Class 12 NCERT Solutions - Mathematics Part I - Chapter 5 Continuity And Differentiability - Exercise 5.5 | Set 2

Exercise 5.5 focuses on the Mean Value Theorem and Rolle's Theorem, which are fundamental concepts in calculus. These theorems provide powerful tools for analyzing functions and their behavior over intervals.

This exercise helps students understand how to apply these theorems to solve various mathematical problems, including proving inequalities, finding function values, and analyzing function characteristics.

Content of this article has been merged with Chapter 5 Continuity And Differentiability - Exercise 5.5 as per the revised syllabus of NCERT.

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

\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. x^y + y^x = 1

Solution:

Given: x^y + y^x = 1

Let us considered

u = x^y and v = y^x

So,

\frac{du}{dx}+\frac{dv}{dx}=0   .........(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

\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 = 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

\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. 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

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 e^{x - 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}

Summary

This set of questions for Exercise 5.5 provides a comprehensive practice on applying the Mean Value Theorem and Rolle's Theorem. It covers a range of applications including verifying the theorems for specific functions, proving inequalities, demonstrating the existence and uniqueness of solutions, and analyzing function behavior. These problems help students deepen their understanding of these important calculus concepts and improve their problem-solving skills.

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