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

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  • Last Updated : 06 Apr, 2021

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. 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

\frac{1}{u}\frac{du}{dx}=x.(\frac{1}{x})+\log x.1

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

\frac{1}{v}.\frac{dv}{dx}=\log2(\cos x)

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

\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)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

\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 = 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

\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)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

\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 xsin 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 = xxcosx and v = \frac{x^2+1}{x^2-1}

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

\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}

Chapter 5 Continuity And Differentiability – Exercise 5.5 | Set 2


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