Class 12 NCERT Solutions- Mathematics Part I – Chapter 5 Continuity And Differentiability – Exercise 5.3
Find of the following:
Question 1. 2x + 3y = sin x
Solution:
On differentiating both sides w.r.t. x, we get
2 + 3 = cos x
3 = cos x – 2
= (cosx – 2)/3
Question 2. 2x + 3y = sin y
Solution:
On differentiating both sides w.r.t. x, we get
2 + 3 = cos y
(cosy – 3) = 2
Question 3. ax + by2 = cos y
Solution:
On differentiating both sides w.r.t. x, we get
a + b * 2y() = -sin y *
(2by + siny) = -a
Question 4. xy + y2 = tan x + y
Solution:
On differentiating both sides w.r.t. x, we get
(x * + y) + 2y = sec2x +
(x + 2y – 1) = sec2x – y
Question 5. x2 + xy + y2 = 100
Solution:
On differentiating both sides w.r.t. x, we get
2x + (x + y) + 2y = 0
(x + 2y) * = -(2x + y)
Question 6. x3 + x2y + xy2 + y3 = 81
Solution:
Differentiate both sides w.r.t. x
3x2+(x2 + y * 2x) + (x * 2y * + y2) + 3y2 * = 0
(x2 + 2xy + 3y2) = -(3x2 + 2xy + y2)
Question 7. Sin2y + cos xy = π
Solution:
Differentiate both sides w.r.t. x
2 sin y * (siny) – sin(xy) * xy = 0
2sin y * cosy – sin(xy)(x * + y) = 0
(2sin cos y – sin (xy) – x)) = y(xy)
Question 8. sin2 x + cos2 y = 1
Solution:
2 sin x * (sin x) + 2 cos y * (cos y) = 0
2 sin x * cos x + 2 cos y*(-sin y) * = 0
2 sin x * cos x – 2 cos x – 2 cos y sin y * = 0
Sin(2x) – sin(2y) – = 0
Question 9. y = sin-1(\frac{2x}{(1 + x2)}
Solution:
Put x = tanθ
θ = tan-1x
y = sin-1(sin 2θ)
y = 2θ
y = 2tan-1x -(1)
On differentiating eq(1), we get
Question 10. , -1/√3 < x < 1/√3
Solution:
Put x = tanθ
θ = tan-1x
y =
y = tan-1(tan 3θ)
y = 3θ
y = 3tan-1x -(1)
On differentiating eq(1), we get
Question 11. , 0 < x < 1
Solution:
Put x = tanθ
θ = tan-1 x
y =
y = cos-1(cos 2θ)
y = 2θ
y = 2tan-1x -(1)
On differentiating eq(1), we get
Question 12. , 0 < x < 1
Solution:
Put x = tanθ
θ = tan-1x
y = sin-1(cos 2θ)
y = sin-1(sin (π/2 – 2θ))
y = π/2 – 2θ
y = π/2 – 2 tan-1x
Question 13. , -1 < x < 1
Solution:
Put x = tanθ
θ = tan-1x
y = cos^{-1}()
y = cos-1(sin 2θ)
y = cos-1(cos (π/2 – 2θ))
y = π/2 – 2θ
y = π/2 – 2tan-1x
Question 14. , -1/√2 < x < 1/√2
Solution:
Put x = sinθ
θ = sin-1 x
y = sin-1(2sinθ√(1 – sin2θ))
y = sin-1(sin 2θ) = 2θ
y = 2sin-1x
Question 15. , 0 < x < 1/√2
Solution:
Put x = tanθ
y = sec-1(1/cos2θ))
y = sec-1(sec2θ) = 2θ
y = 2cos-1x
=
Last Updated :
09 Mar, 2021
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