Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 3 Differentiation Ex 3.4 Questions and Answers.

## Maharashtra State Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.4

1. Find \(\frac{d y}{d x}\) if:

Question 1.

√x + √y = √a

Solution:

√x + √y = √a

Differentiating both sides w.r.t. x, we get

Question 2.

x^{3} + y^{3} + 4x^{3}y = 0

Solution:

x^{3} + y^{3} + 4x^{3}y = 0

Differentiating both sides w.r.t. x, we get

Question 3.

x^{3} + x^{2}y + xy^{2} + y^{3} = 81

Solution:

x^{3} + x^{2}y + xy^{2} + y^{3} = 81

Differentiating both sides w.r.t. x, we get

2. Find \(\frac{d y}{d x}\) if:

Question 1.

y.e^{x} + x.e^{y} = 1

Solution:

y.e^{x} + x.e^{y} = 1

Differentiating both sides w.r.t. x, we get

Question 2.

x^{y} = e^{(x-y)}

Solution:

x^{y} = e^{(x-y)}

∴ log x^{y} = log e^{(x-y)}

∴ y log x = (x – y) log e

∴ y log x = x – y …..[∵ log e = 1]

∴ y + y log x = x

∴ y(1 + log x) = x

∴ y = \(\frac{x}{1+\log x}\)

Question 3.

xy = log(xy)

Solution:

xy = log (xy)

∴ xy = log x + log y

Differentiating both sides w.r.t. x, we get

3. Solve the following:

Question 1.

If x^{5} . y^{7} = (x + y)^{12}, then show that \(\frac{d y}{d x}=\frac{y}{x}\)

Solution:

x^{5} . y^{7} = (x + y)^{12}

∴ log(x^{5} . y^{7}) = log(x + y)^{12}

∴ log x^{5} + log y^{7} = log(x + y)^{12}

∴ 5 log x + 7 log y = 12 log (x + y)

Differentiating both sides w.r.t. x, we get

Question 2.

If log(x + y) = log(xy) + a, then show that \(\frac{d y}{d x}=\frac{-y^{2}}{x^{2}}\)

Solution:

log (x + y) = log (xy) + a

∴ log(x + y) = log x + log y + a

Differentiating both sides w.r.t. x, we get

Question 3.

If e^{x} + e^{y} = e^{(x+y)}, then show that \(\frac{d y}{d x}=-e^{y-x}\).

Solution:

e^{x} + e^{y} = e^{(x+y)} ……….(1)

Differentiating both sides w.r.t. x, we get