# NCERT Solutions - Continuity & Differentiability, Exercise 5.3Maths JEE Notes | EduRev

## JEE : NCERT Solutions - Continuity & Differentiability, Exercise 5.3Maths JEE Notes | EduRev

The document NCERT Solutions - Continuity & Differentiability, Exercise 5.3Maths JEE Notes | EduRev is a part of the JEE Course Mathematics (Maths) Class 12.
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Continuity & Differentiability

Question 1:   Question 2:    Question 3:  Question 4: The given relationship is Differentiating this relationship with respect to x, we obtain    Question 5:

Find The given relationship is x2 +xy +y2 = 100

Differentiating this relationship with respect to x, we obtain [Derivative of constant function is 0] Question 6:  Question 7:  Using chain rule, we obtain Question 8:

Find The given relationship is Differentiating this relationship with respect to x, we obtain Question 9:

Find  Therefore, by quotient rule, we obtain Question 10:    Question 11:  The given relationship is, On comparing L.H.S. and R.H.S. of the above relationship, we obtain  tany/2 = x

Differentiating this relationship with respect to x, we obtain  Question 12:  The given relationship is  From (1), (2), and (3), we obtain  Differentiating this relationship with respect to x, we obtain Question 13:   Question 14:

Find  Differentiating this relationship with respect to x, we obtain Question 15:

Find dy/dx   Differentiating this relationship with respect to x, we obtain  Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

## Mathematics (Maths) Class 12

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