Odisha State Board Elements of Mathematics Class 12 CHSE Odisha Solutions Chapter 7 Continuity and Differentiability Ex 7(e) Textbook Exercise Questions and Answers.

## CHSE Odisha Class 12 Math Solutions Chapter 7 Continuity and Differentiability Exercise 7(e)

Differentiate the following functions by proper substitution.

Question 1.

sin^{-1} 2x\( \sqrt{1-x^2} \)

Solution:

y = sin^{-1} 2x\( \sqrt{1-x^2} \) [Put x = sin θ

= sin^{-1} (2 sin θ . cos θ)

= sin^{-1} sin 2θ = 2θ = 2 sin^{-1} x.

\(\frac{d y}{d x}\) = \(\frac{2}{\sqrt{1-x^2}}\)

Question 2.

tan^{-1} \(\frac{2 x}{1-x^2}\)

Solution:

y = tan^{-1} \(\frac{2 x}{1-x^2}\)

= tan^{-1} \(\frac{2 \tan \theta}{1-\tan ^2 \theta}\) = tan^{-1} (tan 2θ)

= 2θ = 2 tan^{-1} x

∴ \(\frac{d y}{d x}\) = \(\frac{2 x}{1-x^2}\)

Question 3.

tan^{-1} \(\sqrt{\frac{1-t}{1+t}}\)

Solution:

Question 4.

\(\left[\left(\frac{1+t^2}{1-t^2}\right)^2-1\right]^{\frac{1}{2}}\)

Solution:

Question 5.

tan^{-1} \(\left(\frac{\sqrt{x}+\sqrt{a}}{1-\sqrt{x a}}\right)\)

Solution:

Question 6.

sin^{-1} (\(\frac{2 x}{1+x^2}\))

Solution:

y = sin^{-1} \(\frac{2 x}{1+x^2}\) [Put x = tan θ

= sin^{-1} \(\frac{2 \tan \theta}{1+\tan ^2 \theta}\) = sin^{-1} sin θ

= 2θ = 2 tan^{-1} x

∴ \(\frac{d y}{d x}\) = \(\frac{2 x}{1+x^2}\)

Question 7.

sec^{-1} \(\left(\frac{\sqrt{a^2+x^2}}{a}\right)\)

Solution:

Question 8.

sin^{-1} \(\left(\frac{2 \sqrt{t^2-1}}{t^2}\right)\)

Solution:

Question 9.

cos^{-1} \(\left(\frac{1-t^2}{1+t^2}\right)\)

Solution:

y = cos^{-1} \(\left(\frac{1-t^2}{1+t^2}\right)\) [Put t = tan θ

= cos^{-1} \(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\)

= cos^{-1} cos 2θ = 2 tan^{-1} t

∴ \(\frac{d y}{d x}\) = \(\frac{2}{1+t^2}\)

Question 10.

cos^{-1} (2t^{2} – 1)

Solution:

y = cos^{-1} (2t^{2} – 1) [Put t = tan θ

= cos^{-1} (2 cos^{2} θ – 1)

= cos^{-1} cos 2θ = 2θ = 2 cos^{-1} t

∴ \(\frac{d y}{d x}\) = – \(\frac{2}{\sqrt{1-t^2}}\)