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Class 12 NCERT Solutions – Mathematics Part I – Chapter 5 Continuity And Differentiability – Exercise 5.5 | Set 2
  • Last Updated : 05 Apr, 2021

Chapter 5 Continuity And Differentiability – Exercise 5.5 | Set 1

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)

On taking log on both sides, we get

log u = log(x cos x)    

log u = xlog(x cos x)

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

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

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

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



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

\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)))

\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))

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

\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}]  ………(3)

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

\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}]

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

Question 12. xy + yx = 1

Solution:

Given: xy + yx = 1

Let us considered

u = xy and v = yx 

So,

\frac{du}{dx}+\frac{dv}{dx}=0   ………(1)

So first we take u = xy

On taking log on both sides, we get

log u = log(xy)             

log u = y log x

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

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

\frac{1}{u}\frac{du}{dx}=\frac{y}{x}+\frac{dy}{dx}\log x

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

Now we take v = yx

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

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

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

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

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

x^y(\frac{y}{x}+\frac{dy}{dx}\log x)+y^x(\frac{x}{y}\frac{dy}{dx}+\log y)=0

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

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

Question 13. yx = xy  

Solution:

Given: yx = xy  

On taking log on both sides, we get

log(yx) = log(xy)         

xlog y = y log x

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

x\frac{dy}{dx}(\log y)+\log y(\frac{d}{dx}x)=y\frac{d}{dx}\log x+\log x\frac{d}{dx}y

x.\frac{d}{dx}.y+\log y.1=y.\frac{1}{x}+\log x\frac{dy}{dx}

\frac{x}{y}\frac{dy}{dx}+\log y=\frac{y}{x}+\log x\frac{dy}{dx}

(\frac{x}{y}-\log x)\frac{dy}{dx}=(\frac{y}{x}-\log y)

\frac{dy}{dx}=\frac{\frac{y}{x}-\log y}{\frac{x}{y}-\log x}

\frac{dy}{dx}=\frac{y}{x}(\frac{y-x\log y}{x-y\log x})

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

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}

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

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

(\log(\cos x)+x\tan y)\frac{dy}{dx}=\log(\cos y)+y \tan x

\frac{dy}{dx}=\frac{\log(\cos y)+y\tan x}{\log(\cos x)+x\tan y}

Question 15. xy = e(x – y)

Solution:

Given: xy = e(x – y)

On taking log on both sides, we get

log(xy) = log ex – y

log x + log y = x – y

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

\frac{d}{dx}\log x+\frac{d}{dx}\log y=\frac{dx}{dx}-\frac{dy}{dx}

\frac{1}{x}.+\frac{1}{y}.\frac{dy}{dx}=1-\frac{dy}{dx}             

(\frac{1}{y}+1)\frac{dy}{dx}=(1-\frac{1}{x})

\frac{dy}{dx}=\frac{(1-\frac{1}{x})}{(1+\frac{1}{y})}               

\frac{dy}{dx}=\frac{y(x-1)}{x(y+1)}

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

\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}

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})

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

f'(1)=16.(\frac{15}{2})

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

\frac{d}{dx}(u.v)=v\frac{du}{dv}+u\frac{dv}{dx}

\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)

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

\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)

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

\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)}

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

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

Solution:

Let y = u.v.w.

Method 1: Using product Rule 

\frac{dy}{dx}=u\frac{d}{dx}(v.w)+v.w\frac{d}{dx}u

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

\frac{dy}{dx}=u.v.\frac{dw}{dx}+u.w.\frac{dv}{dx}+v.w\frac{du}{dx}

Method 2: Using logarithmic differentiation

Taking log on both sides

log y = log u + log v + log w

Now, Differentiating w.r.t. x

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

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

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

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