Exercise 11.1 Page No: 11.17
Differentiate the following functions from the first principles:
1. e-x
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2. e3x
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3. eax + b
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4. ecos x
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We have to find the derivative of ecos x with the first principle method,
So, let f (x) = ecos x
By using the first principle formula, we get,
Solution:
Exercise 11.2 Page No: 11.37
Differentiate the following functions with respect to x:
1. Sin (3x + 5)
Solution:
Given Sin (3x + 5)
2. tan2 x
Solution:
Given tan2 x
3. tan (xo + 45o)
Solution:
Let y = tan (x° + 45°)
First, we will convert the angle from degrees to radians.
4. Sin (log x)
Solution:
Given sin (log x)
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6. etan x
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7. Sin2 (2x + 1)
Solution:
Let y = sin2 (2x + 1)
On differentiating y with respect to x, we get
8. log7 (2x – 3)
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9. tan 5xo
Solution:
Let y = tan (5x°)
First, we will convert the angle from degrees to radians. We have
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12. logx 3
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18. (log sin x)2
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Let y = (log sin x)2
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21. e3x cos 2x
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22. Sin (log sin x)
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23. etan 3x
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27. tan (esin x)
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30. log (cosec x – cot x)
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33. tan-1 (ex)
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35. sin (2 sin-1 x)
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Let y = sin (2sin–1x)
On differentiating y with respect to x, we get
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Exercise 11.3 Page No: 11.62
Differentiate the following functions with respect to x:
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Let,
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Let,
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7. Sin-1 (2x2 – 1), 0 < x < 1
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8. Sin-1 (1 – 2x2), 0 < x < 1
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Let,
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Exercise 11.4 Page No: 11.74
Find dy/dx in each of the following:
1. xy = c2
Solution:
2. y3 – 3xy2 = x3 + 3x2y
Solution:
Given y3 – 3xy2 = x3 + 3x2y,
Now we have to find dy/dx of given equation, so by differentiating the equation on both sides with respect to x, we get,
3. x2/3 + y2/3 = a2/3
Solution:
Given x2/3 + y2/3 = a2/3,
Now we have to find dy/dx of given equation, so by differentiating the equation on both sides with respect to x, we get,
4. 4x + 3y = log (4x – 3y)
Solution:
Given 4x + 3y = log (4x – 3y),
Now we have to find dy/dx of it, so by differentiating the equation on both sides with respect to x, we get,
Solution:
6. x5 + y5 = 5xy
Solution:
Given x5 + y5 = 5xy
Now we have to find dy/dx of given equation, so by differentiating the equation on both sides with respect to x, we get,
7. (x + y)2 = 2axy
Solution:
Given (x + y)2 = 2axy
Now we have to find dy/dx of given equation, so by differentiating the equation on both sides with respect to x, we get,
8. (x2 + y2)2 = xy
Solution:
Given (x + y)2 = 2axy
Now we have to find dy/dx of given equation, so by differentiating the equation on both sides with respect to x, we get,
9. Tan-1 (x2 + y2)
Solution:
Given tan – 1(x2 + y2) = a,
Now we have to find dy/dx of given function, so by differentiating the equation on both sides with respect to x, we get,
Solution:
11. Sin xy + cos (x + y) = 1
Solution:
Given Sin x y + cos (x + y) = 1
Now we have to find dy/dx of given function, so by differentiating the equation on both sides with respect to x, we get,
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Exercise 11.5 Page No: 11.88
Differentiate the following functions with respect to x:
1. x1/x
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2. xsin x
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3. (1 + cos x)x
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5. (log x)x
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6. (log x)cos x
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Let y = (log x)cos x
Taking log both the sides, we get
7. (Sin x)cos x
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8. ex log x
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9. (Sin x)log x
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10. 10log sin x
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11. (log x)log x
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13. Sin (xx)
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14. (Sin-1 x)x
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16. (tan x)1/x
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18. (i) (xx) √x
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18. (vi) esin x + (tan x)x
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18. (vii) (cos x)x + (sin x)1/x
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19. y = ex + 10x + xx
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20. y = xn + nx + xx + nn
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Exercise 11.6 Page No: 11.98
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Exercise 11.7 Page No: 11.103
Find dy/dx, when
1. x = at2 and y = 2 at
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2. x = a (θ + sin θ) and y = a (1 – cos θ)
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3. x = a cos θ and y = b sin θ
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Given x = a cos θ and y = b sin θ
4. x = a eθ (sin θ – cos θ), y = a eθ (sin θ + cos θ)
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5. x = b sin2 θ and y = a cos2 θ
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6. x = a (1 – cos θ) and y = a (θ + sin θ) at θ = π/2
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9. x = a (cos θ + θ sin θ) and y = a (sin θ – θ cos θ)
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Exercise 11.8 Page No: 11.112
1. Differentiate x2 with respect to x3.
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2. Differentiate log (1 +x2) with respect to tan-1 x.
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3. Differentiate (log x)x with respect to log x.
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4. Differentiate sin-1 √ (1-x2) with respect to cos-1x, if
(i) x ∈ (0, 1)
(ii) x ∈ (-1, 0)
Solution:
(i) Given sin-1 √ (1-x2)
(ii) Given sin-1 √ (1-x2)
Solution:
(i) Let
(ii) Let
(iii) Let
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(i) x ∈ (0, 1/ √2)
(ii) x ∈ (1/√2, 1)
Solution:
(i) Let
(ii) Let
8. Differentiate (cos x)sin x with respect to (sin x)cos x.
Solution:
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