Balbharti12th Maharashtra State Board Maths Solutions BookPdf Chapter 1 Differentiation Ex 1.5 Questions and Answers.
Maharashtra State Board 12th Maths Solutions Chapter 1 Differentiation Ex 1.5


(ii) e2x. tan x
Solution:
Let y = e2x. tan x


(iii) e4x. cos 5x
Solution:
Let y = e4x. cos 5x


(iv) x3. log x
Solution:
Let y = x3. log x

(v) log(log x)
Solution:
Let y = log(log x)


(vi) xx
Solution:
y = xx
log y = log xx= x log x
Differentiating both sides w.r.t. x, we get


(ii) x = 2at2, y = 4at
Solution:
x = 2at2, y = 4at
Differentiating x and y w.r.t. t, we get

(iii) x = sin θ, y = sin3θ at θ =
Solution:
x = sin θ, y = sin3θ
Differentiating x and y w.r.t. θ, we get,


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





(ii) If y = , show that
Solution:
y = ……..(1)

(iii) If x = cos t, y = emt, show that
Solution:
x = cos t, y = emt


(iv) If y = x + tan x, show that
Solution:
y = x + tan x

(v) If y = eax. sin (bx), show that y2– 2ay1+ (a2+ b2)y = 0.
Solution:
y = eax. sin (bx) ………(1)


(vi) If , show that
Solution:



(vii) If 2y = , show that 4(x2– 1)y2+ 4xy1– y = 0.
Solution:
2y = …… (1)
Differentiating both sides w.r.t. x, we get


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


(ix) If y = sin(m cos-1x), then show that
Solution:
y = sin(m cos-1x)
sin-1y = m cos-1x
Differentiating both sides w.r.t. x, we get

(x) If y = log(log 2x), show that xy2+ y1(1 + xy1) = 0.
Solution:
y = log(log 2x)



(xi) If x2+ 6xy + y2= 10, show that
Solution:
x2+ 6xy + y2= 10 …… (1)
Differentiating both sides w.r.t. x, we get



(xii) If x = a sin t – b cos t, y = a cos t + b sin t, Show that
Solution:
x = a sin t – b cos t, y = a cos t + b sin t
Differentiating x and y w.r.t. t, we get





(ii)
Solution:
Let y =

(iii) eax+b
Solution:
Let y = eax+b


(iv) apx+q
Solution:
Let y = apx+q

(v) log(ax + b)
Solution:
Let y = log(ax + b)
Then

(vi) cos x
Solution:
Let y = cos x

(vii) sin(ax + b)
Solution:
Let y = sin(ax + b)


(viii) cos(3 – 2x)
Solution:


(ix) log(2x + 3)
Solution:


(x)
Solution:
Let y =


(xi) y = eax. cos (bx + c)
Solution:
y = eax. cos (bx + c)





(xii) y = e8x. cos (6x + 7)
Solution:



