Class 12 NCERT Solutions – Mathematics Part I – Chapter 5 Continuity And Differentiability – Exercise 5.5

Differentiate the functions given in question 1 to 10 with respect to x.

Question 1. cos x.cos2x.cos3x

Solution:

Let us considered y = cos x.cos2x.cos3x

Now taking log on both sides, we get

log y = log(cos x.cos2x.cos3x)

log y = log(cos x) + log(cos 2x) + log (cos 3x)

Now, on differentiating w.r.t x, we get

= -y(tan x + 2tan 2x + 3 tan 3x)

= -(cos x. cos 2x. cos 3x)(tan x + 2tan 2x + 3tan 3x)

Question 2. 

Solution:

Let us considered y = 

Now taking log on both sides, we get

log y = 

log y = (log(x – 1)(x – 2)(x – 3)(x – 4)(x – 5))

log y = (log(x – 1) + log(x – 2) – log(x – 3) – log(x – 4) – log(x – 5))

Now, on differentiating w.r.t x, we get

Question 3. (log x)^{cos x}

Solution:

Let us considered y = (log x)^{cos x}

Now taking log on both sides, we get

log y = log((log x)^{cos x})

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

Now, on differentiating w.r.t x, we get

Question 4. x^x – 2^{sin x}

Solution:

Given: y = x^x – 2^{sin x}

Let us considered y = u – v 

Where, u = x^x and v = 2^{sin x}

So, dy/dx = du/dx – dv/dx ………(1)

So first we take u = x^x

On taking log on both sides, we get

log u = log x^x 

log u = x log x    

Now, on differentiating w.r.t x, we get

du/dx = u(1 + log x) 

du/dx = x^x(1 + log x) ………(2) 

Now we take v = 2^{sin x}

On taking log on both sides, we get

log v = log (2^sinx)

log v = sin x log2

Now, on differentiating w.r.t x, we get

dv/dx = v(log2cos x)

dv/dx = 2^{sin x}cos xlog2 ………(3) 

Now put all the values from eq(2) and (3) into eq(1)

dy/dx = x^x(1 + log x) – 2^{sin x}cos xlog2

Question 5. (x + 3)².(x + 4)³.(x + 5)⁴

Solution:

Let us considered y = (x + 3)².(x + 4)³.(x + 5)⁴

Now taking log on both sides, we get

log y = log[(x + 3)³.(x + 4)³.(x + 5)⁴]

log y = 2 log(x + 3) + 3 log(x + 4) + 4 log(x + 5)

Now, on differentiating w.r.t x, we get

Question 6. 

Solution:

Given: y = 

Let us considered y = u + v

Where  and 

so, dy/dx = du/dx + dv/dx ………(1)

Now first we take 

On taking log on both sides, we get

log u = 

log u = 

Now, on differentiating w.r.t x, we get

 ………(2)

Now we take 

On taking log on both sides, we get

log v = 

 log v = 

Now, on differentiating w.r.t x, we get

  ………(3)

Now put all the values from eq(2) and (3) into eq(1)

Question 7. (log x)^x + x ^{log x}

Solution:

Given: y = (log x)^x + x ^{log x}

Let us considered y = u + v

Where u = (log x)^x and v = x^{log x}

so, dy/dx = du/dx + dv/dx ………(1)

Now first we take u = (log x)^x

On taking log on both sides, we get

log u = log(log x)^x 

log u = x log(log x)

Now, on differentiating w.r.t x, we get

  ………(2)

Now we take v = x^{log x}

On taking log on both sides, we get

log v = log(x^{log x})

log v = logx log(x)

log v = logx²

Now, on differentiating w.r.t x, we get

   ………(3)

Now put all the values from eq(2) and (3) into eq(1)

Question 8. (sin x)^x + sin^–1√x

Solution:

Given: y = (sin x)^x + sin^–1√x

Let us considered y = u + v

Where u = (sin x)^x and v = sin^–1√x

so, dy/dx = du/dx + dv/dx ………(1)

Now first we take u = (sin x)^x

On taking log on both sides, we get

log u = log(sin x)^x

log u = xlog(sin x)

Now, on differentiating w.r.t x, we get

 ………(2)

Now we take v = sin^–1√x

On taking log on both sides, we get

log v = log sin^–1√x

Now, on differentiating w.r.t x, we get

 ………(3)

Now put all the values from eq(2) and (3) into eq(1)

Question 9. x ^{sin x} + (sin x)^{cos x}

Solution:

Given: y = x ^{sin x} + (sin x)^{cos x}

Let us considered y = u + v

Where u = x ^{sin x} and v = (sin x)^{cos x}

so, dy/dx = du/dx + dv/dx ………(1)

Now first we take u = x ^{sin x}

On taking log on both sides, we get

log u = log x^{sin x} 

log u = sin x(log x)

Now, on differentiating w.r.t x, we get

 ………(2)

Now we take v =(sin x)^{cos x}

On taking log on both sides, we get

log v = log(sin x)^{cos x}

log v = cosx log(sinx)

Now, on differentiating w.r.t x, we get

 ………(3)

Now put all the values from eq(2) and (3) into eq(1)

Question 10. 

Solution:

Given: y = 

Let us considered y = u + v

Where u = x^xcosx and v = 

so, dy/dx = du/dx + dv/dx ………(1)

Now first we take u = x^xcosx

On taking log on both sides, we get

log u = log (x ^xcosx)   

log u = x.cosx.logx 

Now, on differentiating w.r.t x, we get

 ………(2)

Now we take v =

On taking log on both sides, we get

log v = log

log v = log(x² + 1) – log(x² – 1)

Now, on differentiating w.r.t x, we get

  ………(3)

Now put all the values from eq(2) and (3) into eq(1)

Question 11. Differentiate the function with respect to x.

(x cos x)^x + (x sin x)^1/x

Solution:

Given: (x cos x)^x + (x sin x)^1/x

Let us considered y = u + v 

Where, u = (x cos x)^x and v = (x sin x)^1/x

So, dy/dx = du/dx + dv/dx ………(1)

So first we take u = (x cos x)^x 

On taking log on both sides, we get

log u = log(x cos x)^x 

log u = xlog(x cos x)

Now, on differentiating w.r.t x, we get

 ………(2)

Now we take u =(x sin x)^1/x

On taking log on both sides, we get

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)

Now, on differentiating w.r.t x, we get

 ………(3)

Now put all the values from eq(2) and (3) into eq(1)

Find dy/dx of  the function given in questions 12 to 15

Question 12. x^y + y^x = 1

Solution:

Given: x^y + y^x = 1

Let us considered

u = x^y and v = y^x

So,

………(1)

So first we take u = x^y

On taking log on both sides, we get

log u = log(x^y)             

log u = y log x

Now, on differentiating w.r.t x, we get

 ………(2)

Now we take v = y^x

On taking log on both sides, we get

log v = log(y)^x

log v = x log y

Now, on differentiating w.r.t x, we get

  ………(3)

Now put all the values from eq(2) and (3) into eq(1)

Question 13. y^x = x^y  

Solution:

Given: y^x = x^y  

On taking log on both sides, we get

log(yx) = log(x^y)         

xlog y = y log x

Now, on differentiating w.r.t x, we get

Question 14. (cos x)^y = (cos y)^x

Solution:

Given: (cos x)^y = (cos y)^x

On taking log on both sides, we get

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

Now, on differentiating w.r.t x, we get

Question 15. xy = e^{(x – y)}

Solution:

Given: xy = e^{(x – y)}

On taking log on both sides, we get

log(xy) = log e^{x – y}

log x + log y = x – y

Now, on differentiating w.r.t x, we get

Question 16. Find the derivative of the function given by f(x) = (x + 1)(x + x²)(1 + x⁴)(1 + x⁸) and hence find f'(1).

Solution:

Given: f(x) = (x + 1)(x + x²)(1 + x⁴)(1 + x⁸)

Find: f'(1)

On taking log on both sides, we get

log(f(x)) = log(1 + x) + log(1 + x²) + log(1 + x⁴) + log(1 + x⁸)

Now, on differentiating w.r.t x, we get

∴ f'(1) = 2.2.2.2.

f'(1) = 120

Question 17. Differentiate (x⁵ – 5x + 8)(x³ + 7x + 9) in three ways mentioned below 

(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? 

Solution:

(i) By using product rule

dy/dx = (3x⁴ – 15x³ + 24x² + 7x² – 35x + 56) + (2x⁴ + 14x² + 18x – 5x³ – 35x – 45)

dy/dx = 5x⁴ – 20x³ + 45x² – 52x + 11

(ii) By expansion 

y = (x² – 5x + 8)(x³ + 7x + 9)

y = x⁵ + 7x³ + 9x² – 5x⁴ – 35x² – 45x + 8x³ + 56x + 72

y = x⁵ – 5x⁴ + 15x³ – 26x² + 11x + 72

dy/dx = 5x⁴ – 20x³ + 45x² – 52x + 11

(iii) By logarithmic expansion 

Taking log on both sides 

log y = log(x² – 5x + 8) + log(x³ + 7x + 9)

Now on differentiating w.r.t. x, we get

dy/dx = 2x⁴ + 14x² + 18x – 5x³ – 35x – 45 + 3x⁴ – 15x³ + 24x² + 7x² – 35x + 56

dy/dx = 5x⁴ – 20x³ + 45x² – 52x + 11

Answer is always same what-so-ever method we use.

Question 18. If u, v and w are function of x, then show that

Solution:

Let y = u.v.w.

Method 1: Using product Rule 

Method 2: Using logarithmic differentiation

Taking log on both sides

log y = log u + log v + log w

Now, Differentiating w.r.t. x