# Class 12 NCERT Solutions – Mathematics Part I – Chapter 5 Continuity And Differentiability – Exercise 5.5 | Set 2

### Chapter 5 Continuity And Differentiability – Exercise 5.5 | Set 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 }= 1Let 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 + x^{2})(1 + x^{4})(1 + x^{8}) and hence find f'(1).

**Solution:**

Given: f(x) = (x + 1)(x + x

^{2})(1 + x^{4})(1 + x^{8})Find: f'(1)

On taking log on both sides, we get

log(f(x)) = log(1 + x) + log(1 + x

^{2}) + log(1 + x^{4}) + log(1 + x^{8})Now, on differentiating w.r.t x, we get

∴ f'(1) = 2.2.2.2.

f'(1) = 120

### Question 17. Differentiate (x^{5 }– 5x + 8)(x^{3 }+ 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 ruledy/dx = (3x

^{4 }– 15x^{3 }+ 24x^{2 }+ 7x^{2 }– 35x + 56) + (2x^{4 }+ 14x^{2 }+ 18x – 5x^{3 }– 35x – 45)dy/dx = 5x

^{4 }– 20x^{3 }+ 45x^{2 }– 52x + 11

(ii)By expansiony = (x

^{2 }– 5x + 8)(x^{3 }+ 7x + 9)y = x

^{5 }+ 7x^{3 }+ 9x^{2 }– 5x^{4 }– 35x^{2 }– 45x + 8x^{3 }+ 56x + 72y = x

^{5 }– 5x^{4 }+ 15x^{3 }– 26x^{2 }+ 11x + 72dy/dx = 5x

^{4 }– 20x^{3 }+ 45x^{2 }– 52x + 11

(iii)By logarithmic expansionTaking log on both sides

log y = log(x

^{2 }– 5x + 8) + log(x^{3 }+ 7x + 9)Now on differentiating w.r.t. x, we get

dy/dx = 2x

^{4 }+ 14x^{2 }+ 18x – 5x^{3 }– 35x – 45 + 3x^{4 }– 15x^{3 }+ 24x^{2 }+ 7x^{2 }– 35x + 56dy/dx = 5x

^{4 }– 20x^{3 }+ 45x^{2 }– 52x + 11Answer 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 differentiationTaking log on both sides

log y = log u + log v + log w

Now, Differentiating w.r.t. x