# Class 12 NCERT Solutions – Mathematics Part I – Chapter 5 Continuity And Differentiability – Exercise 5.5 | Set 1

• Last Updated : 06 Apr, 2021

### 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. xx – 2sin x

Solution:

Given: y = xx – 2sin x

Let us considered y = u – v

Where, u = xx and v = 2sin x

So, dy/dx = du/dx – dv/dx ………(1)

So first we take u = xx

On taking log on both sides, we get

log u = log x

log u = x log x

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

du/dx = u(1 + log x)

du/dx = xx(1 + log x) ………(2)

Now we take v = 2sin x

On taking log on both sides, we get

log v = log (2sinx)

log v = sin x log2

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

dv/dx = v(log2cos x)

dv/dx = 2sin xcos xlog2 ………(3)

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

dy/dx = xx(1 + log x) – 2sin xcos 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 = xlog 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)

log u = x log(log x)

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

………(2)

Now we take v = xlog x

On taking log on both sides, we get

log v = log(xlog 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 xsin 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 = xxcosx and v =

so, dy/dx = du/dx + dv/dx ………(1)

Now first we take u = xxcosx

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)

### Chapter 5 Continuity And Differentiability – Exercise 5.5 | Set 2

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