Differentiation Class 12 Commerce Maths 1 Chapter 3 Exercise 3.4 Answers Maharashtra Board
Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 3 Differentiation Ex 3.4 Questions and Answers.
Std 12 Maths 1 Exercise 3.4 Solutions Commerce Maths
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