Maharashtra State Board Class 12 (HSC Board) Mathematics & Statistics Solutions Chapter 1 Differentiation Ex 1.3 - Digest Answers | Maharashtra Board Solutions

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Maharashtra State Board 12th Maths Solutions Chapter 1 Differentiation Ex 1.3

Question 1 Maharashtra Board Solution

Differentiate the following w.r.t. x: (i)

Solution & Step-by-Step Answer:

Let y = Then, log y = log {(x+1)^{2}}{(x+2)^{3}(x+3)^{4}} = log (x + 1)2 – log (x + 2)3 – log (x + 3)4 = 2 log (x +1) – 3 log (x + 2) – 4 log (x + 3) Differentiating both sides w.r.t. x, we get

(ii)
Solution:
Let y =
Then log y = log [3]{{4 x-1}{(2 x+3)(5-2 x)^{2}}}
Differentiating both sides w.r.t. x, we get

(iii)
Solution:
Let y =
Then log y = log ≤ft(x^{2}+3)^{} ^{3} 2 x 2^{x^{2}}
Differentiating both sides w.r.t. x, we get

(iv)
Solution:
Let y =
Then log y = log {≤ft(x^{2}+2 x+2)^{}}{(+3)^{3}( x)^{x}}
Differentiating both sides w.r.t. x, we get

(v)
Solution:
Let y =
Then log y = log {x^{5} ^{3} 4 x}{ ^{2} 3 x}
= log x5+ log tan34x – log sin23x
= 5 log x+ 3 log (tan 4x) – 2 log (sin 3x)
Differentiating both sides w.r.t. x, we get

(vi)
Solution:
Let y =
Then log y = log () = (tan^-1x)(log x)
Differentiating both sides w.r.t. x, we get

(vii) (sin x)^x
Solution:
Let y = (sin x)^x
Then log y = log (sin x)^x= x. log (sin x)
Differentiating both sides w.r.t. x, we get

(viii) sin x^x
Solution:
Let y = (sin x^x)
Then
……. (1)
Let u = x^x
Then log u = log x^x= x. log x
Differentiating both sides w.r.t. x, we get

Question 2 Maharashtra Board Solution

Differentiate the following w.r.t. x: (i) xe + xx + ex + ee

Solution & Step-by-Step Answer:

Let y = xe + xx + ex + ee Let u = xx Then log u = log xx = x log x Differentiating both sides w.r.t. x, we get

(ii)
Solution:
Let y =
Put u = and v =
Then y = u + v
∴
Take u =
log u = log = x^x. log x
Differentiating both sides w.r.t. x, we get

(iii) (log x)^x– (cos x)^{cot x}
Solution:
Let y = (log x)^x– (cos x)^{cot x}
Put u = (log x)^xand v = (cos x)^{cot x}
Then y = u – v
∴ ……..(1)
Take u = (log x)^x
∴ log u = log (log x)^x= x. log (log x)
Differentiating both sides w.r.t. x, we get

(iv)
Solution:
Let y =
Put u = and v = (log x)^{sin x}
Then y = u + v
∴ ……….(1)
Take u =
∴ log u = log = e^x. log x
Differentiating both sides w.r.t. x, we get

Also, v = (log x)^{sin x}
∴ log v = log (log x)^{sin x}= (sin x). (log log x)
Differentiating both sides w.r.t. x, we get

=

(v)
Solution:
Let y =
Put u = (log x)^{tan x}
∴ log u =log(log x)^{tan x}= (tan x).(log log x)
Differentiating both sides w.r.t. x, we get

(vi) (sin x)^{tan x}+ (cos x)^{cot x}
Solution:
Let y = (sin x)^{tan x}+ (cos x)^{cot x}
Put u = (sin x)^{tan x}and v = (cos x)^{cot x}
Then y = u + v
∴ ………(1)
Take u = (sin x)^{tan x}
∴ log u = log (sin x)^{tan x}= (tan x). (log sin x)
Differentiating both sides w.r.t. x, we get

(vii)
Solution:
Let y =
Put u = , v = and w =
Then y = u + v + w
∴ ………(1)

(viii) at x =
Solution:
Let y =
∴ log y = log ≤ft[( x)^{ x}]^{ x}
= tan x. log(tan x)tan x
= tan x. tan x log (tan x)
= (tan x)2. log (tan x)
Differentiating both sides w.r.t. x, we get

Question 3 Maharashtra Board Solution

Find if (i) √x + √y = √a

Solution & Step-by-Step Answer:

√x + √y = √a Differentiating both sides w.r.t. x, we get

(ii) x√x + y√y = a√a
Solution:
x√x + y√y = a√a
∴
Differentiating both sides w.r.t. x, we get

(iii) x + √xy + y = 1
Solution:
x + √xy + y = 1
Differentiating both sides w.r.t. x, we get

(iv) x³+ x²y + xy²+ y³= 81
Solution:
x³+ x²y + xy²+ y³= 81
Differentiating both sides w.r.t. x, we get

(v) x²y²– tan^-1() = cot^-1()
Solution:
x²y²– tan^-1() = cot^-1()
∴ x²y²= tan^-1() + cot^-1()
∴ x²y²= …….[∵ ]
Differentiating both sides w.r.t. x, we get

(vi) xe^y+ ye^x= 1
Solution:
xe^y+ ye^x= 1
Differentiating both sides w.r.t. x, we get

(vii) e^x+y= cos (x – y)
Solution:
e^x+y= cos (x – y)
Differentiating both sides w.r.t. x, we get

(viii) cos (xy) = x + y
Solution:
cos (xy) = x + y
Differentiating both sides w.r.t. x, we get

(ix)
Solution:

∴ e^x-y= log() …….[e^x= y ⇒ x = log y]
∴ e^x-y= log x – log y
Differentiating both sides w.r.t. x, we get

Question 4 Maharashtra Board Solution

Show that in the following, where a and p are constants. (i) x7y5 = (x + y)12

Solution & Step-by-Step Answer:

x7y5 = (x + y)12 (log x7y5) = log(x + y)12 log x7 + log y5 = log(x + y)12 7 log x + 5 log y = 12 log (x + y) Differentiating both sides w.r.t. x, we get

(ii) x^py⁴= (x + y)^p+4, p∈N
Solution:
x^py⁴= (x + y)^p+4
Taking log
log (x^py⁴) = log(x + y)^p+4
log x^p+ log y⁴= (p + 4) log(x + y)
p log x + 4 log y = (p + 4) log(x + y)
Differentiating both sides w.r.t. x, we get

(iii)
Solution:

Differentiating both sides w.r.t. x, we get

Differentiating both sides w.r.t. x, we get

(iv)
Solution:

Differentiating both sides w.r.t. x, we get

(v)
Solution:

(vi)
Solution:

Differentiating both sides w.r.t. x, we get

(vii)
Solution:

Differentiating both sides w.r.t. x, we get

(viii)
Solution:

Differentiating both sides w.r.t. x, we get

Question 5 Maharashtra Board Solution

(i) If log (x + y) = log (xy) + p, where p is a constant, then prove that .

Solution & Step-by-Step Answer:

log (x + y) = log (xy) + p ∴ log (x + y) = log x + log y + p Differentiating both sides w.r.t. x, we get

(ii) If , show that
Solution:

(iii) If , show that
Solution:

Differentiating both sides w.r.t. x, we get

(iv) If e^x+ e^y= e^x+y, then show that
Solution:
e^x+ e^y= e^x+y……(1)
Differentiating both sides w.r.t. x, we get

(v) If , show that
Solution:

2x⁵– 2y⁵= x⁵+ y⁵
3y⁵= x⁵
Differentiating both sides w.r.t. x, we get

∴

(vi) If x^y= e^x-y, then show that
Solution:
x^y= e^x-y
log x^y= log e^x-y
y log x = (x – y) log e
y log x = (x – y) ….. [∵ log e = 1]
y + y log x = x – y
y + y log x = x
y(1 + log x) = x
y =

(vii) If , then show that
Solution:

y²= cos x +
y²= cos x + y
Differentiating both sides w.r.t. x, we get

(viii) If , then show that
Solution:

(ix) If , then show that
Solution:

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

Chapter 1 Differentiation Ex 1.3 Solutions