Chapter 1 Differentiation Ex 1.1 Solutions

Balbharti12th Maharashtra State Board Maths Solutions BookPdf Chapter 1 Differentiation Ex 1.1 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 1 Differentiation Ex 1.1

(ii)
Solution:
Let y =
Differentiating w.r.t. x, we get

(iii)
Solution:

(iv)
Solution:
Let y =
Differentiating w.r.t. x, we get

(v)
Solution:
Let y =
Differentiating w.r.t. x, we get

(vi)
Solution:
Let y =
Differentiating w.r.t. x, we get

(ii)
Solution:
Let y =
Differentiating w.r.t. x, we get

(iii) log[tan()]
Solution:
Let y = log[tan()]
Differentiating w.r.t. x, we get

(iv)
Solution:
Let y =
Differentiating w.r.t. x, we get

(v) cot³[log (x³)]
Solution:
Let y = cot³[log (x³)]
Differentiating w.r.t. x, we get

(vi) 5^{sin3x+ 3}
Solution:
Let y = 5^{sin3x+ 3}
Differentiating w.r.t. x, we get

(vii) cosec ()
Solution:
Let y = cosec ()
Differentiating w.r.t. x, we get

(viii) log[cos (x³– 5)]
Solution:
Let y = log[cos (x³– 5)]
Differentiating w.r.t. x, we get

(ix) e^{3 sin2x – 2 cos2x}
Solution:
Let y = e^{3 sin2x – 2 cos2x}
Differentiating w.r.t. x, we get

(x) cos²[log (x²+ 7)]
Solution:
Let y = cos²[log (x²+ 7)]
Differentiating w.r.t. x, we get

(xi) tan[cos (sinx)]
Solution:
Let y = tan[cos (sinx)]
Differentiating w.r.t. x, we get

(xii) sec[tan (x⁴+ 4)]
Solution:
Let y = sec[tan (x⁴+ 4)]
Differentiating w.r.t. x, we get

= sec[tan(x⁴+ 4)]∙tan[tan(x⁴+ 4)]∙sec²(x⁴+ 4)(4x³+ 0)
= 4x³sec²(x⁴+ 4)∙sec[tan(x⁴+ 4)]∙tan[tan(x⁴+ 4)].

(xiii) e^{log[(logx)2 – logx2]}
Solution:
Let y = e^{log[(logx)2 – logx2]}
= (log x)²– log x²…[∵ e^{log x}= x]
Differentiating w.r.t. x, we get

(xiv) sin
Solution:
Let y = sin
Differentiating w.r.t. x, we get

(xv) log[sec(e^x2)]
Solution:
Let y = log[sec(e^x2)]
Differentiating w.r.t. x, we get

(xvi) log_e2(logx)
Solution:
Let y = log_e2(logx) =

(xvii) [log{log(logx)}]²
Solution:
let y = [log{log(logx)}]²
Differentiating w.r.t. x, we get

(xviii) sin²x²– cos²x²
Solution:
Let y = sin²x²– cos²x²
Differentiating w.r.t. x, we get

= 2sinx²∙cosx²× 2x + 2sinx²∙cosx²× 2x
= 4x(2sinx²∙cosx²)
= 4xsin(2x²).

(iii)
Solution:
Let y =
Differentiating w.r.t. x, we get

(iv)
Solution:
Let y =
Differentiating w.r.t. x, we get
=

(v) (1 + sin²x)²(1 + cos²x)³
Solution:
Let y = (1 + sin²x)²(1 + cos²x)³
Differentiating w.r.t. x, we get

= 3(1 + sin²x)²(1 + cos²x)²∙[2cosx(-sinx)] + 2 (1 + sin²x)(1 + cos²x)³∙[2sinx-cosx]
= 3 (1 + sin²x)²(1 + cos²x)²(-sin 2x) + 2(1 + sin²x)(1 + cos²x)³(sin 2x)
= sin2x (1 + sin²x) (1 + cos²x)²[-3(1 + sin²x) + 2(1 + cos²x)]
= sin2x (1 + sin²x)(1 + cos²x)²(-3 – 3sin²x + 2 + 2cos²x)
= sin2x (1 + sin²x)(1 + cos²x)²[-1 – 3 sin²x + 2 (1 – sin²x)]
= sin 2x(1 + sin²x)(1 + cos²x)²(-1 – 3 sin²x + 2 – 2 sin²x)
= sin2x (1 + sin²x)(1 + cos²x)²(1 – 5 sin²x).

(vi)
Solution:
Let y =
Differentiating w.r.t. x, we get
=

(vii) log(sec 3x+ tan 3x)
Solution:
Let y = log(sec 3x+ tan 3x)
Differentiating w.r.t. x, we get

(viii)
Solution:

(ix) cot – log
Solution:
Let y = cot – log
Differentiating w.r.t. x, we get

(x)
Solution:

(xi)
Solution:
let y =
Differentiating w.r.t. x, we get

(xii) log[tan³x·sin⁴x·(x²+ 7)⁷]
Solution:
Let y = log [tan³x·sin⁴x·(x²+ 7)⁷]
= log tan³x + log sin⁴x + log (x²+ 7)⁷
= 3 log tan x + 4 log sin x + 7 log (x²+ 7)
Differentiating w.r.t. x, we get

= 6cosec2x + 4 cotx +

(xiii) log
Solution:

(xiv) log
Solution:
Using log = log a – log b
log a^b= b log a


cosec

(xv) log
Solution:

(xvi) log
Solution:

(xvii) log
Solution:

(xviii) log
Solution:

(xix) y= (25)^log5(secx)− (16)^log4(tanx)
Solution:
y = (25)^log5(secx)− (16)^log4(tanx)
= 5^2log5(secx)– 4^2log4(tanx)
= 5^log5(sec5x)– 4^log4(tan2x)
= sec²x – tan²x … [∵ = x]
∴ y = 1
Differentiating w.r.t. x, we get
= (1) = 0

(xx)
Solution:
Let y =
Differentiating w.r.t. x, we get

(ii) If R(x) = g[3 + f(x)] find R’ (4).
Solution:
R(x) = g[3 + f(x)]
∴ R'(x) = {g[3+f(x)]}
= g'[3 + f(x)]∙[3 + f(x)]
= g'[3 +f(x)]∙[0 + f'(x)]
= g'[3 + f(x)]∙f'(x)
∴ R'(4) = g'[3 + f(4)]∙f'(4)
= g'[3 + 3]∙f'(4) … [∵ f(x) = 3, when x = 4]
= g'(6)∙f'(4)
= 7 × 5 … [From the table]
= 35.

(iii) If s(x) = f[9− f(x)] find s’ (4).
Solution:
s(x) = f[9− f(x)]
∴ s'(x) = {f[9 – f(x)]}
= f'[9 – f(x)]∙[0 – f(x)]
= f'[9 – f(x)]∙[0 – f'(x)]
= -f'[9 – f(x)] – f'(x)
∴ s'(4) = -f'[9 – f(4)] – f'(4)
= -f'[9 – 3] – f'(4) … [∵ f(x) = 3, when x = 4]
= -f'(6) – f'(4)
= -(-4)(5) … [From the table]
= 20.

(iv) If S(x) = g[g(x)] find S’ (6)
Solution:
S(x) = g[g(x)]
∴ S'(x) = g[g(x)]
= g'[g(x)]∙[g(x)]
= g'[g(x)]∙g'(x)
∴ S ‘(6) = g'[g'(6)]∙g'(6)
= g'(2)∙g'(6) … [∵ g (x) = 2, when x = 6]
= 4 × 7 … [From the table]
= 28.