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
$\frac{1}{y}.\frac{dy}{dx}=\frac{1}{\cos x}.\frac{d}{dx}\cos x+\frac{1}{\cos2x}.\frac{d}{dx}\cos2x+\frac{1}{\cos3x}.\frac{d}{dx}\cos3x$
$\frac{1}{y}.\frac{dy}{dx}=\frac{1}{\cos x}.(-\sin x)+\frac{1}{\cos 2x}(-\sin2x).\frac{d}{dx}.2x+\frac{1}{\cos3x}.(-\sin3x).\frac{d}{dx}3x$
$\frac{1}{y}.\frac{dy}{dx}=\frac{-\sin x}{\cos x}-\frac{2\sin2x}{\cos2x}-\frac{3\sin 3x}{\cos3x}$
$\frac{dy}{dx}$ = -y(tan x + 2tan 2x + 3 tan 3x)
$\frac{dy}{dx}$ = -(cos x. cos 2x. cos 3x)(tan x + 2tan 2x + 3tan 3x)

Question 2. $\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}$

Solution:

Let us considered y = $\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}$
Now taking log on both sides, we get
log y = $\log (\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)})^{\frac{1}{2}}$
log y = $\frac{1}{2}$ (log(x – 1)(x – 2)(x – 3)(x – 4)(x – 5))
log y = $\frac{1}{2}$ (log(x – 1) + log(x – 2) – log(x – 3) – log(x – 4) – log(x – 5))
Now, on differentiating w.r.t x, we get
$\frac{1}{y}.\frac{dy}{dx}=\frac{1}{2}[(\frac{1}{x-1})+(\frac{1}{x-2})+(\frac{1}{x-3})+(\frac{1}{x-4})+(\frac{1}{x-5})]$
$\frac{dy}{dx}=\frac{y}{2}[(\frac{1}{x-1})+(\frac{1}{x-2})+(\frac{1}{x-3})+(\frac{1}{x-4})+(\frac{1}{x-5})]$
$\frac{dy}{dx}=\frac{1}{2}\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}[(\frac{1}{x-1})+(\frac{1}{x-2})-(\frac{1}{x-3})-(\frac{1}{x-4})-(\frac{1}{x-5})]$

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
$\frac{1}{y}.\frac{dy}{dx}=\cos x(\frac{1}{\log x}).\frac{d}{dx}\log x+\log(\log x).(-\sin x)$
$\frac{1}{y}.\frac{dy}{dx}=\frac{\cos x}{\log x}.\frac{1}{x}-\sin x \log(\log x)$
$\frac{dy}{dx}=y(\frac{\cos x}{\log x}.\frac{1}{x}-\sin x\log (\log x))$
$\frac{dy}{dx}=(\log x)^{\cos x}(\frac{\cos x}{\log x}.\frac{1}{x}-\sin x \log(\log x))$

Question 4. x^x– 2^{sin x}

Solution:

Given: y = x^x– 2^{sin x}
Let us considered y = u – v
Where, u = x^xand 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
$\frac{1}{u}\frac{du}{dx}=x.(\frac{1}{x})+\log x.1$
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
$\frac{1}{v}.\frac{dv}{dx}=\log2(\cos x)$
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
$\frac{1}{y}.\frac{dy}{dx}=\frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5}$
$\frac{dy}{dx}=y(\frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5})$
$\frac{dy}{dx}=(x+3)^2(x+4)^3(x+5)^4(\frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5})$

Question 6. $(x+\frac{1}{x})^x+x(1+\frac{1}{x})$

Solution:

Given: y = $(x+\frac{1}{x})^x+x(1+\frac{1}{x})$
Let us considered y = u + v
Where $u=(x+\frac{1}{x})^x$ and $v=x^{1+\frac{1}{x}}$
so, dy/dx = du/dx + dv/dx ………(1)
Now first we take $u=(x+\frac{1}{x})^x$
On taking log on both sides, we get
log u = $\log(x+\frac{1}{x})^x$
log u = $xlog(x+\frac{1}{x})$
Now, on differentiating w.r.t x, we get
$\frac{1}{u}.\frac{du}{dx}=x\frac{1}{(x+\frac{1}{x})}.\frac{d}{dx}(x+\frac{1}{x})+\log(x+\frac{1}{x})$
$\frac{1}{u}.\frac{du}{dx}=(\frac{x^2}{x^2+1})(\frac{x^2-1}{x^2})+\log(x+\frac{1}{x})$
$\frac{dy}{dx}=u[\frac{x^2-1}{x^2+1}+\log(x+\frac{1}{x})]$
$\frac{du}{dx}=(x+\frac{1}{x})^x(\frac{x^2-1}{x^2+1}+\log(x+\frac{1}{x}))$ ………(2)
Now we take $v=x^{1+\frac{1}{x}}$
On taking log on both sides, we get
log v = $log x^{(1 + \frac{1}{x})}$
log v = $(1 + \frac{1}{x})log x$
Now, on differentiating w.r.t x, we get
$\frac{1}{v}\frac{dv}{dx}=(1+\frac{1}{x})\frac{d}{dx}(\log x)\log x\frac{d}{dx}(1+\frac{1}{x})$
$\frac{dv}{dx}=v[(\frac{x+1}{x}).\frac{1}{x}+\log x(\frac{-1}{x^2})]$
$\frac{dv}{dx}=x^{1+\frac{1}{x}}.[\frac{1+1-\log x}{x^2}]$ ………(3)
Now put all the values from eq(2) and (3) into eq(1)
$\frac{dy}{dx}=(x+\frac{1}{x})^2(\frac{x^2-1}{x^2+1}+\log(x+\frac{1}{x}))+x^{(1+\frac{1}{x}).[\frac{x+1-\log x}{x^2}]}$

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)^xand 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
$\frac{1}{u}\frac{du}{dx}=x.\frac{d}{dx}\log(\log x)+\log(\log x).\frac{d}{dx}.x$
$\frac{1}{u}\frac{du}{dx}=u(\frac{x}{\log x}.\frac{1}{x}+\log(\log x))$
$\frac{du}{dx}=(\log x)^x(\frac{1}{\log x}+\log(\log x))$ ………(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
$\frac{1}{v}.\frac{dv}{dx}=\frac{d}{dx}(\log^2x)$
$\frac{1}{v}.\frac{dv}{dx}=2\log x\frac{d}{dx}(\log x)$
$\frac{dv}{dx}=v.(2\log x.\frac{1}{x})$
$\frac{dv}{dx}=x^{\log x}.\frac{2\log x }{x}$ ………(3)
Now put all the values from eq(2) and (3) into eq(1)
$\frac{dy}{dx}=(\log x)^x(\frac{1}{\log x}+\log(\log x))+x^{\log x}.\frac{2\log x}{x}$

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)^xand 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
$\frac{1}{u}.\frac{du}{dx}=x\frac{d}{dx}(\log(\sin x))+\log(\sin x).\frac{d}{dx}x$
$\frac{1}{u}.\frac{du}{dx}=x.\frac{1}{\sin x}.\frac{d}{dx}.\sin x+\log(\sin x)$
$\frac{du}{dx}=u(\frac{x\cos x}{\sin x}+\log(\sin x))$
$\frac{du}{dx}=(\sin x)^x.(x\cot x+\log(\sin x))$ ………(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
$\frac{dv}{dx}=\frac{1}{\sqrt{1-(\sqrt{x}^2)}}.\frac{d}{dx}\sqrt{x}$
$\frac{dv}{dx}=\frac{1}{2\sqrt{x-x^2}}$ ………(3)
Now put all the values from eq(2) and (3) into eq(1)
$\frac{dy}{dx}=(\sin x)^x.(x\cot x+\log(\sin x))+\frac{1}{2\sqrt{x-x^2}}$

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
$\frac{1}{u}.\frac{du}{dx}=\sin x\frac{d}{dx}\log x+\log x\frac{d}{dx}\sin x$
$\frac{du}{dx}=u.[\sin x.\frac{1}{x}+\log x(\cos x)]$
$\frac{du}{dx}=x^{\sin x}(\frac{\sin x}{x}+\log x(\cos x))$ ………(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
$\frac{1}{v}\frac{dv}{dx}=\cos x\frac{d}{dx}\log(\sin x)+\log(\sin x).\frac{d}{dx}\cos x$
$\frac{1}{v}\frac{dv}{dx}=(\cos x.\frac{1}{\sin x}.\frac{d}{dx}\sin x+\log(\sin x).(-\sin x))$
$\frac{dv}{dx}=\sin x^{\cos x}(\cot x.\cos x-\sin x\log(\sin x))$ ………(3)
Now put all the values from eq(2) and (3) into eq(1)
$\frac{dy}{dx}=x^{\sin x}(\frac{\sin x}{x}+\log X(\cos x)+\sin x^{\cos x}(\cot x\cos x-\sin x\log(\sin x)))$

Question 10. $x^{x\cos x}+\frac{x^2+1}{x^2-1}$

Solution:

Given: y = $x^{x\cos x}+\frac{x^2+1}{x^2-1}$
Let us considered y = u + v
Where u = x^xcosxand v = $\frac{x^2+1}{x^2-1}$
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
$\frac{1}{u}\frac{du}{dx}=\frac{d}{dx}(x).cosx.logx+x.\frac{d}{dx}(cosx).logx+x.cosx.\frac{d}{dx}(logx)$
$\frac{du}{dx}=u[1.cosx.logx+x.(-sinx).logx+x.cosx.\frac{1}{x}]$
$\frac{du}{dx}=x^{xcosx}[cosx(1+logx)-xsinxlogx]$ ………(2)
Now we take v = $\frac{x^2+1}{x^2-1}$
On taking log on both sides, we get
log v = log $\frac{x^2+1}{x^2-1}$
log v = log(x² + 1) – log(x²– 1)
Now, on differentiating w.r.t x, we get
$\frac{1}{v}\frac{dv}{dx}=\frac{1}{x^2+1}.\frac{d}{dx}(x^2+1)-\frac{1}{x^2-1}.\frac{d}{dx}(x^2-1)$
$\frac{fv}{dx}=v(\frac{2x}{x^2+1}-\frac{2x}{x^2-1})$
$\frac{dv}{dx}=\frac{(x^2+1)}{(x^2-1)}(2x).(\frac{(x^2-1)-(x^2+1)}{(x^2+1)(x^2-1)})$
$\frac{dv}{dx}=\frac{(2x)(-2)}{(x^2-1)^2}$
$\frac{dv}{dx}=\frac{-4x}{(x^2-1)^2}$ ………(3)
Now put all the values from eq(2) and (3) into eq(1)
$\frac{dy}{dx}=x^{xcosx}[cosx(1+logx)-xsinxlogx]-\frac{4x}{(x^2-1)^2}$