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Class 12 RD Sharma Solutions – Chapter 11 Differentiation – Exercise 11.3 | Set 3

Last Updated : 08 May, 2021
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Question 33. Differentiatey=tan^{-1}\left(\frac{x^{\frac{1}{3}}+a^\frac{1}{3}}{1-(ax)^{\frac{1}{3}}}\right)  with respect to x.

Solution:

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}})}

Question 34. Differentiatey=sin^{-1}\left(\frac{2^{x+1}}{1+4^x}\right)  with respect to x.

Solution:

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

On putting 2x = 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 (2x)

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}

Question 35. Ify=sin^{-1}\left(\frac{2x}{1+x^2}\right)+sec^{-1}\left(\frac{1+x^2}{1-x^2}\right)  , 0 < x < 1, prove that\frac{dy}{dx}=\frac{4}{1+x^2}  .

Solution:

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.

Question 36. Ify=sin^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right)+cos^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right)  , 0 < x < ∞, prove that\frac{dy}{dx}=\frac{2}{1+x^2}  .

Solution:

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.

Question 37 Differentiate the following with respect to x :

(i) cos−1 (sin x)

Solution:

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)

Solution:

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}

Question 38. Differentiatey=cot^{-1}\left[\frac{\sqrt{1+sinx}+\sqrt{1-sinx}}{\sqrt{1+sinx}-\sqrt{1-sinx}}\right]  , 0 < x < π/2 with respect to x.

Solution:

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}

Question 39. Ify=tan^{-1}\left(\frac{2x}{1-x^2}\right)+sec^{-1}\left(\frac{1+x^2}{1-x^2}\right)  , x > 0, prove that\frac{dy}{dx}=\frac{4}{1+x^2}  .

Solution:

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.

Question 40. Ify=sec^{-1}\left(\frac{x+1}{x-1}\right)+sin^{-1}\left(\frac{x-1}{x+1}\right)  , x > 0, find\frac{dy}{dx}  .

Solution:

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

Question 41. Ify=sin\left[2tan^{-1}\sqrt{(\frac{1-x}{1+x})}\right]  , find\frac{dy}{dx}  .

Solution:

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}}

Question 42. If y=cos^{-1}(2x)+2cos^{-1}(\sqrt{1-4x^2})  , 0 < x < 1/2, find\frac{dy}{dx}  .

Solution:

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}}

Question 43. If the derivative of tan−1 (a + bx) takes the value of 1 at x = 0, prove that 1 + a2 = b.

Solution:

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.

Question 44. Ify=cos^{-1}(2x)+2cos^{-1}(\sqrt{1-4x^2})  , −1/2 < x < 0, find\frac{dy}{dx}  .

Solution:

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}}

Question 45. Ify=tan^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)  , find\frac{dy}{dx}  .

Solution:

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}}

Question 46. If y=cos^{-1}\left(\frac{2x-3\sqrt{1-x^2}}{\sqrt{13}}\right)  , find\frac{dy}{dx}  .

Solution:

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)

LetcosØ=\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}}

Question 47. Differentiatey=sin^{-1}\left[\frac{2^{x+1}×3^x}{1+(36)^x}\right]  with respect to x.

Solution:

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 6x = 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 (6x)

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}



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