Students can download 11th Business Maths Chapter 6 Applications of Differentiation Ex 6.4 Questions and Answers, Notes, Samcheer Kalvi 11th Business Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

## Tamilnadu Samacheer Kalvi 11th Business Maths Solutions Chapter 6 Applications of Differentiation Ex 6.4

### Samacheer Kalvi 11th Business Maths Applications of Differentiation Ex 6.4 Text Book Back Questions and Answers

Question 1.

If z = (ax + b) (cy + d), then find \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\).

Solution:

Given, z = (ax + b) (cy + d)

Differentiating partially with respect to x we get,

\(\frac{\partial z}{\partial x}\) = (cy + d) \(\frac{\partial}{\partial x}\) (ax + b) [∵ (cy + d) is constant]

= (cy + d) (a + 0)

= a(cy + d)

Differentiating partially with respect to y we get,

\(\frac{\partial z}{\partial y}\) = (ax + b) \(\frac{\partial}{\partial y}\) (cy + d)

= (ax + b)(c + 0)

= c(ax + b)

Question 2.

If u = e^{xy}, then show that \(\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}\) = u(x^{2} + y^{2}).

Solution:

Given, u = e^{xy}

Differentiating partially with respect to x, we get,

\(\frac{\partial u}{\partial x}\) = yexy (Treating y as constant)

\(\frac{\partial^{2} u}{\partial x^{2}}=\frac{\partial}{\partial x}\left(y e^{x y}\right)\)

= \(y \frac{\partial}{\partial x}\left(e^{xy}\right)\)

= y(ye^{xy})

= y^{2}e^{xy} ……… (1)

We have u = e^{xy}

Differentiating partially with respect to y,

\(\frac{\partial u}{\partial y}=x e^{x y}\)

Again differentiating partially with respect to x, we get,

\(\frac{\partial^{2} u}{\partial y^{2}}=\frac{\partial}{\partial y}\left(x e^{x y}\right)\)

= \(x \frac{\partial}{\partial y}\left(e^{x y}\right)\)

= x^{2}e^{xy} ……… (2)

Adding (1) and (2) we get,

\(\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}\) = e^{xy}(x^{2} + y^{2})

= u(x + y ) [∵ u = e^{xy}]

Question 3.

Let u = x cos y + y cos x. Verify \(\frac{\partial^{2} u}{\partial x \partial y}=\frac{\partial^{2} u}{\partial y \partial x}\)

Solution:

u = x cos y + y cos x

Differentiating partially with respect to y, we get,

\(\frac{\partial u}{\partial y}\) = \(\frac{\partial}{\partial y}\) (xcosy) + \(\frac{\partial}{\partial y}\) (y cos x)

= x \(\frac{\partial}{\partial y}\)(cos y) + cos x \(\frac{\partial}{\partial y}\)(y)

= x(-sin y) + cos x

Again differentiating partially with respect to x, we get

\(\frac{\partial}{\partial x}\left(\frac{\partial u}{\partial y}\right)\) = \(\frac{\partial}{\partial x}\)(-x sin y) + \(\frac{\partial}{\partial x}\) (cos x)

= \(\frac{\partial}{\partial x}\) (-x sin y) + \(\frac{\partial}{\partial x}\) (cos x)

= -sin y \(\frac{\partial}{\partial x}(x)\) + (-sin x)

= -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

\(\frac{\partial u}{\partial x}\) = cos y \(\frac{\partial}{\partial x}\) (x) + y \(\frac{\partial}{\partial x}\) (cos x)

= cos y (1) + y(-sin x)

= cos y – y sin x

Again differentiating partially with respect to y we get,

\(\frac{\partial}{\partial y}\left(\frac{\partial u}{\partial x}\right)\) = \(\frac{\partial}{\partial y}\) (cos y – y sin x)

= \(\frac{\partial}{\partial y}\) (cos y) – \(\frac{\partial}{\partial y}\) (y sin x)

= -sin y – sin x \(\frac{\partial}{\partial y}\) (y)

= -sin y – sin x (1)

= -sin y – sin x ………(2)

From (1) and (2), \(\frac{\partial^{2} u}{\partial x \partial y}=\frac{\partial^{2} u}{\partial y \partial x}\)

Hence verified.

Question 4.

Verify Euler’s theorem for the function u = x^{3} + y^{3} + 3xy^{2}.

Solution:

u = x^{3} + y^{3} + 3xy^{2}

i.e., u(x, y) = x^{3} + y^{3} + 3xy^{2}

u(tx, ty) = (tx)^{3} + (ty)^{3} + 3(tx) (ty)^{2}

= t^{3}x^{3} + t^{3}y^{3} + 3tx (t^{2}y^{2})

= t^{3}(x^{3} + y^{3} + 3xy^{2})

= t^{3}u

∴ u is a homogeneous function in x and y of degree 3.

∴ By Euler’s theorem, \(x \cdot \frac{\partial u}{\partial x}+y \cdot \frac{\partial u}{\partial y}=3 u\)

Verification:

u = x^{3} + y^{3} + 3xy^{2}

\(\frac{\partial u}{\partial x}\) = 3x^{2} + 0 + 3y^{2} \(\frac{\partial}{\partial x}\) (x)

= 3x^{2} + 3y^{2}(1)

= 3x^{2} + 3y^{2} …….. (1)

∴ x . \(\frac{\partial u}{\partial x}\) = 3x^{3} + 3xy^{2}

\(\frac{\partial u}{\partial y}\) = 0 + 3y^{2} + 3x (2y) = 3y^{2} + 6xy

y . \(\frac{\partial u}{\partial y}\) = 3y^{3} + 6xy^{2} ……… (2)

∴ (1) + (2) gives

\(x \cdot \frac{\partial u}{\partial x}+y \cdot \frac{\partial u}{\partial y}\) = 3x^{3} + 3y^{3} + 9xy^{2}

= 3(x^{3} + y^{3} + 3xy^{2})

= 3u

Hence Euler’s theorem is verified.

Question 5.

Let u = x^{2}y^{3} cos(\(\frac{x}{y}\)). By using Euler’s theorem show that \(x \cdot \frac{\partial u}{\partial x}+y \cdot \frac{\partial u}{\partial y}\)

Solution:

Given, u = x^{2}y^{3} cos(\(\frac{x}{y}\))

i.e., u(tx, ty) = (tx)^{2} (ty)^{3} cos(\(\frac{tx}{ty}\))

= t^{2}x^{2}t^{3}y^{3} cos(\(\frac{x}{y}\))

= t^{5}x^{2}y^{3} cos(\(\frac{x}{y}\))

= t^{5} u

∴ u is a homogeneous function in x and y of degree 5.

∴ By Euler’s theorem, \(x \cdot \frac{\partial u}{\partial x}+y \cdot \frac{\partial u}{\partial y}=5 u\)

Hence Proved.