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# 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

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## Mathematics (Maths) Class 12

209 videos|222 docs|124 tests

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