sqrt{2}}\right)$相对于x为-3π/ 4 ＜x ＜π/ 4。

解决方案：

We have,$y=sin^{-1}\left(\frac{sinx+cosx}{\sqrt{2}}\right)$, −3π/4 < x < π/4

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

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

Now, −3π/4 < x < π/4

=> −π/2 < (x+π/4) < π/2

So, y =$x+\frac{π}{4}$

Differentiating with respect to x, we get,

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

= 1 + 0

= 1

问题11.区分$y=cos^{-1}\left(\frac{sinx+cosx}{\sqrt{2}}\right)$相对于x为-π/ 4 ＜x ＜π/ 4。

解决方案：

We have,$y=cos^{-1}\left(\frac{sinx+cosx}{\sqrt{2}}\right)$, −π/4 < x < π/4

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

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

Now, −π/4 < x < π/4

=> −π/2 < (x−π/4) < 0

So, y =$-(x-\frac{π}{4})$

=$-x+\frac{π}{4}$

Differentiating with respect to x, we get,

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

= −1 + 0

= −1

问题12.区分$y=tan^{-1}\left(\frac{x}{1+\sqrt{1-x^2}}\right)$ ，相对于x，-1 ＜x ＜1。

解决方案：

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

On putting x = sin θ, we get,

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

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

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

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

Now, −1 < x < 1

=> −1 < sin θ < 1

=> −π/2 < θ < π/2

=> −π/4 < θ/2 < π/4

So, y =$\frac{1}{2}sin^{-1}x$

Differentiating with respect to x, we get,

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

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

问题13.区分$y=tan^{-1}\left(\frac{x}{a+\sqrt{a^2-x^2}}\right)$ ，相对于x，-a ＜x ＜a。

解决方案：

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

On putting x = a sin θ, we get,

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

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

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

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

Now, −a < x < a

=> −1 < x/a < 1

=> −π/2 < θ < π/2

=> −π/4 < θ/2 < π/4

So, y =$\frac{1}{2}sin^{-1}(\frac{x}{a})$

Differentiating with respect to x, we get,

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

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

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

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

问题14.区分$y=sin^{-1}\left(\frac{x+\sqrt{1-x^2}}{\sqrt{2}}\right)$ ，相对于x，-1 ＜x ＜1。

解决方案：

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

On putting x = sin θ, we get,

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

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

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

Now, −1 < x < 1

=> −1 < sin θ < 1

=> −π/2 < θ < π/2

=> −π/2 < (θ+π/4) < 3π/4

So, y =$sin^{-1}x+\frac{π}{4}$

Differentiating with respect to x, we get,

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

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

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

问题15.区分$y=cos^{-1}\left(\frac{x+\sqrt{1-x^2}}{\sqrt{2}}\right)$ ，相对于x，-1 ＜x ＜1。

解决方案：

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

On putting x = sin θ, we get,

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

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

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

Now, −1 < x < 1

=> −1 < sin θ < 1

=> −π/2 < θ < π/2

=> −3π/4 < (θ−π/4) < π/4

So, y =$-(sin^{-1}x-\frac{π}{4})$

=$-sin^{-1}x+\frac{π}{4}$

Differentiating with respect to x, we get,

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

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

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

问题16.区分$y=tan^{-1}\left(\frac{4x}{1-4x^{2}}\right)$ ，相对于x为-1/2 ＜x ＜1/2。

解决方案：

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

On putting 2x = tan θ, we get,

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

=$tan^{-1}\left(tan2θ\right)$

Now, −1/2 < x < 1/2

=> −1 < 2x < 1

=> −1 < tan θ < 1

=> −π/4 < θ < π/4

=> −π/2 < 2θ < π/2

Therefore, y = 2 tan⁻¹ (2x)

Differentiating with respect to x, we get,

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

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

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