Exercise 6.4 - Chapter 6 - Application of Differentiation - 11th Business Maths Guide Samacheer Kalvi Solutions

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Exercise 6.4

Text Book Back Questions and Answers

Question 1.
If z = (ax + b) (cy + d), then find 

Solution:
Given, z = (ax + b) (cy + d)
Differentiating partially with respect to x we get,

= (cy + d) (a + 0)
= a(cy + d)
Differentiating partially with respect to y we get,

= (cy + d) (a + 0)
= a(cy + d)
Differentiating partially with respect to y we get,

= (ax + b)(c + 0)
= c(ax + b)

Question 2.
If u = e^xy, then show that 

Solution:
Given, u = e^xy
Differentiating partially with respect to x, we get,

= y(ye^xy)
= y²e^xy ……… (1)
We have u = e^xy
Differentiating partially with respect to y,

Again differentiating partially with respect to x, we get,

Question 3.
Let u = x cos y + y cos x. Verify

Solution:
u = x cos y + y cos x
Differentiating partially with respect to y, we get,

= x(-sin y) + cos x
Again differentiating partially with respect to x, we get

= -sin y (1) + (-sin x)
= -sin y – sin x ……… (1)
Now u = x cos y + y cos x
Differentiating partially with respect to x we get

= cos y (1) + y(-sin x)
= cos y – y sin x
Again differentiating partially with respect to y we get,

Question 4.
Verify Euler’s theorem for the function u = x³ + y³ + 3xy².
Solution:
u = x³ + y³ + 3xy²
i.e., u(x, y) = x³ + y³ + 3xy²
u(tx, ty) = (tx)³ + (ty)³ + 3(tx) (ty)²
= t³x³ + t³y³ + 3tx (t²y²)
= t³(x³ + y³ + 3xy²)
= t³u
∴ u is a homogeneous function in x and y of degree 3.
∴ By Euler’s theorem,

Verification:
u = x³ + y³ + 3xy²

= 3x² + 3y²(1)
= 3x² + 3y² …….. (1)

= 3(x³ + y³ + 3xy²)
= 3u
Hence Euler’s theorem is verified.

Question 5.
Let  By using Euler’s theorem show that

Solution:
Given,

∴ u is a homogeneous function in x and y of degree 5.
∴ By Euler’s theorem,

Hence Proved.