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# Class 12 RD Sharma Solutions – Chapter 11 Differentiation – Exercise 11.5 | Set 2

• Last Updated : 26 May, 2021

### 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 + sinâˆ’1 âˆšx.

Solution:

We have,

=> y = (sin x)x + sinâˆ’1 âˆšx

=>

=>

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 (Ï€/4).

Solution:

We have,

=> y = (tan x)log x + cos2 (Ï€/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 x13y7 = log (x+y)20

=> log x13 + log y7 = 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 = (x2 + 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 Ã— yx = 1, prove that .

Solution:

We have,

=> xy Ã— yx = 1

On taking log of both the sides, we get,

=> log (xy Ã— yx) = log 1

=> log xy + 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 = eyâˆ’x, prove that .

Solution:

We have,

=> yx = eyâˆ’x

On taking log of both the sides, we get,

=> log yx = log eyâˆ’x

=> x log y = (y âˆ’ x) log e

=> x log y = y âˆ’ x

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

=>

=>

=>

=>

=>

=>

=>

=>

Hence proved.

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