NCERT Solutions Class 12 maths Chapter-5 (Continuity And Differentiability)Exercise 5.5

Jolly September 06, 2022

NCERT Solutions Class 12 maths Chapter-5 (Continuity And Differentiability)Exercise 5.5

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-5 (Continuity And Differentiability)Exercise 5.5 This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.
We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.
Solutions Class 12 maths Chapter-5 (Continuity And Differentiability)Exercise 5.5








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. 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 xx      

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)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}

Exercise 5.5
Set 2

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. 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}

Chapter-5 (Continuity And Differentiability)

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