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NCERT Solutions Exercise 5.4: Continuity & Differentiability

Q1: Differentiate the following w.r.t. x: NCERT Solutions Exercise 5.4: Continuity & Differentiability
Ans: Let y = NCERT Solutions Exercise 5.4: Continuity & Differentiability
By using the quotient rule, we obtain
NCERT Solutions Exercise 5.4: Continuity & Differentiability

Q2: Differentiate the following w.r.t. x: esin-1x
Ans: Let y = esin-1x
By using the chain rule, we obtain
NCERT Solutions Exercise 5.4: Continuity & Differentiability

Q3: Differentiate the following w.r.t. x: NCERT Solutions Exercise 5.4: Continuity & Differentiability
Ans: Let y = NCERT Solutions Exercise 5.4: Continuity & Differentiability
By using the chain rule, we obtain
NCERT Solutions Exercise 5.4: Continuity & Differentiability

Q4: Differentiate the following w.r.t. x: sin(tan-1e-x)
Ans: Let y = sin(tan-1e-x)
By using the chain rule, we obtain
NCERT Solutions Exercise 5.4: Continuity & Differentiability

Q5: Differentiate the following w.r.t. x: log(cosex)
Ans: Let y = log(cosex)
By using the chain rule, we obtain
NCERT Solutions Exercise 5.4: Continuity & Differentiability

Q6: Differentiate the following w.r.t. x:
NCERT Solutions Exercise 5.4: Continuity & Differentiability
Ans:
NCERT Solutions Exercise 5.4: Continuity & Differentiability

Q7: Differentiate the following w.r.t. x: NCERT Solutions Exercise 5.4: Continuity & Differentiability
Ans: Let
NCERT Solutions Exercise 5.4: Continuity & Differentiability

Q8: Differentiate the following w.r.t. x:
NCERT Solutions Exercise 5.4: Continuity & Differentiability
Ans: Let y = NCERT Solutions Exercise 5.4: Continuity & Differentiability
By using the chain rule, we obtain
NCERT Solutions Exercise 5.4: Continuity & Differentiability

Q9: Differentiate the following w.r.t. x:
NCERT Solutions Exercise 5.4: Continuity & Differentiability
Ans: Let y = NCERT Solutions Exercise 5.4: Continuity & Differentiability
By using the quotient rule, we obtain
NCERT Solutions Exercise 5.4: Continuity & Differentiability

Q10: Differentiate the following w.r.t. x: NCERT Solutions Exercise 5.4: Continuity & Differentiability, x>0
Ans: Let y = NCERT Solutions Exercise 5.4: Continuity & Differentiability
By using the chain rule, we obtain
NCERT Solutions Exercise 5.4: Continuity & Differentiability 

The document NCERT Solutions Exercise 5.4: Continuity & Differentiability is a part of the JEE Course Mathematics (Maths) Class 12.
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FAQs on NCERT Solutions Exercise 5.4: Continuity & Differentiability

1. What is the difference between continuity and differentiability in calculus?
Ans. Continuity refers to the smoothness of a function where there are no breaks, jumps, or holes. Differentiability, on the other hand, refers to the existence of a derivative at a point, indicating the rate at which the function changes at that point.
2. How do you determine if a function is continuous at a point?
Ans. To determine if a function is continuous at a point, check if the function is defined at that point, the limit of the function exists at that point, and the value of the function equals the limit at that point.
3. Can a function be continuous but not differentiable?
Ans. Yes, a function can be continuous but not differentiable at a point if there is a sharp corner, cusp, or vertical tangent at that point. In such cases, the function is continuous but the derivative does not exist.
4. What is the significance of continuity and differentiability in real-world applications?
Ans. Continuity and differentiability are crucial in real-world applications such as physics, engineering, and economics. They help in analyzing rates of change, determining maximum and minimum values, and understanding the behavior of functions in various scenarios.
5. How can we determine if a function is differentiable at a point using limits?
Ans. To determine if a function is differentiable at a point using limits, calculate the limit of the difference quotient as it approaches zero. If the limit exists, the function is differentiable at that point.
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