2nd PUC Basic Maths Question Bank Chapter 18 Differential Calculus Ex 18.3

Students can Download Basic Maths Exercise 18.3 Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Basic Maths Question Bank Chapter 18 Differential Calculus Ex 18.3

Part – A

2nd PUC Basic Maths Differential Calculus Ex 18.3 One or Two Marks Questions and Answers

Question 1.
3x2 + 4y2 = 10
Answer:
Given 3x2 + 4y2 = 10
Diff w.r.t x
6x + 8y \(\frac{d y}{d x}\) = 0
\(\Rightarrow \quad \frac{d y}{d x}=\frac{-8 y}{6 x}=\frac{-4 y}{3 x}\)

Question 2.
\(\sqrt{x}+\sqrt{y}=3\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 18 Differential Calculus Ex 18.3 - 1

KSEEB Solutions

Question 3.
y2 = 4ax.
Answer:
Given y2 = 4ax.
Differentiate with respect to x
2y \(\frac{d y}{d x}\) = 4a.1 ⇒ \(\frac{d y}{d x}\) = \(\frac{4 a}{2 y}=\frac{2 a}{y}\)

Question 4.
\(x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 18 Differential Calculus Ex 18.3 - 2

Question 5.
x2 = 4ay
Answer:
Given x2 = 4ay
Differentiate with respect to x, 2x = 4a \(\frac{d y}{d x}\) ⇒ \(\frac{d y}{d x}\) = \(\frac{2 x}{4 a}=\frac{x}{2 a}\)

Question 6.
\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 18 Differential Calculus Ex 18.3 - 3

Question 7.
x3 + y3 = 3axy
Answer:
Given x3 + y3 = 3axy
3x2 + 3y2 \(\frac{d y}{d x}\) = 3a \(\left(x \cdot \frac{d y}{d x}+y \cdot 1\right)\)
\(\frac{d y}{d x}\) (3y2 – 3ax) = 3ay – 3x2 = \(\frac{d y}{d x}=\frac{a y-x^{2}}{y^{2}-a x}\)

Question 8.
x – y = 0
Answer:
Given x – y = 0
Differentiate with respect to x,
1 – \(\frac{d y}{d x}\) = 0 ⇒ \(\frac{d y}{d x}\) = 1

KSEEB Solutions

Question 9.
x2 – y2 = a2
Answer:
Given x2 – y2 = a2
Differentiate with respect to x we get,
2x – 2y. \(\frac{d y}{d x}\) = 0 ⇒ \(\frac{d y}{d x}\) = \(\frac{2 x}{2 y}=\frac{x}{y}\)

Question 10.
x + \(\sqrt{x y}\) = x2.
Answer:
2nd PUC Basic Maths Question Bank Chapter 18 Differential Calculus Ex 18.3 - 4

Part-B

2nd PUC Basic Maths Differential Calculus Ex 18.3 Three marks Questions and Answers

Question 1.
log(xy) = x2 + y2
Answer:
Given log(xy) = x2 + y2
log x + log y = x2 + y2 differentiate w.r.t x
2nd PUC Basic Maths Question Bank Chapter 18 Differential Calculus Ex 18.3 - 5

Question 2.
2x + 2y = 2x+y
Answer:
Given 2x + 2y = 2x+y
Differentiate w.r.t. x we get
2x log 2 + 2y log 2 \(\frac{d y}{d x}\)
2nd PUC Basic Maths Question Bank Chapter 18 Differential Calculus Ex 18.3 - 6

KSEEB Solutions

Question 3.
xy = yx.
Answer:
Given xy = yx., taking logm both sides
y log x = x log y differentiate
Both sides w.r.t x
2nd PUC Basic Maths Question Bank Chapter 18 Differential Calculus Ex 18.3 - 7

Question 4.
sin xy = cos(x + y).
Answer:
Given sin xy = cos(x + y), diff w.r.t x.
cos(xy) \(\left[\mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}\right]\)
\(\frac{d y}{d x}\) [sin(x + y) + xcos (xy)]
= -sin(x + y) – y cos (xy)
\(\frac{d y}{d x}=\frac{-[\sin (x+y)+\cos x y]}{(\sin (x+y)+x \cos x y)}\)

Question 5.
y = 4x+y
Answer:
Given y = 4x+y, diff. w r.t. x
\(\frac{d y}{d x}\) = 4x+y log 4(1 + \(\frac{d y}{d x}\)) = 4x+y
\(\frac{d y}{d x}\)(1 – 4x+y log 4) = 4x+y log 4
∴ \(\frac{d y}{d x}=\frac{4^{x+y} \cdot \log 4}{1-4^{x+y} \cdot \log 4}\)

KSEEB Solutions

Part-C

2nd PUC Basic Maths Differential Calculus Ex 18.3 Five Marks Questions and Answers.

Question 1.
If \(\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}}\) = a, Prove that x . \(\frac{d y}{d x}\) = y.
Answer:
2nd PUC Basic Maths Question Bank Chapter 18 Differential Calculus Ex 18.3 - 8

Question 2.
If xy = ey – x, show that \(\frac{d y}{d x}\) = \(\frac{2-\log x}{(1-\log x)^{2}}\)
Answer:
Given xy = ey – x . Taking log both sides
y log x = (y – x)log ee
x = y (1 – log x) ∵ log ee = 1
2nd PUC Basic Maths Question Bank Chapter 18 Differential Calculus Ex 18.3 - 9

KSEEB Solutions

Question 3.
If cos y = x cos(a + y). show that \(\frac{d y}{d x}\) = \(\frac{\cos ^{2}(a+y)}{\sin a}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 18 Differential Calculus Ex 18.3 - 10

Question 4.
If ex = yx show that \(\frac{d y}{d x}\) = \(\frac{(\log y)^{2}}{\log y-1}\)
Answer:
Given ex = yx Taking logm both sides
y log ee = x log y
y = x log y differentiate w.r.t x
2nd PUC Basic Maths Question Bank Chapter 18 Differential Calculus Ex 18.3 - 11

Question 5.
If ex+y = xy show that \(\frac{d y}{d x}\) = \(\frac{y(1-x)}{x(y-1)}\)
Answer:
Given yex+y = xy
Taking log m both sides
(x + y) loge = log(xy)
x+y = log x + log y diff w.r.t x
2nd PUC Basic Maths Question Bank Chapter 18 Differential Calculus Ex 18.3 - 12

KSEEB Solutions

Question 6.
If yx = xy show that \(\frac{d y}{d x}\) = \(\frac{y(y=x \log y)}{x(x-y \log x)}\)
Answer:
Given yx = xy, Taking logm both sides
x log y = y log x, diff w.r.t x
2nd PUC Basic Maths Question Bank Chapter 18 Differential Calculus Ex 18.3 - 13

Leave a Comment

error: Content is protected !!