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

• Last Updated : 05 Apr, 2021

### (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)

On taking log on both sides, we get

log u = log(x cos 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) ### Question 12. xy + yx = 1

Solution:

Given: xy + yx = 1

Let us considered

u = xy and v = yx

So, ………(1)

So first we take u = xy

On taking log on both sides, we get

log u = log(xy)

log u = y log x

Now, on differentiating w.r.t x, we get   ………(2)

Now we take v = yx

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. yx = xy

Solution:

Given: yx = xy

On taking log on both sides, we get

log(yx) = log(xy)

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 ex – 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

### 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.

### 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   My Personal Notes arrow_drop_up
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