Step-by-step solution provided.

Question 9: Diffrentiate the following w. r. t. x. (i) cos-1

Step-by-step solution provided.

Maharashtra State Board Class 12 (HSC Board) Mathematics & Statistics Solutions Chapter 1 Differentiation Ex 1.2 - Digest Answers | Maharashtra Board Solutions

Balbharti12th Maharashtra State Board Maths Solutions BookPdf Chapter 1 Differentiation Ex 1.2 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 1 Differentiation Ex 1.2

Question 1 Maharashtra Board Solution

Find the derivative of the function y = f (x) using the derivative of the inverse function x = f-1( y) in the following (i) y =

Solution & Step-by-Step Answer:

y = … (1) We have to find the inverse function of y = f(x), i.e. x in terms of y. From (1), y2 = x ∴ x = y2

(ii) y =
Solution:
y = …(1)
We have to find the inverse function of y = f(x), i.e. x in terms of y.
From (1),

(iii) y =
Solution:
y = ….(1)
We have to find the inverse function of y = f(x), i.e. x in terms of y.
From (1),

(iv) y = log (2x – 1)
Solution:
y = log (2x – 1) …(1)
We have to find the inverse function of y = f(x), i.e. x in terms of y.
From (1),

(v) y = 2x + 3
Solution:
y = 2x + 3 ….(1)
We have to find the inverse function of y = f(x), i.e. x in terms of y.
From (1),

(vi) y = e^x– 3
Solution:
y = e^x– 3 ….(1)
We have to find the inverse function of y = f(x), i.e. x in terms of y.
From (1),
e^x= y + 3
∴ x = log(y + 3)
∴ x = f^-1(y) = log(y + 3)

(vii) y = e^{2x – 3}
Solution:
y = e^{2x – 3}….(1)
We have to find the inverse function of y = f(x), i.e. x in terms of y.
From (1),
2x – 3 = log y ∴ 2x = log y + 3

(viii) y = log₂
Solution:
y = log₂ …(1)
We have to find the inverse function of y = f(x), i.e. x in terms of y.
From (1),
= 2^y∴ x = 2∙2^y= 2^y+1
∴ x = f^-1(y) = 2^y+1

Question 2 Maharashtra Board Solution

Find the derivative of the inverse function of the following (i) y = x2·ex

Solution & Step-by-Step Answer:

y = x2·ex Differentiating w.r.t. x, we get

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

(iii) y = x·7^x
Solution:
y = x·7^x
Differentiating w.r.t. x, we get

(iv) y = x²+ logx
Solution:
y = x²+ logx
Differentiating w.r.t. x, we get

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

Question 3 Maharashtra Board Solution

Find the derivative of the inverse of the following functions, and also fid their value at the points indicated against them. (i) y = x5 + 2x3 + 3x, at x = 1

Solution & Step-by-Step Answer:

y = x5 + 2x3 + 3x Differentiating w.r.t. x, we get = (x5 + 2x3 + 3x) = 5x4 + 2 × 3x2 + 3 × 1 = 5x4 + 6x2 + 3 The derivative of inverse function of y = f(x) is given by

(ii) y = e^x+ 3x + 2, at x = 0
Solution:
y = e^x+ 3x + 2
Differentiating w.r.t. x, we get
= (e^x+ 3x + 2)
The derivative of inverse function of y = f(x) is given by

(iii) y = 3x²+ 2 log x³, at x = 1
Solution:
y = 3x²+ 2 log x³
= 3x²+ 6 log x
Differentiating w.r.t. x, we get

The derivative of inverse function of y = f(x) is given by

(iv) y = sin (x – 2) + x², at x = 2
Solution:
y = sin (x – 2) + x²
Differentiating w.r.t. x, we get

Question 4 Maharashtra Board Solution

If f(x) = x3 + x – 2, find (f-1)’ (0). Question is modified. If f(x) = x3 + x – 2, find (f-1)’ (-2).

Solution & Step-by-Step Answer:

f(x) = x3 + x – 2 ….(1) Differentiating w.r.t. x, we get

Question 5 Maharashtra Board Solution

Using derivative prove (i) tan-1x + cot-1x =

Solution & Step-by-Step Answer:

let f(x) = tan-1x + cot-1x Differentiating w.r.t. x, we get Since, f'(x) = 0, f(x) is a constant function. Let f(x) = k. For any value of x, f(x) = k Let x = 0. Then f(0) = k ….(2) From (1), f(0) = tan-1(0) + cot-1(0) = 0 +

(ii) sec^-1x + cosec^-1x = ... [for |x| ≥ 1]
Solution:
Let f(x) = sec^-1x + cosec^-1x for |x| ≥ 1 ….(1)
Differentiating w.r.t. x, we get

Since, f'(x) = 0, f(x) is a constant function.
Let f(x) = k.
For any value of x, f(x) = k, where |x| > 1
Let x = 2.
Then, f(2) = k ……(2)

Question 6 Maharashtra Board Solution

Diffrentiate the following w. r. t. x. (i) tan-1(log x)

Solution & Step-by-Step Answer:

Let y = tan-1(log x) Differentiating w.r.t. x, we get

(ii) cosec^-1(e^-x)
Solution:
Let y = cosec^-1(e^-x)
Differentiating w.r.t. x, we get

(iii) cot^-1(x³)
Solution:
Let y = cot^-1(x³)
Differentiating w.r.t. x, we get

(iv) cot^-1(4^x
Solution:
Let y = cot^-1(4^x
Differentiating w.r.t. x, we get

(v) tan^-1()
Solution:
Let y = tan^-1()
Differentiating w.r.t. x, we get

(vi) sin^-1
Solution:
Let y = sin^-1
Differentiating w.r.t. x, we get

(vii) cos^-1(1 – x²)
Solution:
Let y = cos^-1(1 – x²)
Differentiating w.r.t. x, we get

(viii) sin^-1
Solution:
Let y = sin^-1
Differentiating w.r.t. x, we get

(ix) cos³[cos^-1(x³)]
Solution:
Let y = cos³[cos^-1(x³)]
= [cos(cos^-1x³)]³
= (x³)³= x⁹
Differentiating w.r.t. x, we get
= (x⁹) = 9x⁸.

(x) sin⁴[sin^-1()]
Solution:
Let y = sin⁴[sin^-1()]
= {sin[sin^-1()]}⁸
= ()⁴= x²
Differentiating w.r.t. x, we get
= (x²) = 2x.

Question 7 Maharashtra Board Solution

Diffrentiate the following w. r. t. x. (i) cot-1[cot (ex2)]

Solution & Step-by-Step Answer:

Let y = cot-1[cot (ex2)] = ex2

(ii) cosec^-1
Solution:

(iii) cos^-1
Solution:

(iv) cos^-1
Solution:

(v) tan^-1
Solution:

(vi) cosec^-1
Solution:

(vii) tan^-1
Solution:

(viii) cot^-1
Solution:
Let y = cot^-1

(ix) tan^-1
Solution:

(x) tan^-1
Solution:
Let y = tan^-1

(xi) tan^-1(cosec x + cot x)
Solution:
Let y = tan^-1(cosec x + cot x)

(xii) cot^-1
Solution:

Question 8 Maharashtra Board Solution

(i)

Solution & Step-by-Step Answer:

(ii)
Solution:

(iii)
Solution:

(iv)
Solution:

(v)
Solution:

= e^x.

(vi)
Solution:

y = sin^-1[sin(2^x)∙cosα – cos(2^x)∙sinα]
= sin^–[sin(2^x– α)]
= 2^x– α, where α is a constant
Differentiating w.r.t. x, we get
= (2^x– α)
= (2^x) – (α)
= 2^x∙log2 – 0
= 2^x∙log2