Balbharati Maharashtra State Board12th Commerce Maths Solution Book PdfChapter 3 Differentiation Ex 3.4 Questions and Answers.
Maharashtra State Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.4
1. Find if:
Question 1
Maharashtra Board Solution
√x + √y = √a
Solution & Step-by-Step Answer:
√x + √y = √a Differentiating both sides w.r.t. x, we get
Question 2
Maharashtra Board Solution
x3 + y3 + 4x3y = 0
Solution & Step-by-Step Answer:
x3 + y3 + 4x3y = 0 Differentiating both sides w.r.t. x, we get
Question 3
Maharashtra Board Solution
x3 + x2y + xy2 + y3 = 81
Solution & Step-by-Step Answer:
x3 + x2y + xy2 + y3 = 81 Differentiating both sides w.r.t. x, we get
2. Find if:
Question 1
Maharashtra Board Solution
y.ex + x.ey = 1
Solution & Step-by-Step Answer:
y.ex + x.ey = 1 Differentiating both sides w.r.t. x, we get
Question 2
Maharashtra Board Solution
xy = e(x-y)
Solution & Step-by-Step Answer:
xy = e(x-y) ∴ log xy = 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 =
Question 3
Maharashtra Board Solution
xy = log(xy)
Solution & Step-by-Step Answer:
xy = log (xy) ∴ xy = log x + log y Differentiating both sides w.r.t. x, we get
3. Solve the following:
Question 1
Maharashtra Board Solution
If x5. y7 = (x + y)12, then show that
Solution & Step-by-Step Answer:
x5. y7 = (x + y)12 ∴ log(x5. y7) = log(x + y)12 ∴ log x5 + log y7 = 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
Maharashtra Board Solution
If log(x + y) = log(xy) + a, then show that
Solution & Step-by-Step Answer:
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
Maharashtra Board Solution
If ex + ey = e(x+y), then show that .
Solution & Step-by-Step Answer:
ex + ey = e(x+y) ……….(1) Differentiating both sides w.r.t. x, we get
