Chapter 1 Differentiation Ex 1.4 Solutions

Balbharti12th Maharashtra State Board Maths Solutions BookPdf Chapter 1 Differentiation Ex 1.4 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 1 Differentiation Ex 1.4

(ii) x = a cot θ, y = b cosec θ
Solution:
x = a cot θ, y = b cosec θ
Differentiating x and y w.r.t. θ, we get

(iii) x = , y = log (a²+ m²)
Solution:
x = , y = log (a²+ m²)
Differentiating x and y w.r.t. m, we get

(iv) x = sin θ, y = tan θ
Solution:
x = sin θ, y = tan θ
Differentiating x and y w.r.t. θ, we get

(v) x = a(1 – cos θ), y = b(θ – sin θ)
Solution:
x = a(1 – cos θ), y = b(θ – sin θ)
Differentiating x and y w.r.t. θ, we get

(vi) x = , y = , where a > 0, a ≠ 1 and t ≠ 0
Solution:
x = , y = ………(1)
Differentiating x and y w.r.t. t, we get

(vii) x = , y =
Solution:
x = , y =
Put t = tan θ Then θ = tan^-1t

(viii) x = cos^-1(4t³– 3t), y =
Solution:
x = cos^-1(4t³– 3t), y =
Put t = cos θ. Then θ = cos^-1t
x = cos^-1(4cos³θ – 3cos θ)

(ii) x = a cos³θ, y = a sin³θ at θ =
Solution:
x = a cos³θ, y = a sin³θ
Differentiating x and y w.r.t. θ, we get

(iii) x = t²+ t + 1, y = sin() + cos() at t = 1
Solution:
x = t²+ t + 1, y = sin() + cos()
Differentiating x and y w.r.t. t, we get

(iv) x = 2 cos t + cos 2t, y = 2 sin t – sin 2t at t =
Solution:
x = 2 cos t + cos 2t, y = 2 sin t – sin 2t
Differentiating x and y w.r.t. t, we get

(v) x = t + 2 sin(πt), y = 3t – cos(πt) at t =
Solution:
x = t + 2 sin(πt), y = 3t – cos(πt)
Differentiating x and y w.r.t. t, we get

(ii) If x = , y = , then show that
Solution:
x = , y =
log x = log , log y = log
log x = (sin 3t)(log e), log y = (cos 3t)(log e)
log x = sin 3t, log y = cos 3t ….. (1) [∵ log e = 1]
Differentiating both sides w.r.t. t, we get

(iii) If x = , y = , then show that y2 – = 0.
Solution:
x = , y =

(iv) If x = a cos³t, y = a sin³t, then show that
Solution:
x = a cos³t, y = a sin³t
Differentiating x and y w.r.t. t, we get

(v) If x = 2 cos⁴(t + 3), y = 3 sin⁴(t + 3), show that
Solution:
x = 2 cos⁴(t + 3), y = 3 sin⁴(t + 3)

(vi) If x = log (1 + t²), y = t – tan^-1t, show that
Solution:
x = log (1 + t²), y = t – tan^-1t
Differentiating x and y w.r.t. t, we get

(vii) If x = , y = , show that sin x + = 0
Solution:
x = , y =
Differentiating x and y w.r.t. t, we get

(viii) If x = , y = , show that
Solution:
x = , y =

(ii) Differentiate w.r.t
Solution:
Let u = and v =
Then we want to find

(iii) Differentiate w.r.t
Solution:

(iv) Differentiate w.r.t. tan^-1x
Solution:
Let u = and v = tan^-1x
Then we want to find
Put x = tan θ. Then θ = tan^-1x.

(v) Differentiate 3x w.r.t. log_x3.
Solution:
Let u = 3x and v = log_x3.
Then we want to find
Differentiating u and v w.r.t. x, we get

(vi) Differentiate w.r.t. sec^-1x.
Solution:
Let u = and v = sec^-1x
Then we want to find .
Differentiating u and v w.r.t. x, we get

(vii) Differentiate x^xw.r.t. x^{sin x}.
Solution:
Let u = x^xand v = x^{sin x}
Then we want to find .
Take, u = x^x
log u = log x^x= x log x
Differentiating both sides w.r.t. x, we get

(viii) Differentiate w.r.t.
Solution:
Let u = and v =
Then we want to find
u =
Put x = tan θ. Then θ = tan^-1x and