Exercise 5.6 - Chapter 5 - Differential Calculus - 11th Business Maths Guide Samacheer Kalvi Solutions

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Exercise 5.6

Text Book Back Questions and Answers

Question 1.

(i) xy – tan(xy)
(ii) x² – xy + y² = 1
(iii) x³ + y³ + 3axy = 1
Solution:
(i) Given xy = tan(xy)
Differentiating both sides with respect to x, we get

(ii) x² – xy + y² = 7
Differentiating both side with respect to x,

(iii) x³ + y³ + 3axy = 1
Differentiating both sides with respect to x,

Question 2.

Solution:
Given

Squaring both sides we get
⇒ x² (1 + y) = y² (1 + x)
⇒ x² + x²y = y² + y²x
⇒ x² – y² + x²y – y²x = 0
⇒ (x + y) (x – y) + xy(x – y) = 0
⇒ (x – y) [(x + y) + xy] = 0
∴ x – y = 0 (or) x + y + xy = 0
x = y (or) x + y + xy = 0
Given that x ≠ y
x + y + xy = 0
⇒ y + xy = -x
⇒ y(1 + x) = -x

Hence proved.