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Class 12 RD Sharma Solutions – Chapter 11 Differentiation – Exercise 11.3 | Set 1

  • Last Updated : 08 May, 2021

Question 1. Differentiatey=cos^{-1}(2x\sqrt{1-x^2})  , 1/√2 < x < 1 with respect to x.

Solution:

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}}

Question 2. Differentiatey=cos^{-1}\left(\sqrt{\frac{1+x}{2}}\right)  ,−1 < x < 1 with respect to x.

Solution:

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}}

Question 3. Differentiatey=sin^{-1}\left(\sqrt{\frac{1-x}{2}}\right)  , 0 < x < 1 with respect to x.

Solution:

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}}

Question 4. Differentiatey=sin^{-1}(\sqrt{1-x^2})  , 0 < x < 1 with respect to x.

Solution:

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}}

Question 5. Differentiatey=tan^{-1}\left(\frac{x}{\sqrt{a^2-x^2}}\right)  , −a < x < a with respect to x.

Solution:



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}}

Question 6. Differentiatey=sin^{-1}\left(\frac{x}{\sqrt{a^2+x^2}}\right)  with respect to x.

Solution:

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}}

Question 7. Differentiatey=sin^{-1}\left(2x^2-1\right)  , 0 < x < 1 with respect to x.

Solution:

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}}

Question 8. Differentiatey=sin^{-1}\left(1-2x^2\right)  , 0 < x < 1 with respect to x.

Solution:

We havey=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}}

Question 9. Differentiatey=cos^{-1}\left(\frac{x}{\sqrt{a^2+x^2}}\right)  with respect to x.

Solution:

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}}

Question 10. Differentiatey=sin^{-1}\left(\frac{sinx+cosx}{\sqrt{2}}\right)  , −3π/4 < x < π/4 with respect to x.

Solution:

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

Question 11. Differentiatey=cos^{-1}\left(\frac{sinx+cosx}{\sqrt{2}}\right)  , −π/4 < x < π/4 with respect to x.

Solution:

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

Question 12. Differentiatey=tan^{-1}\left(\frac{x}{1+\sqrt{1-x^2}}\right)  , −1 < x < 1 with respect to x.

Solution:

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}}

Question 13. Differentiatey=tan^{-1}\left(\frac{x}{a+\sqrt{a^2-x^2}}\right)  , −a < x < a with respect to x.

Solution:

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}}

Question 14. Differentiatey=sin^{-1}\left(\frac{x+\sqrt{1-x^2}}{\sqrt{2}}\right)  , −1 < x < 1 with respect to x.

Solution:

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}}

Question 15. Differentiatey=cos^{-1}\left(\frac{x+\sqrt{1-x^2}}{\sqrt{2}}\right)  , −1 < x < 1 with respect to x.

Solution:

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}}

Question 16. Differentiatey=tan^{-1}\left(\frac{4x}{1-4x^{2}}\right)  , −1/2 < x < 1/2 with respect to x.

Solution:

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}}

Attention reader! Don’t stop learning now. Participate in the Scholorship Test for First-Step-to-DSA Course for Class 9 to 12 students.




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