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

### 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 = logx

^{2}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}√xLet us considered y = u + v

Where u = (sin x)

^{x }and v = sin^{–1}√xso, 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}√xOn taking log on both sides, we get

log v = log sin

^{–1}√xNow, 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(x

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

………(3)

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