# RD Sharma Class 12 Ex 11.5 Solutions Chapter 11 Differentiatio

Here we provide RD Sharma Class 12 Ex 11.5 Solutions Chapter 11 Differentiation for English medium students, Which will very helpful for every student in their exams. Students can download the RD Sharma Class 12 Ex 11.5 Solutions Chapter 11 Differentiation book pdf download. Now you will get step-by-step solutions to each question.

## RD Sharma Class 12 Ex 11.5 Solutions Chapter 11 Differentiatio

### Question 1. Differentiate y = x1/x with respect to x.

Solution:

We have,

=> y = x1/x

On taking log of both the sides, we get,

=> log y = log x1/x

=> log y = (1/x) (log x)

On differentiating both sides with respect to x, we get,

=>

=>

=>

=>

=>

=>

### Question 2. Differentiate y = xsin x with respect to x.

Solution:

We have,

=> y = xsin x

On taking log of both the sides, we get,

=> log y = log xsin x

=> log y = sin x log x

On differentiating both sides with respect to x, we get,

=>

=>

=>

=>

=>

### Question 3. Differentiate y = (1 + cos x)x with respect to x.

Solution:

We have,

=> y = (1 + cos x)x

On taking log of both the sides, we get,

=> log y = log (1 + cos x)x

=> log y = x log (1 + cos x)

On differentiating both sides with respect to x, we get,

=>

=>

=>

=>

=>

=>

### Question 4. Differentiate  with respect to x.

Solution:

We have,

=>

On taking log of both the sides, we get,

=> log y = log

=> log y = cos−1 x log x

On differentiating both sides with respect to x, we get,

=>

=>

=>

=>

=>

### Question 5. Differentiate y = (log x)x with respect to x.

Solution:

We have,

=> y = (log x)x

On taking log of both the sides, we get,

=> log y = log (log x)x

=> log y = x log (log x)

On differentiating both sides with respect to x, we get,

=>

=>

=>

=>

=>

### Question 6. Differentiate y = (log x)cos x with respect to x.

Solution:

We have,

=> y = (log x)cos x

On taking log of both the sides, we get,

=> log y = log (log x)cos x

=> log y = cos x log (log x)

On differentiating both sides with respect to x, we get,

=>

=>

=>

=>

=>

### Question 7. Differentiate y = (sin x)cos x with respect to x.

Solution:

We have,

=> y = (sin x)cos x

On taking log of both the sides, we get,

=> log y = log (sin x)cos x

=> log y = cos x log (sin x)

On differentiating both sides with respect to x, we get,

=>

=>

=>

=>

=>

=>

### Question 8. Differentiate y = ex log x with respect to x.

Solution:

We have,

=> y=ex log x

=> y =

=> y = xx

On taking log of both the sides, we get,

=> log y = log xx

=> log y = x log x

On differentiating both sides with respect to x, we get,

=>

=>

=>

=>

=>

### Question 9. Differentiate y = (sin x)log x with respect to x.

Solution:

We have,

=> y = (sin x)log x

On taking log of both the sides, we get,

=> log y = log (sin x)log x

=> log y = log x log (sin x)

On differentiating both sides with respect to x, we get,

=>

=>

=>

=>

=>

### Question 10. Differentiate y = 10log sin x with respect to x.

Solution:

We have,

=> y = 10log sin x

On taking log of both the sides, we get,

=> log y = log 10log sin x

=> log y = log (sin x) log 10

On differentiating both sides with respect to x, we get,

=>

=>

=>

=>

=>

=>

### Question 11. Differentiate y = (log x)log x with respect to x.

Solution:

We have,

=> y = (log x)log x

On taking log of both the sides, we get,

=> log y = log (log x)log x

=> log y = log x log (log x)

On differentiating both sides with respect to x, we get,

=>

=>

=>

=>

=>

=>

### Question 12. Differentiate  with respect to x.

Solution:

We have,

=>

On taking log of both the sides, we get,

=> log y = log

=> log y = 10x log 10

On differentiating both sides with respect to x, we get,

=>

=>

=>

=>

=>

=>

### Question 13. Differentiate y = sin xx with respect to x.

Solution:

We have,

=> y = sin xx

=> sin−1 y = xx

On taking log of both the sides, we get,

=> log (sin−1 y) = log xx

=> log (sin−1 y) = x log x

On differentiating both sides with respect to x, we get,

=>

=>

=>

=>

=>

=>

=>

=>

### Question 14. Differentiate y = (sin−1x)x with respect to x.

Solution:

We have,

=> y = (sin−1x)x

On taking log of both the sides, we get,

=> log y = (sin−1x)x

=> log y = x log (sin−1x)

On differentiating both sides with respect to x, we get,

=>

=>

=>

=>

=>

### Question 15. Differentiate  with respect to x.

Solution:

We have,

=>

On taking log of both the sides, we get,

=> log y = log

=> log y = sin−1x log x

On differentiating both sides with respect to x, we get,

=>

=>

=>

=>

=>

### Question 16. Differentiate  with respect to x.

Solution:

We have,

=>

On taking log of both the sides, we get,

=> log y = log

=> log y =

On differentiating both sides with respect to x, we get,

=>

=>

=>

=>

=>

### Question 17. Differentiate  with respect to x.

Solution:

We have,

=>

On taking log of both the sides, we get,

=> log y = log

=> log y = tan−1 x log x

On differentiating both sides with respect to x, we get,

=>

=>

=>

=>

=>

### (i) y = xx √x

Solution:

We have,

=> y = xx √x

On taking log of both the sides, we get,

=> log y = log (xx √x)

=> log y = log xx + log √x

=> log y = x log x +

On differentiating both sides with respect to x, we get,

=>

=>

=>

=>

=>

### (ii)

Solution:

We have,

=>

=>

=>

On differentiating both sides with respect to x, we get,

=>

=>

=>

=>

=>

### (iii)

Solution:

We have,

=>

=>

=>

On differentiating both sides with respect to x, we get,

=>

=>

=>

=>

=>

### (iv) y = (x cos x)x + (x sin x)1/x

Solution:

We have,

=> y=(x cos x)x + (x sin x)1/x

=>

=>

=>

On differentiating both sides with respect to x, we get,

=>

=>

=>

=>

### (v)

Solution:

We have,

=>

=>

=>

On differentiating both sides with respect to x, we get,

=>

=>

=>

=>

=>

### (vi) y = esin x + (tan x)x

Solution:

We have,

=> y = esin x + (tan x)x

=>

=>

On differentiating both sides with respect to x, we get,

=>

=>

=>

### (vii) y = (cos x)x + (sin x)1/x

Solution:

We have,

=> y = (cos x)x + (sin x)1/x

=>

=>

On differentiating both sides with respect to x, we get,

=>

=>

=>

### (viii) , for x > 3

Solution:

We have,

=>

=>

=>

On differentiating both sides with respect to x, we get,

=>

=>

=>

### Question 19. Find dy/dx when y = ex + 10x + xx.

Solution:

We have,

=> y = ex + 10x + xx

=>

=>

On differentiating both sides with respect to x, we get,

=>

=>

=>

=>

### Question 20. Find dy/dx when y = xn + nx + xx + nn.

Solution:

We have,

=> y = xn + nx + xx + nn

=>

=>

On differentiating both sides with respect to x, we get,

=>

=>

=>

=>

### Question 21. Find dy/dx when .

Solution:

We have,

=>

=>

On taking log of both the sides, we get,

=>

=>

=>

On differentiating both sides with respect to x, we get,

=>

=>

=>

=>

=>

### Question 22. Find dy/dx when .

Solution:

We have,

=>

=>

On taking log of both the sides, we get,

=>

=>

=>

On differentiating both sides with respect to x, we get,

=>

=>

=>

=>

=>

### Question 23. Find dy/dx when y = e3x sin 4x 2x.

Solution:

We have

=> y = e3x sin 4x 2x.

On taking log of both the sides, we get,

=> log y = log (e3x sin 4x 2x)

=> log y = log e3x + log (sin 4x) + log 2x

=> log y = 3x log e + log (sin 4x) + x log 2

=> log y = 3x + log (sin 4x) + x log 2

On differentiating both sides with respect to x, we get,

=>

=>

=>

=>

=>

### Question 24. Find dy/dx when y = sin x sin 2x sin 3x sin 4x.

Solution:

We have,

=> y = sin x sin 2x sin 3x sin 4x

On taking log of both the sides, we get,

=> log y = log (sin x sin 2x sin 3x sin 4x)

=> log y = log sin x + log sin 2x + log sin 3x + log sin 4x

On differentiating both sides with respect to x, we get,

=>

=>

=> = cotx + 2cot2x + 3cot3x + 4cot4x

=> = y(cotx + 2cot2x + 3cot3x + 4cot4x)

=> = (sinxsin2x sin3xsin4x)(cotx + 2cot2x + 3cot3x + 4cot4x)

### Question 25. Find dy/dx when y = xsin x + (sin x)x.

Solution:

We have,

=> y = xsin x + (sin x)x

Let u = xsin x and v = (sin x)x. Therefore, y = u + v.

Now, u = xsin x

On taking log of both the sides, we get,

=> log u = log xsin x

=> log u = sin x log x

On differentiating both sides with respect to x, we get,

=>

=>

=>

=>

=>

Also, v = (sin x)x

On taking log of both the sides, we get,

=> log v = log (sin x)x

=> log v = x log sin x

On differentiating both sides with respect to x, we get,

=>

=>

=>

=>

=>

Now we have, y = u + v.

=>

=>

### Question 26. Find dy/dx when y = (sin x)cos x + (cos x)sin x.

Solution:

We have,

=> y = (sin x)cos x + (cos x)sin x

=>

=>

On differentiating both sides with respect to x, we get,

=>

=>

=>  = (sinx)cosx[cosxcotx – sinxlog(sinx)] + (cosx)sinx[-tanxsinx + cosxlog(cosx)]

=>  = (sinx)cosx[cosxcotx – sinxlog(sinx)] + (cosx)sinx[cosxlog(cosx) – tanxsinx]

### Question 27. Find dy/dx when y = (tan x)cot x + (cot x)tan x.

Solution:

We have,

=> y = (tan x)cot x + (cot x)tan x

=>

=>

On differentiating both sides with respect to x, we get,

=>

=>

=>

=>  = (tanx)cotx[cosec2x – log(tanx)(cosec2x)] + (cotx)tanx[-sec2x + log(cotx)(sec2x)]

=>  = (tanx)cotx[cosec2x – cosec2xlog(tanx)] + (cotx)tanx[sec2xlog(cotx) – sec2x]

### Question 28. Find dy/dx when y = (sin x)x + sin−1 √x.

Solution:

We have,

=> y = (sin x)x + sin−1 √x

=>

=>

On differentiating both sides with respect to x, we get,

=>

=>

=>

=>

### (i) y = xcos x + (sin x)tan x

Solution:

We have,

=> y = xcos x + (sin x)tan x

=>

=>

On differentiating both sides with respect to x, we get,

=>

=>

=>

=>

### (ii) y = xx + (sin x)x

Solution:

We have,

=> y = xx + (sin x)x

=>

=>

On differentiating both sides with respect to x, we get,

=>

=>

=>

=>

### Question 30. Find dy/dx when y = (tan x)log x + cos2 (π/4).

Solution:

We have,

=> y = (tan x)log x + cos2 (π/4)

=>

=>

On differentiating both sides with respect to x, we get,

=>

=>

=>

=>

### Question 31. Find dy/dx when .

Solution:

We have,

=>

=>

=>

On differentiating both sides with respect to x, we get,

=>

=>

=>

=>

=>

### Question 32. Find dy/dx when y = (log x)x+ xlogx.

Solution:

We have,

=> y = (log x)x+ xlogx

Let u = (log x)x and v = xlogx. Therefore, y = u + v.

Now, u = (log x)x

On taking log of both the sides, we get,

=> log u = log (log x)x

=> log u = x log (log x)

On differentiating both sides with respect to x, we get,

=>

=>

=>

=>

=>

=>

=>

Also, v = xlogx

On taking log of both the sides, we get,

=> log v = log xlogx

=> log v = log x (log x)

=> log v = (log x)2

On differentiating both sides with respect to x, we get,

=>

=>

=>

=>

=>

Now, y = u + v

=>

=>

### Question 33. If x13y7 = (x+y)20, prove that .

Solution:

We have,

=> x13y7 = (x+y)20

On taking log of both the sides, we get,

=> log x13y= log (x+y)20

=> log x13 + log y= log (x+y)20

=> 13 log x + 7 log y = 20 log (x+y)

On differentiating both sides with respect to x, we get,

=>

=>

=>

=>

=>

=>

=>

=>

Hence proved.

### Question 34. If x16y9 = (x2 + y)17, prove that .

Solution:

We have,

=> x16y9 = (x+ y)17

On taking log of both the sides, we get,

=> log x16y9 = log (x2 + y)17

=> log x16 + log y9 = log (x2 +y)17

=> 16 log x + 9 log y = 17 log (x2 + y)

On differentiating both sides with respect to x, we get,

=>

=>

=>

=>

=>

=>

=>

=>

=>

Hence proved.

### Question 35. If y = sin xx, prove that .

Solution:

We have,

=> y = sin xx

Let u = xx. Now y = sin u.

On taking log of both the sides, we get,

=> log u = log xx

=> log u = x log x

On differentiating both sides with respect to x, we get,

=>

=>

=>

=>

Now, y = sin u

=>

=>

=>

Hence proved.

### Question 36. If xx + yx = 1, prove that .

Solution:

We have,

=> xx + yx = 1

=>

=>

On differentiating both sides with respect to x, we get,

=>

=>

=>

=>

=>

=>

Hence proved.

### Question 37. If xy × yx = 1, prove that .

Solution:

We have,

=> xy × yx = 1

On taking log of both the sides, we get,

=> log (x× yx) = log 1

=> log x+ log yx = log 1

=> y log x + x log y = log 1

On differentiating both sides with respect to x, we get,

=>

=>

=>

=>

=>

=>

Hence proved.

### Question 38. If xy + yx = (x+y)x+y, find dy/dx.

Solution:

We have,

=> xy + yx = (x+y)x+y

=>

=>

On differentiating both sides with respect to x, we get,

=>

=>

=>

=>

=>

### Question 39. If xm yn = 1, prove that .

Solution:

We have,

=> xm yn = 1

On taking log of both the sides, we get,

=> log (xm yn)= log 1

=> log xm + log yn = log 1

=> m log x + n log y = log 1

On differentiating both sides with respect to x, we get,

=>

=>

=>

=>

Hence proved.

### Question 40. If yx = ey−x, prove that .

Solution:

We have,

=> yx = ey−x

On taking log of both the sides, we get,

=> log yx = log ey−x

=> x log y = (y − x) log e

=> x log y = y − x

On differentiating both sides with respect to x, we get,

=>

=>

=>

=>

=>

=>

=>

=>

Hence proved.

Question 41. If (sin x)y = (cos y)x, prove that .
Solution:

We have,
=> (sin x)y = (cos y)x
On taking log of both the sides, we get,
=> log (sin x)y = log (cos y)x
=> y log (sin x) = x log (cos y)
On differentiating both sides with respect to x, we get,
=>
=>
=>
=>
Hence proved.
Question 42. If (cos x)y = (tan y)x, prove that \frac{dy}{dx}=\frac{log tany+ytanx}{logcosx-xsecycosecy}     .
Solution:
We have, (cos x)y = (tan y)x
On taking log of both the sides, we get,
=> log (cos x)y = log (tan y)x
=> y log (cos x) = x log (tan y)
On differentiating both sides with respect to x, we get,
=>
=>
=>
=>
=>
=>
=>
=>
=>
=>
Hence proved.
Question 43. If ex + ey = ex+y, prove that .
Solution:
We have,
=> ex + ey = ex+y
On differentiating both sides with respect to x, we get,
=>
=>
=>
=>
=>
=>
=>
=>
Hence proved.
Question 44. If ey = yx, prove that .
Solution:
We have,
=> ey = yx
On taking log of both the sides, we get,
=> log ey = log yx
=> y log e = x log y
=> y = x log y
On differentiating both sides with respect to x, we get,
=>
=>
=>
=>
=>
=>
=>
=>
=>
=>
Hence proved.
Question 45. If ex+y − x = 0, prove that .
Solution:
We have,
=> ex+y − x = 0
On differentiating both sides with respect to x, we get,
=>
=>
Now, we know ex+y − x = 0
=> ex+y = x
So, we get,
=>
=>
=>
=>
=>
Hence proved.
Question 46. If y = x sin (a+y), prove that .
Solution:
We have,
=> y = x sin (a+y)
On differentiating both sides with respect to x, we get,
=>
=>
=>
=>
Now we know, y = x sin (a+y)
=>
So, we get,
=>
=>
=>
=>
Hence proved.
Question 47. If x sin (a+y) + sin a cos (a+y) = 0, prove that .
Solution:
We have,
=> x sin (a+y) + sin a cos (a+y) = 0
On differentiating both sides with respect to x, we get,
=>
=>
=>
=>
Now we know, x sin (a+y) + sin a cos (a+y) = 0
=>
So, we get,
=>
=>
=>
=>
=>
=>
=>
Hence proved.
Question 48. If (sin x)y = x + y, prove that .
Solution:
We have,
=> (sin x)y = x + y
On taking log of both the sides, we get,
=> log (sin x)y = log (x + y)
=> y log sin x = log (x + y)
On differentiating both sides with respect to x, we get,
=>
=>
=>
=>
=>
=>
=>
Hence proved.
Question 49. If xy log (x+y) = 1, prove that .
Solution:
We have,
=> xy log (x+y) = 1
On differentiating both sides with respect to x, we get,
=>
=>
=>
Now, we know, xy log (x+y) = 1.
=>
So, we get,
=>
=>
=>
=>
=>
=>
=>
=>
Hence proved.
Question 50. If y = x sin y, prove that .
Solution:
We have,
=> y = x sin y
On differentiating both sides with respect to x, we get,
=>
=>
=>
=>
Now, we know y = x sin y
=>
So, we get,
=>
=>
=>
Hence proved.
Question 51. Find the derivative of the function f(x) given by,
f(x) = (1+x) (1+x2) (1+x4) (1+x8) and hence find f'(1).
Solution:
Here we are given,
=> f(x) = (1+x) (1+x2) (1+x4) (1+x8)
On differentiating both sides with respect to x, we get,
=>
=>
Now, the value of f'(x) at 1 is,
=> f'(1) = (1 + 1) (1 + 1) (1 + 1) (8) + (1 + 1) (1 + 1) (1 + 1) (4) + (1 + 1) (1 + 1) (1 + 1) (2) + (1 + 1) (1 + 1) (1 + 1) (1)
=> f'(1) = (2) (2) (2) (8) + (2) (2) (2) (4) + (2) (2) (2) (2) + (2) (2) (2) (1)
=> f'(1) = 64 + 32 + 16 + 8
=> f'(1) = 120
Therefore, the value of f'(1) is 120.
Question 52. If , find .
Solution:
We are given,
=>
On differentiating both sides with respect to x, we get,
=>
=>
=>
=>
=>
=>
=>
=>
=>
=>
Question 53. If y = (sin x − cos x)sin x−cos x, π/4 < x < 3π/4, find .
Solution:
We have,
=> y = (sin x − cos x)sin x−cos x
On taking log of both the sides, we get,
=> log y = log (sin x − cos x)sin x−cos x
=> log y = (sin x − cos x) log (sin x−cos x)
On differentiating both sides with respect to x, we get,
=>
=> = (1)(cosx + sinx) + (cosx + sinx)log (sin x − cos x)
=> = cosx + sinx + (cosx + sinx)log (sin x − cos x)
=> = (cosx + sinx)(1 + log (sin x − cos x))
=> = y(cosx + sinx)(1 + log (sin x − cos x))
=> = (sinx – cosx)sinx-cosx(cosx + sinx)(1 + log (sin x − cos x))
Question 54. Find dy/dx of function xy = ex-y.
Solution:
We have,
=> xy = ex-y
On taking log of both the sides, we get,
=> log xy = log ex-y
=> log x + log y = (x − y) log e
=> log x + log y = x − y
On differentiating both sides with respect to x, we get,
=>
=>
=>
=>
=>
Question 55. Find dy/dx of function yx + xy + xx = ab.
Solution:
We have,
=> yx + x+ xx = ab
=>
=>
On differentiating both sides with respect to x, we get,
=>
=>
=>
=>
=>
=>
Question 56. If (cos x)y = (cos y)x, find dy/dx.
Solution:
We have,
=> (cos x)y = (cos y)x
On taking log of both the sides, we get,
=> log (cos x)= log (cos y)x
=> y log (cos x) = x log (cos y)
On differentiating both sides with respect to x, we get,
=>
=>
=>
=>
=>
Question 57. If cos y = x cos (a+y), where cos a ≠ ±1, prove that .
Solution:
We have,
=> cos y = x cos (a+y)
On differentiating both sides with respect to x, we get,
=>
=>
=>
=>
=>
=>
=>
=>
=>
=>
Hence proved.
Question 58. If , prove that .
Solution:
We have,
=>
On differentiating both sides with respect to x, we get,
=>
=>
=>
=>
=>
=>
=>
Hence proved.
Question 59. If , prove that .
Solution:
We have,
=>
On taking log of both the sides, we get,
=> log x = log
=>
=>
=>
On differentiating both sides with respect to x, we get,
=>
=>
We know,
=>
So, we get,
=>
=>
=>
=>
=>
Hence proved.
Question 60. If , find dy/dx.
Solution:
We have,
=>
=>
=>
On differentiating both sides with respect to x, we get,
=>
=>
=>
=>
Question 61. If , find dy/dx.
Solution:
We are given,
=>
Now we know,
If  then,
In the given expression, we have 1/x instead of x.
So, using the above theorem, we get,
=>

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment in the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share rdsharmasolutions.in to your friends.