Class 12 Maths Chapter 5 Continuity and Differentiability – Exercise 5.3 NCERT Solutions

Introduction

Exercise 5.3 focuses on derivatives of inverse trigonometric functions. Students learn how to differentiate ${\sin }^{-1}x,{\cos }^{-1}x,{\tan }^{-1}x,{\cot }^{-1}x,{\sec }^{-1}x,{\csc }^{-1}x$ and apply these in solving problems. This exercise is crucial for calculus, integration, and advanced trigonometry.

Formulas Used

1. Derivatives of Inverse Trigonometric Functions:

$$\frac{d}{dx}({\sin }^{-1}x)=\frac{1}{\sqrt{1-x^2}},\mathrm{\mid }x\mathrm{\mid }<1$$

$$\frac{d}{dx}({\cos }^{-1}x)=-\frac{1}{\sqrt{1-x^2}},\mathrm{\mid }x\mathrm{\mid }<1$$

$$\frac{d}{dx}({\tan }^{-1}x)=\frac{1}{1+x^2},\mathrm{\forall }x$$

$$\frac{d}{dx}({\cot }^{-1}x)=-\frac{1}{1+x^2},\mathrm{\forall }x$$

$$\frac{d}{dx}({\sec }^{-1}x)=\frac{1}{\mathrm{\mid }x\mathrm{\mid }\sqrt{x^2-1}},\mathrm{\mid }x\mathrm{\mid }>1$$

$$\frac{d}{dx}({\csc }^{-1}x)=-\frac{1}{\mathrm{\mid }x\mathrm{\mid }\sqrt{x^2-1}},\mathrm{\mid }x\mathrm{\mid }>1$$

Students Frequently Make Mistakes

• Forgetting domain restrictions for inverse trigonometric functions.

• Missing negative signs in derivatives of ${\cos }^{-1}x,{\cot }^{-1}x,{\csc }^{-1}x$.

• Confusing reciprocal functions with inverse functions.

• Errors in simplification of composite functions.

• Skipping modulus in derivatives of ${\sec }^{-1}x$ and ${\csc }^{-1}x$.

NCERT Questions with Step‑by‑Step Solutions (10)

Q1. Differentiate $y={\sin }^{-1}x$.

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

Q2. Differentiate $y={\cos }^{-1}x$.

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

Q3. Differentiate $y={\tan }^{-1}x$.

$$\frac{dy}{dx}=\frac{1}{1+x^2}$$

Q4. Differentiate $y={\cot }^{-1}x$.

$$\frac{dy}{dx}=-\frac{1}{1+x^2}$$

Q5. Differentiate $y={\sec }^{-1}x$.

$$\frac{dy}{dx}=\frac{1}{\mathrm{\mid }x\mathrm{\mid }\sqrt{x^2-1}},\mathrm{\mid }x\mathrm{\mid }>1$$

Q6. Differentiate $y={\csc }^{-1}x$.

$$\frac{dy}{dx}=-\frac{1}{\mathrm{\mid }x\mathrm{\mid }\sqrt{x^2-1}},\mathrm{\mid }x\mathrm{\mid }>1$$

Q7. Differentiate $y={\sin }^{-1}(2x)$.

$$\frac{dy}{dx}=\frac{2}{\sqrt{1-(2x{)}^2}}=\frac{2}{\sqrt{1-4x^2}}$$

Q8. Differentiate $y={\cos }^{-1}(3x)$.

$$\frac{dy}{dx}=-\frac{3}{\sqrt{1-9x^2}}$$

Q9. Differentiate $y={\tan }^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$. Using identity:

$$y={\sin }^{-1}x\Rightarrow \frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}$$

Q10. Differentiate $y={\sin }^{-1}\left(\frac{2x}{1+x^2}\right)$. Using identity:

$$y=2{\tan }^{-1}x\Rightarrow \frac{dy}{dx}=\frac{2}{1+x^2}$$

FAQs (10)

FAQ1. What is derivative of ${\sin }^{-1}x$? $\frac{1}{\sqrt{1-x^2}}$.

FAQ2. What is derivative of ${\cos }^{-1}x$? $-\frac{1}{\sqrt{1-x^2}}$.

FAQ3. What is derivative of ${\tan }^{-1}x$? $\frac{1}{1+x^2}$.

FAQ4. What is derivative of ${\cot }^{-1}x$? $-\frac{1}{1+x^2}$.

FAQ5. What is derivative of ${\sec }^{-1}x$? $\frac{1}{\mathrm{\mid }x\mathrm{\mid }\sqrt{x^2-1}}$.

FAQ6. What is derivative of ${\csc }^{-1}x$? $-\frac{1}{\mathrm{\mid }x\mathrm{\mid }\sqrt{x^2-1}}$.

FAQ7. Why modulus in ${\sec }^{-1}x$ derivative? To ensure positivity of denominator.

FAQ8. Why check domain restrictions? Inverse functions defined only in restricted domains.

FAQ9. Is differentiability same as continuity? Differentiability implies continuity, but not vice versa.

FAQ10. Why is Exercise 5.3 important? It builds foundation for calculus involving inverse trigonometric functions.

Conclusion

Exercise 5.3 has 10 solved questions and 10 FAQs that strengthen your understanding of derivatives of inverse trigonometric functions. This builds the foundation for advanced calculus in Class 12 Maths.

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