RD Sharma Class 12 Ex 11.3 Solutions Chapter 11 Differentiation

Leave a Comment / Class 12 Maths / By

Here we provide RD Sharma Class 12 Ex 11.3 Solutions Chapter 11 Differentiation for English medium students, Which will very helpful for every student in their exams. Students can download the RD Sharma Class 12 Ex 11.3 Solutions Chapter 11 Differentiation book pdf download. Now you will get step-by-step solutions to each question.

TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter11
Exercise11.3
CategoryRD Sharma Solutions

Table of Contents

RD Sharma Class 12 Ex 11.3 Solutions Chapter 11 Differentiation

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}}

Question 17. Differentiatey=tan^{-1}\left(\frac{2^{x+1}}{1-4^x}\right) , −∞ < x < 0 with respect to x.

Solution:

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

On putting 2x = tan θ, we get,

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

=tan^{-1}\left(tan2θ\right)

Now, −∞ < x < 0

=> 0 < 2x < 1

=> 0 < θ < π/4

=> 0 < 2θ < π/2

So, y = 2θ

= 2 tan−1 (2x)

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}

Question 18. Differentiatey=tan^{-1}\left(\frac{2a^{x}}{1-a^{2x}}\right) , a > 1, −∞ < x < 0 with respect to x.

Solution:

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

On putting ax = tan θ, we get,

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

=tan^{-1}\left(tan2θ\right)

Now, −∞ < x < 0

=> 0 < ax < 1

=> 0 < θ < π/4

=> 0 < 2θ < π/2

So, y = 2θ

= 2 tan−1 (ax)

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}}

Question 19. Differentiatey=sin^{-1}\left(\frac{\sqrt{1+x}+\sqrt{1-x}}{2}\right) , 0 < x < 1 with respect to x.

Solution:

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}}

Question 20. Differentiatey=tan^{-1}\left(\frac{\sqrt{1+a^2x^2}-1}{ax}\right) , x ≠ 0 with respect to x.

Solution:

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)}

Question 21. Differentiatey=tan^{-1}\left(\frac{sinx}{1+cosx}\right) , −π < x < π with respect to x.

Solution:

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}

Question 22. Differentiatey=sin^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right) with respect to x.

Solution:

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}

Question 23. Differentiatey=cos^{-1}\left(\frac{1-x^{2n}}{1+x^{2n}}\right) , 0 < x < ∞ with respect to x.

Solution:

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}}

Question 24. Differentiatey=sin^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)+sec^{-1}\left(\frac{1+x^{2}}{1-x^{2}}\right) , x ∈ R with respect to x.

Solution:

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

Question 25. Differentiatey=tan^{-1}\left(\frac{a+x}{1-ax}\right) with respect to x.

Solution:

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}

Question 26. Differentiatey=tan^{-1}\left(\frac{\sqrt{x}+\sqrt{a}}{1-\sqrt{xa}}\right) with respect to x.

Solution:

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)}

Question 27. Differentiatey=tan^{-1}\left(\frac{a+btanx}{b-atanx}\right) with respect to x.

Solution:

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

Question 28. Differentiatey=tan^{-1}\left(\frac{a+bx}{b-ax}\right) with respect to x.

Solution:

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}

Question 29. Differentiatey=tan^{-1}\left(\frac{x-a}{x+a}\right) with respect to x.

Solution:

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}

Question 30. Differentiatey=tan^{-1}\left(\frac{x}{1+6x^2}\right) with respect to x.

Solution:

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}

Question 31. Differentiatey=tan^{-1}\left(\frac{5x}{1-6x^2}\right) with respect to x.

Solution:

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}

Question 32. Differentiatey=tan^{-1}\left(\frac{cosx+sinx}{cosx-sinx}\right) , −π/4 < x < π/4 with respect to x.

Solution:

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

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}

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment in the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share rdsharmasolutions.in to your friends.

Related Posts

Leave a Comment

Your email address will not be published. Required fields are marked *