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

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

[Tex]\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[/Tex]

[Tex]\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[/Tex]

[Tex]\frac{1}{y}.\frac{dy}{dx}=\frac{-\sin x}{\cos x}-\frac{2\sin2x}{\cos2x}-\frac{3\sin 3x}{\cos3x}[/Tex]

[Tex]\frac{dy}{dx}    [/Tex]= -y(tan x + 2tan 2x + 3 tan 3x)

[Tex]\frac{dy}{dx}    [/Tex]= -(cos x. cos 2x. cos 3x)(tan x + 2tan 2x + 3tan 3x)

Question 2. [Tex]\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}[/Tex]

Solution:

Let us considered y = [Tex]\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}[/Tex]

Now taking log on both sides, we get

log y = [Tex]\log (\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)})^{\frac{1}{2}}[/Tex]

log y = [Tex]\frac{1}{2}    [/Tex](log(x – 1)(x – 2)(x – 3)(x – 4)(x – 5))

log y = [Tex]\frac{1}{2}    [/Tex](log(x – 1) + log(x – 2) – log(x – 3) – log(x – 4) – log(x – 5))

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

[Tex]\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})][/Tex]

[Tex]\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})][/Tex]

[Tex]\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})][/Tex]

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

[Tex]\frac{1}{y}.\frac{dy}{dx}=\cos x(\frac{1}{\log x}).\frac{d}{dx}\log x+\log(\log x).(-\sin x)[/Tex]

[Tex]\frac{1}{y}.\frac{dy}{dx}=\frac{\cos x}{\log x}.\frac{1}{x}-\sin x \log(\log x)[/Tex]

[Tex]\frac{dy}{dx}=y(\frac{\cos x}{\log x}.\frac{1}{x}-\sin x\log (\log x))[/Tex]

[Tex]\frac{dy}{dx}=(\log x)^{\cos x}(\frac{\cos x}{\log x}.\frac{1}{x}-\sin x \log(\log x))[/Tex]

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

[Tex]\frac{1}{u}\frac{du}{dx}=x.(\frac{1}{x})+\log x.1 [/Tex]

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

[Tex]\frac{1}{v}.\frac{dv}{dx}=\log2(\cos x)[/Tex]

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

[Tex]\frac{1}{y}.\frac{dy}{dx}=\frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5}[/Tex]

[Tex]\frac{dy}{dx}=y(\frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5})[/Tex]

[Tex]\frac{dy}{dx}=(x+3)^2(x+4)^3(x+5)^4(\frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5})[/Tex]

Question 6. [Tex](x+\frac{1}{x})^x+x(1+\frac{1}{x})[/Tex]

Solution:

Given: y = [Tex](x+\frac{1}{x})^x+x(1+\frac{1}{x})[/Tex]

Let us considered y = u + v

Where [Tex]u=(x+\frac{1}{x})^x   [/Tex] and [Tex]v=x^{1+\frac{1}{x}}[/Tex]

so, dy/dx = du/dx + dv/dx ………(1)

Now first we take [Tex]u=(x+\frac{1}{x})^x[/Tex]

On taking log on both sides, we get

log u = [Tex]\log(x+\frac{1}{x})^x    [/Tex]           

log u = [Tex]xlog(x+\frac{1}{x}) [/Tex]

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

[Tex]\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})[/Tex]

[Tex]\frac{1}{u}.\frac{du}{dx}=(\frac{x^2}{x^2+1})(\frac{x^2-1}{x^2})+\log(x+\frac{1}{x})[/Tex]

[Tex]\frac{dy}{dx}=u[\frac{x^2-1}{x^2+1}+\log(x+\frac{1}{x})][/Tex]

[Tex]\frac{du}{dx}=(x+\frac{1}{x})^x(\frac{x^2-1}{x^2+1}+\log(x+\frac{1}{x}))   [/Tex] ………(2)

Now we take [Tex]v=x^{1+\frac{1}{x}}[/Tex]

On taking log on both sides, we get

log v = [Tex]log x^{(1 + \frac{1}{x})} [/Tex]

 log v = [Tex](1 + \frac{1}{x})log x[/Tex]

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

 [Tex]\frac{1}{v}\frac{dv}{dx}=(1+\frac{1}{x})\frac{d}{dx}(\log x)\log x\frac{d}{dx}(1+\frac{1}{x})[/Tex]

[Tex]\frac{dv}{dx}=v[(\frac{x+1}{x}).\frac{1}{x}+\log x(\frac{-1}{x^2})][/Tex]

[Tex]\frac{dv}{dx}=x^{1+\frac{1}{x}}.[\frac{1+1-\log x}{x^2}]   [/Tex]  ………(3)

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

[Tex]\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}]}[/Tex]

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

[Tex]\frac{1}{u}\frac{du}{dx}=x.\frac{d}{dx}\log(\log x)+\log(\log x).\frac{d}{dx}.x    [/Tex] 

[Tex]\frac{1}{u}\frac{du}{dx}=u(\frac{x}{\log x}.\frac{1}{x}+\log(\log x))[/Tex]

[Tex]\frac{du}{dx}=(\log x)^x(\frac{1}{\log x}+\log(\log x))   [/Tex]  ………(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

[Tex]\frac{1}{v}.\frac{dv}{dx}=\frac{d}{dx}(\log^2x)[/Tex]

[Tex]\frac{1}{v}.\frac{dv}{dx}=2\log x\frac{d}{dx}(\log x)[/Tex]

[Tex]\frac{dv}{dx}=v.(2\log x.\frac{1}{x})[/Tex]

[Tex]\frac{dv}{dx}=x^{\log x}.\frac{2\log x }{x}   [/Tex]   ………(3)

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

[Tex]\frac{dy}{dx}=(\log x)^x(\frac{1}{\log x}+\log(\log x))+x^{\log x}.\frac{2\log x}{x}[/Tex]

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

[Tex]\frac{1}{u}.\frac{du}{dx}=x\frac{d}{dx}(\log(\sin x))+\log(\sin x).\frac{d}{dx}x    [/Tex] 

[Tex]\frac{1}{u}.\frac{du}{dx}=x.\frac{1}{\sin x}.\frac{d}{dx}.\sin x+\log(\sin x)[/Tex]

[Tex]\frac{du}{dx}=u(\frac{x\cos x}{\sin x}+\log(\sin x))[/Tex]

[Tex]\frac{du}{dx}=(\sin x)^x.(x\cot x+\log(\sin x))   [/Tex] ………(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

[Tex]\frac{dv}{dx}=\frac{1}{\sqrt{1-(\sqrt{x}^2)}}.\frac{d}{dx}\sqrt{x}[/Tex]

[Tex]\frac{dv}{dx}=\frac{1}{2\sqrt{x-x^2}}   [/Tex] ………(3)

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

[Tex]\frac{dy}{dx}=(\sin x)^x.(x\cot x+\log(\sin x))+\frac{1}{2\sqrt{x-x^2}}[/Tex]

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

[Tex]\frac{1}{u}.\frac{du}{dx}=\sin x\frac{d}{dx}\log x+\log x\frac{d}{dx}\sin x    [/Tex] 

[Tex]\frac{du}{dx}=u.[\sin x.\frac{1}{x}+\log x(\cos x)][/Tex]

[Tex]\frac{du}{dx}=x^{\sin x}(\frac{\sin x}{x}+\log x(\cos x))   [/Tex] ………(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

[Tex]\frac{1}{v}\frac{dv}{dx}=\cos x\frac{d}{dx}\log(\sin x)+\log(\sin x).\frac{d}{dx}\cos x[/Tex]

[Tex]\frac{1}{v}\frac{dv}{dx}=(\cos x.\frac{1}{\sin x}.\frac{d}{dx}\sin x+\log(\sin x).(-\sin x))[/Tex]

[Tex]\frac{dv}{dx}=\sin x^{\cos  x}(\cot x.\cos x-\sin x\log(\sin x))   [/Tex] ………(3)

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

[Tex]\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)))[/Tex]

Question 10. [Tex]x^{x\cos x}+\frac{x^2+1}{x^2-1}[/Tex]

Solution:

Given: y = [Tex]x^{x\cos x}+\frac{x^2+1}{x^2-1}[/Tex]

Let us considered y = u + v

Where u = x^xcosx and v = [Tex]\frac{x^2+1}{x^2-1}[/Tex]

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

[Tex]\frac{1}{u}\frac{du}{dx}=\frac{d}{dx}(x).cosx.logx+x.\frac{d}{dx}(cosx).logx+x.cosx.\frac{d}{dx}(logx)[/Tex]

[Tex]\frac{du}{dx}=u[1.cosx.logx+x.(-sinx).logx+x.cosx.\frac{1}{x}][/Tex]

[Tex]\frac{du}{dx}=x^{xcosx}[cosx(1+logx)-xsinxlogx]   [/Tex] ………(2)

Now we take v =[Tex]\frac{x^2+1}{x^2-1}[/Tex]

On taking log on both sides, we get

log v = log[Tex]\frac{x^2+1}{x^2-1}[/Tex]

log v = log(x² + 1) – log(x² – 1)

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

[Tex]\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)[/Tex]

[Tex]\frac{fv}{dx}=v(\frac{2x}{x^2+1}-\frac{2x}{x^2-1})[/Tex]

[Tex]\frac{dv}{dx}=\frac{(x^2+1)}{(x^2-1)}(2x).(\frac{(x^2-1)-(x^2+1)}{(x^2+1)(x^2-1)})[/Tex]

[Tex]\frac{dv}{dx}=\frac{(2x)(-2)}{(x^2-1)^2}    [/Tex]                       

[Tex]\frac{dv}{dx}=\frac{-4x}{(x^2-1)^2}   [/Tex]  ………(3)

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

[Tex]\frac{dy}{dx}=x^{xcosx}[cosx(1+logx)-xsinxlogx]-\frac{4x}{(x^2-1)^2}[/Tex]

Question 11. Differentiate the function with respect to x.

(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)^x 

On taking log on both sides, we get

log u = log(x cos x)^x 

log u = xlog(x cos x)

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

[Tex]\frac{1}{u}\frac{du}{dx}=x\frac{d}{dx}(\log x+\log(\cos x))+\log x+\log \cos x[/Tex]

[Tex]\frac{1}{u}\frac{du}{dx}=x(\frac{1}{x}+\frac{1}{\cos x}\frac{d}{dx}\cos x)+\log x+\log\cos x[/Tex]

[Tex]\frac{du}{dx}=u(x(\frac{1}{x}+\frac{-\sin x}{\cos x})+\log x+\log(\cos x))[/Tex]

[Tex]\frac{du}{dx}=(x\cos x)^x(1-x\tan x+\log x+\log(\cos x)) [/Tex] ………(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

[Tex]\frac{1}{v}\frac{dv}{dx}=\frac{1}{x}\frac{d}{dx}(\log x+\log(\sin x)+\frac{d}{dx}(\frac{1}{x}).(\log x+\log(\sin x)))[/Tex]

[Tex]\frac{1}{v}.\frac{dv}{dx}=\frac{1}{x}(\frac{1}{x}+\frac{1}{\sin x}.\frac{d}{dx}\sin x)+(\frac{-1}{x^2})(\log x+\log(\sin x))[/Tex]

[Tex]\frac{dv}{dx}=v(\frac{1}{x}(\frac{1}{x}+\frac{\cos x}{\sin x})\frac{-1}{x^2}(\log x+\log(\sin x)))[/Tex]

[Tex]\frac{dv}{dx}=(x\sin x)^{1/2}.[(\frac{1}{x^2}+\frac{\cot x}{x})-\frac{\log x}{x^2}-\frac{\log(\sin x)}{x^2}] [/Tex] ………(3)

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

[Tex]\frac{dy}{dx}=(x\cos)^x(1-x\tan x+\log x+\log(\cos x))+(x\sin x)^{\frac{1}{x}}.[\frac{xcotx+1-log(xsinx)}{x^2}][/Tex]

Find dy/dx of  the function given in questions 12 to 15

Question 12. x^y + y^x = 1

Solution:

Given: x^y + y^x = 1

Let us considered

u = x^y and v = y^x

So,

[Tex]\frac{du}{dx}+\frac{dv}{dx}=0   [/Tex]………(1)

So first we take u = x^y

On taking log on both sides, we get

log u = log(x^y)             

log u = y log x

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

[Tex]\frac{1}{u}.\frac{du}{dx}=y.\frac{d}{dx}\log x+\frac{dy}{dx}.\log x[/Tex]

[Tex]\frac{1}{u}\frac{du}{dx}=\frac{y}{x}+\frac{dy}{dx}\log x[/Tex]

[Tex]\frac{du}{dx}=x^4(\frac{y}{x}+\frac{dy}{dx}\log x)   [/Tex] ………(2)

Now we take v = y^x

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

[Tex]\frac{1}{v}.\frac{dv}{dx}=x\frac{d}{dx}(\log x)+\log y\frac{d}{dx}x[/Tex]

[Tex]\frac{dv}{dx}=v(x.\frac{1}{y}.\frac{dy}{dx}+\log y)[/Tex]

[Tex]\frac{dv}{dx}=y^x(\frac{x}{y}\frac{dy}{dx}+\log x)   [/Tex]  ………(3)

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

[Tex]x^y(\frac{y}{x}+\frac{dy}{dx}\log x)+y^x(\frac{x}{y}\frac{dy}{dx}+\log y)=0[/Tex]

[Tex](x^y.\log x+xy^{x-1})\frac{dy}{dx}=-(yx^{y-1}+y^x\log y)[/Tex]

[Tex]\frac{dy}{dx}=\frac{-yx^{y-1}+y^x\log y}{x^y\log x+xy^{x-1}}[/Tex]

Question 13. y^x = x^y  

Solution:

Given: y^x = x^y  

On taking log on both sides, we get

log(yx) = log(x^y)         

xlog y = y log x

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

[Tex]x\frac{dy}{dx}(\log y)+\log y(\frac{d}{dx}x)=y\frac{d}{dx}\log x+\log x\frac{d}{dx}y[/Tex]

[Tex]x.\frac{d}{dx}.y+\log y.1=y.\frac{1}{x}+\log x\frac{dy}{dx}[/Tex]

[Tex]\frac{x}{y}\frac{dy}{dx}+\log y=\frac{y}{x}+\log x\frac{dy}{dx}[/Tex]

[Tex](\frac{x}{y}-\log x)\frac{dy}{dx}=(\frac{y}{x}-\log y)[/Tex]

[Tex]\frac{dy}{dx}=\frac{\frac{y}{x}-\log y}{\frac{x}{y}-\log x}[/Tex]

[Tex]\frac{dy}{dx}=\frac{y}{x}(\frac{y-x\log y}{x-y\log x})[/Tex]

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

[Tex]y\frac{d}{dx}\log(\cos x)+\log(\cos x).\frac{dy}{dx}=x\frac{d}{dx}\log (\cos y)+\log(\cos y)\frac{dx}{dx}[/Tex]

[Tex]y\frac{1}{\cos x}\frac{d}{dx}\cos x+\log(\cos x)\frac{dy}{dx}=x\frac{1}{\cos y}\frac{d}{dx}\cos y+\log(\cos y).1[/Tex]

[Tex]\frac{y}{\cos x}.-\sin x+\log(\cos x).\frac{dy}{dx}=\frac{x}{\cos y}.(-\sin y).\frac{dy}{dx}+\log(\cos y)[/Tex]

[Tex](\log(\cos x)+x\tan y)\frac{dy}{dx}=\log(\cos y)+y \tan x[/Tex]

[Tex]\frac{dy}{dx}=\frac{\log(\cos y)+y\tan x}{\log(\cos x)+x\tan y}[/Tex]

Question 15. xy = e^{(x – y)}

Solution:

Given: xy = e^{(x – y)}

On taking log on both sides, we get

log(xy) = log e^{x – y}

log x + log y = x – y

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

[Tex]\frac{d}{dx}\log x+\frac{d}{dx}\log y=\frac{dx}{dx}-\frac{dy}{dx}[/Tex]

[Tex]\frac{1}{x}.+\frac{1}{y}.\frac{dy}{dx}=1-\frac{dy}{dx}   [/Tex]          

[Tex](\frac{1}{y}+1)\frac{dy}{dx}=(1-\frac{1}{x})[/Tex]

[Tex]\frac{dy}{dx}=\frac{(1-\frac{1}{x})}{(1+\frac{1}{y})}   [/Tex]            

[Tex]\frac{dy}{dx}=\frac{y(x-1)}{x(y+1)}[/Tex]

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

[Tex]\frac{1}{f(x)}.\frac{d}{dx}f{x}=\frac{1}{1+x}\frac{d}{dx}(1+x)+\frac{1}{1+x^2}(1+x)^2+\frac{1}{1+x^4}.\frac{d}{dx}(1+x^4)+\frac{1}{1+x^8}\frac{d}{dx}(1+x^8)\frac{f'(x)}{f(x)}=\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\frac{8x^7}{1+x^8}[/Tex]

[Tex]f'(x)=(1+x)(1+x^2)(1+x^4)(1+x^8)(\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{4x^3}{1+x^2}+\frac{8x^7}{1+x^8})[/Tex]

∴ f'(1) = 2.2.2.2.[Tex](\frac{1}{2}+\frac{2}{2}+\frac{4}{2}+\frac{8}{2})[/Tex]

[Tex]f'(1)=16.(\frac{15}{2})[/Tex]

f'(1) = 120

Question 17. Differentiate (x⁵ – 5x + 8)(x³ + 7x + 9) in three ways mentioned below 

(i) By using product rule

(ii) By expanding the product to obtain a single polynomial

(iii) By logarithmic differentiation.

Do they all give the same answer? 

Solution:

(i) By using product rule

[Tex]\frac{d}{dx}(u.v)=v\frac{du}{dv}+u\frac{dv}{dx}[/Tex]

[Tex]\frac{dy}{dx}=(x^2-5x+8)\frac{d}{dx}(x^3+7x+9)+(x^3+7x+9).\frac{d}{dx}(x^2-5x+8)[/Tex]

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

[Tex]\frac{1}{y}.\frac{dy}{dx}=\frac{1}{x^2-5x+8}.\frac{d}{dx}(x^2-5x+8)+\frac{1}{x^3+7x+9}\frac{d}{dx}(x^3+7x+9)[/Tex]

[Tex]\frac{1}{y}.\frac{dy}{dx}=\frac{2x-5}{x^2-5x+8}+\frac{3x^2+7}{x^3+7x+9}[/Tex]

[Tex]\frac{1}{(x^2-5x+8)(x^3+7x+9)}\frac{dy}{dx}=\frac{(2x-5)(x^3+7x+9)+(3x^2+7)(x^2-5x+8)}{(x^2-5x+8)(x^3+7x+9)}[/Tex]

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.

Question 18. If u, v and w are function of x, then show that

[Tex]\frac{d}{dx}(u.v.w)=\frac{du}{dx}v.w+u.\frac{dv}{dx}.w+u.v.\frac{dw}{dx}[/Tex]

Solution:

Let y = u.v.w.

Method 1: Using product Rule 

[Tex]\frac{dy}{dx}=u\frac{d}{dx}(v.w)+v.w\frac{d}{dx}u[/Tex]

[Tex]\frac{dy}{dx}=u.[v.\frac{dw}{dx}+w\frac{du}{dx}]+v.w.\frac{du}{dx}[/Tex]

[Tex]\frac{dy}{dx}=u.v.\frac{dw}{dx}+u.w.\frac{dv}{dx}+v.w\frac{du}{dx}[/Tex]

Method 2: Using logarithmic differentiation

Taking log on both sides

log y = log u + log v + log w

Now, Differentiating w.r.t. x

[Tex]\frac{1}{y}\frac{dy}{dx}=\frac{1}{u}\frac{du}{dx}+\frac{1}{v}\frac{dv}{dx}+\frac{1}{w}.\frac{dw}{dx}  [/Tex] 

[Tex]\frac{dy}{dx}=(u.v.w)(\frac{1}{u}\frac{du}{dx}+\frac{1}{v}\frac{dv}{dx}+\frac{1}{w}\frac{dw}{dx})[/Tex]

[Tex]\frac{dy}{dx}=v.w\frac{du}{dx}+uw\frac{dv}{dx}+uv\frac{dw}{dx}[/Tex]