NCERT Solutions for Chapter 5 Continuity and Differentiability Class 12 Maths 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. 
xx - 2sin x 

Solution

Let y = xx - 2sin x

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

v = 2^{sin 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 + x^{log x}

Solution

Let y = (log x)^x + x^{log x} 

Also, let u = (log x)^x and v = x^{log 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 = x^{log x} 
⇒ log v = log (x^{log 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. 

x^{sin x} + (sin x)^{cos x}

Solution

Let y = x^{sin x} + (sin x)^{cos x} 
Also, let u = x^{sin x} and v = (sin x)^{cos x} 
∴ y = u + v 
⇒ dy/dx = du/dx + dv/dx  ...(1) 
u = x^{sin x} 
⇒ log u = log(x^{sin 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. 

x^{x cos x}  + (x² + 1)/(x² - 1)

Solution

Let y = x^{x cos x}  + (x² + 1)/(x² - 1) 

Also, let u = x^{x cos x}  and v = (x² + 1)/(x² - 1) 
∴ y = u + v 
⇒  dy/dx = du/dx + dv/dx  ...(1)
u = x^{x cos x}  
⇒ log u = log (x^{x 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 = x^{x 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 

x^y + y^x = 1

Solution

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

13. Find dy/dx of function y^x = x^y 

Solution

The given function is y^x = x^y 

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 (e^{x - 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