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Class 12 Maths NCERT Solutions for Chapter 5 Continuity and Differentiability Exercise 5.5

Class 12 Maths NCERT Solutions for Chapter 5 Continuity and Differentiability Exercise 5.5

Continuity and Differentiability Exercise 5.5 Solutions

1. Differentiate the function with respect to x.
cos x . cos 2x . cos 3x 

Solution

Let y = cos x . cos 2x .cos 3x
Taking logarithm on both the sides, we obtain 
log y = log(cos x . cos 2x . cos 3x) 
⇒ log y = log(cos x) + log(cos 2x) + log(cos 3x) 
Differentiating both sides with respect to x, we obtain 


2. Differentiate the function with respect to x. 

Solution

Let y = 
Taking logarithm on both the sides, we obtain 


3. Differentiate the function with respect to x.
(log x)cos x

Solution

Let y = (log x)cos x
Taking logarithm on both the sides, we obtain
log y = cos x. log (log x) 
Differentiating both sides with respect to x, we obtain 

4. Differentiate the function with respect to x. 
x- 2sin x 

Solution

Let y = x- 2sin x

Also, let xx = u and 2sin x = v
∴ y = u - v 
⇒ dy/dx = (du/dx ) - dv/dx 
u = xx 
Taking logarithm on both the sides, we obtain 
log u = x log x
Differentiating both sides with respect to x, we obtain 

v = 2sin x 
Taking logarithm on both the sides with respect to x, we obtain 
log v = sin x. log 2 
Differentiating both sides with respect to x, we obtain 


5. Differentiate the function with respect to x.
(x +3)2 . (x + 4)3 . (x + 5)4    

Solution

Let y = (x +3)2 . (x + 4)3 . (x + 5)4    
Taking logarithm on both the sides, we obtain  
log y = log(x +3)2 + log(x + 4)3 + log(x + 5)4   
⇒ log y = 2log(x + 3) + 3 log(x + 4) + 4 log(x + 5) 
Differentiating both sides with respect to x, we obtain 

6. Differentiate the function with respect to x. 
(x + 1/x)x + x(1 + 1/x)
Solution
Let y = (x + 1/x)x + x(1 + 1/x) 
Also, let u = (x + 1/x)x and v = x(1 + 1/x) 
∴ y = u + v 
⇒ dy/dx = du/dx + dv/dx  ...(1)
Then u = (x + 1/x)x 
⇒ log u = log (x + 1/x)x 
⇒ log u = x log (x + 1/x) 
Differentiating both sides with respect to x, we obtain 
v = x(1 + 1/x) 
⇒ log v = log [x(1 + 1/x) )
⇒ log v = (1 + 1/x) log x 
Differentiating both sides with respect to x, we obtain 

7. Differentiate the function with respect to x. 
(log x)x + xlog x
Solution
Let y = (log x)x + xlog x 
Also, let u = (log x)x and v = xlog x 
∴ y = u + v 
⇒ dy/dx = du/dx + dv/dx  ...(1) 
u = (log x)x 
⇒ log u = log [(log x)x ]
⇒ log u =  x log (log x) 
Differentiating both sides with respect to x, we obtain 

v = xlog x 
⇒ log v = log (xlog x )
⇒ log v = logx . logx = (log x)2 
Differentiating both sides with respect to x, we obtain 

8. Differentiate the function with respect to x. 
(sin x)x + sin-1 √x
Solution
Let y = (sin x)x + sin-1 √x
Also, let u = (sin x)x and v = sin-1 √x
∴ y = u + v 
⇒ dy/dx = du/dx + dv/dx ...(1) 
u = (sin x)x 
⇒ log u = log (sin x)x 
⇒ log u = x log(sin x) 
Differentiating both sides with respect to x, we obtain 

9. Differentiate the function with respect to x. 
xsin x + (sin x)cos x
Solution
Let y = xsin x + (sin x)cos x 
Also, let u = xsin x and v = (sin x)cos x 
∴ y = u + v 
⇒ dy/dx = du/dx + dv/dx  ...(1) 
u = xsin x 
⇒ log u = log(xsin x )
⇒ log u = sin x logx 
Differentiating both sides with respect to x, we obtain 

v = (sin x)cos x 
⇒ log v = log(sin x)cos x 
⇒ log v = cos x log (sin x) 
Differentiating both sides with respect to x, we obtain 

10. Differentiate the function with respect to x. 
xx cos x  + (x2 + 1)/(x2 - 1)
Solution
Let y = xx cos x  + (x2 + 1)/(x2 - 1) 
Also, let u = xx cos x  and v = (x2 + 1)/(x2 - 1) 
∴ y = u + v 
⇒  dy/dx = du/dx + dv/dx  ...(1)
u = xx cos x  
⇒ log u = log (xx cos x  ) 
⇒ log u = x cos x log x 
Differentiating both sides with respect to x, we obtain 
From (1), (2), and (3), we obtain 
dy/dx = xx cos x  [cos x(1 + log x) - x sin x log x] - 4x/(x2 - 1)2  

11. Differentiate the function with respect to x. 
(x cos x)x +  (x sin x)1/x 
Solution
Let y =  (x cos x)x +  (x sin x)1/x 
Also, let u = (x cos x)x and v =  (x sin x)1/x 
∴ y = u + v 
⇒ dy/dx = du/dx + dv/dx  ...(1) 
u = (x cos x)x 
⇒ log u = log (x cos x)x 
⇒ log u = x log(x cos x) 
⇒ log u = x[log x + log cos x] 
⇒ log u = x log x + x log cos x 
Differentiating both sides with respect to x, we obtain 

v = (x sin x)1/x 
⇒ log v = log (x sin x)1/x 
⇒ log v = (1/x) log(x sin x) 
⇒ log v = (1/x)[log x + log sin x]
⇒ log v = (1/x) log x + (1/x) log sin x
Differentiating both sides with respect to x, we obtain 

12. Find dy/dx of function 
xy + yx = 1
Solution
The given function is xy + yx = 1 
Let xy = u and yx = v 
Then, the function becomes u + v = 1 
∴ du/dx + dv/dx = 0  ...(1)
u = xy 
⇒ log u = log (xy )
⇒ log u = y log x
Differentiating both sides with respect to x, we obtain 

13. Find dy/dx of function yx = xy 
Solution
The given function is yx = xy 
Taking logarithm on both the sides, we obtain 
x log y = y log x
Differentiating both sides with respect to x, we obtain 

14. Find dy/dx of function 
(cos x)y + (cos y)x 
Solution
The given function is (cos x)y + (cos y)x 
Taking logarithm on both the sides, we obtain 
y log cos x = x log cos y
differentiating both sides, we obtain 

15. Find dy/dx of function 
xy = e(x - y)
Solution
The given function is xy = e(x - y) 
Taking logarithm on both the sides, we obtain 
log (xy) = log (ex - y )
⇒ log x + log y = (x - y) log e
⇒ log x + log y = (x - y) × 1
⇒ log x + log y = x - y 
Differentiating both sides with respect to x, we obtain 

16. Find the derivative of the function given by f(x) = (1 + x)(1 + x2 )(1 + x4 )(1 + x8 ) and hence find f '(1). 
Solution
The given relationship is f (x) = (1 + x)(1 + x2 )(1 + x4 )(1 + x8 )
Taking logarithm on both the sides, we obtain 
log f(x) = log (1 + x) + log(1 + x2 ) + log (1 + x4 ) + log(1 + x8 )
Differentiating both sides with respect to x, we obtain

17. Differentiate (x2 – 5x + 8) (x3 + 7x + 9) in three ways mentioned
(i) by using product rule
(ii) by expanding the product to obtain a single polynomial.
(iii) by logarithmic differentiation.
Do they all give the same answer?d below:
Solution
(i) Let x2 - 5x + 8 = u and x3 + 7x +9 = v 
∴ y = uv 

(ii) y = (x2 - 5x + 8)(x3 + 7x + 9)
= x2 (x3 + 7x + 9) - 5x(x3 + 7x + 9) + 8(x2 + 7x + 9) 
= x5 + 7x3 + 9x2 - 5x4 - 35x2 - 75x + 8x3 + 56x + 72 
= x5 - 5x4 + 15x3 - 26x2 + 11x + 72 
∴ dy/dx = d/dx (x5 - 5x4 + 15x3 - 26x2 + 11x + 72)
= 5x4 - 5 × 4x3 + 15 × 3x2 - 26 × 2x + 11 × 1 + 0 
= 5x4 - 20x3 + 45x2 - 52x + 11 

(iii) y = (x2 - 5x + 8)(x3 + 7x + 9) 
Taking logarithm on both the sides, we obtain 
log y = log(x2 - 5x + 8) + log(x3 + 7x + 9) 
Differentiating both sides with respect to x, we obtain

18. If u, v and w are functions of x, then show that 
d/dx(u. v. w) = du/dx (u.w) + u . (dv/dx) . w + u.v . dw/dx 
in two ways - first by repeated application of product rule, second by logarithmic differentiation. 
Solution
Let y = u.v.w = u.(v.w)
By applying product rule, we obtain 
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