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

\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. 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

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

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

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

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

\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 = 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

\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 = 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

\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 x^{sin 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 = x^xcosx and v = \frac{x^2+1}{x^2-1}

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

\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}

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}