Differentiate the following functions with respect to x :

\displaystyle \text{Question 1: } x^3 \sin x 

Answer:

\displaystyle \text{Let } u = x^3 ; \hspace{1.0cm}  v = \sin x

\displaystyle \text{Then, } u' = 3x^2  ; \hspace{1.0cm}  v' = \cos x

\displaystyle \text{Using the product rule: }

\displaystyle \frac{d}{dx} ( uv) = uv'+vu'

\displaystyle \frac{d}{dx} (x^3 \sin x ) = x^3 \cos x + \sin x ( 3x^2) = x^2( x \cos x + 3 \sin x) 

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\displaystyle \text{Question 2: } x^3 e^x 

Answer:

\displaystyle \text{Let } u = x^3  ; \hspace{1.0cm}  v = e^x

\displaystyle \text{Then, } u' = 3x^2 ; \hspace{1.0cm}  v' = e^x

\displaystyle \text{Using the product rule: }

\displaystyle \frac{d}{dx} ( uv) = uv'+vu'

\displaystyle \frac{d}{dx} (x^3 e^x) = x^3 e^x + e^x( 3x^2) = x^2 e^x ( x+3)

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\displaystyle \text{Question 3: } x^2 e^x \log x 

Answer:

\displaystyle \text{Let } u = x^2 ; \hspace{1.0cm}  v = e^x ; \hspace{1.0cm}  w = \log x

\displaystyle \text{Then, } u' =2x  ; \hspace{1.0cm}  v' = e^x  ; \hspace{1.0cm}  w' = \frac{1}{x}

\displaystyle \text{Using the product rule: }

\displaystyle \frac{d}{dx} ( uvw) = u'vw+ uv'w + uvw'

\displaystyle \frac{d}{dx} (x^2 e^x \log x) =  2x e^x \log x + x^2 e^x \log x + x^2 e^x \frac{1}{x}

\displaystyle =  2x e^x \log x + x^2 e^x \log x + x e^x 

\displaystyle = xe^x ( 2 \log x + x \log x + 1)

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\displaystyle \text{Question 4: } x^n \tan x 

Answer:

\displaystyle \text{Let } u = x^n ; \hspace{1.0cm}  v = \tan x

\displaystyle \text{Then, } u' = nx^{n-1} ; \hspace{1.0cm}  v' = \sec^2 x

\displaystyle \text{Using the product rule: }

\displaystyle \frac{d}{dx} ( uv) = uv'+vu'

\displaystyle \frac{d}{dx} (x^n \tan x) =  x^n \sec^2 x + \tan x ( n x^{n-1}) = x^{n-1} ( x \sec^2 x + n \tan x)

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\displaystyle \text{Question 5: } x^n \log_a x 

Answer:

\displaystyle \text{Let } u = x^n ; \hspace{1.0cm}  v =\log_a x = \frac{\log x}{\log a}

\displaystyle \text{Then, } u' = n x^{n-1} ; \hspace{1.0cm}  v' =\frac{1}{x \log a}

\displaystyle \text{Using the product rule: }

\displaystyle \frac{d}{dx} ( uv) = uv'+vu'

\displaystyle \frac{d}{dx} (x^n \log_a x) =x^n  \frac{1}{x \log a} + \log_a x \ (nx^{n-1}) \\ \\ {\hspace{2.5cm} = x^{n-1} \frac{1}{\log a}  + \log_a x ( n x^{n-1})} \\ \\ {\hspace{2.5cm} = x^{n-1} \Big( \frac{1}{\log a} + n \log_a x \Big)}

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\displaystyle \text{Question 6: } (x^3 + x^2 + 1) \sin x 

Answer:

\displaystyle \text{Let } u = x^3 + x^2 + 1 ; \hspace{1.0cm}  v =\sin x

\displaystyle \text{Then, } u' = 3x^2 + 2x ; \hspace{1.0cm}  v' = \cos x

\displaystyle \text{Using the product rule: }

\displaystyle \frac{d}{dx} ( uv) = uv'+vu'

\displaystyle \frac{d}{dx} ((x^3 + x^2 + 1) \sin x) =  (x^3 + x^2 + 1) \cos x + (3x^2 + 2x) \sin x 

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\displaystyle \text{Question 7: }  \sin x \cos x

Answer:

\displaystyle \text{Let } u = \sin x ; \hspace{1.0cm}  v = \cos x

\displaystyle \text{Then, } u' = \cos x ; \hspace{1.0cm}  v' = - \sin x

\displaystyle \text{Using the product rule: }

\displaystyle \frac{d}{dx} ( uv) = uv'+vu'

\displaystyle \frac{d}{dx} (\sin x \cos x) = \sin x ( - \sin x ) + \cos x \cdot \cos  = - \sin^2 x + \cos^2 x = \cos 2x

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\displaystyle \text{Question 8: } \frac{2^x \cot x}{\sqrt{x}} 

Answer:

\displaystyle \frac{2^x \cot x}{\sqrt{x}} = 2^x \cdot \cot x \cdot ( x^{\frac{-1}{2} } )

\displaystyle \text{Let } u = 2^x ; \hspace{1.0cm}  v = \cot x ; \hspace{1.0cm}  w = x^{\frac{-1}{2} }

\displaystyle \text{Then, } u' = 2^x \log 2 ; \hspace{1.0cm}  v' = - \mathrm{cosec}^2 x ; \hspace{1.0cm}  w' = \frac{-1}{2} x^{\frac{-3}{2} }

\displaystyle \text{Using the product rule: }

\displaystyle \frac{d}{dx} ( uvw) = u'vw+ uv'w + uvw'

\displaystyle \frac{d}{dx} (\frac{2^x \cot x}{\sqrt{x}}) = 2^x \log 2 \cdot \cot x \cdot  x^{\frac{-1}{2} } + 2^x ( - \mathrm{cosec}^2 x) x^{\frac{-1}{2} } + 2^x \cdot \cot x \cdot \frac{-1}{2} x^{\frac{-3}{2} }

\displaystyle = 2^x \log 2 \cdot \cot x \cdot  \frac{1}{\sqrt{x}} + 2^x ( - \mathrm{cosec}^2 x) \frac{1}{\sqrt{x}} + 2^x \cdot \cot x \cdot \frac{-1}{2} \frac{1}{x\sqrt{x}}

\displaystyle = \frac{2^x}{\sqrt{x}}  ( \log 2 \cdot \cot x - \mathrm{cosec}^2 x - \frac{\cot x}{2x} )

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\displaystyle \text{Question 9: } x^2 \sin x \log x 

Answer:

\displaystyle \text{Let } u = x^2 ; \hspace{1.0cm}  v = \sin x ; \hspace{1.0cm}  w =\log x

\displaystyle \text{Then, } u' = 2x ; \hspace{1.0cm}  v' = \cos x ; \hspace{1.0cm}  w' = \frac{1}{x}

\displaystyle \text{Using the product rule: }

\displaystyle \frac{d}{dx} ( uvw) = u'vw+ uv'w + uvw'

\displaystyle \frac{d}{dx} (x^2 \sin x \log x) =  2x \cdot \sin x \cdot \log x + x^2 \cdot \cos \cdot \log x + x^2 \cdot \sin x \cdot \frac{1}{x}

\displaystyle  = 3x \sin x \log x + x^2 \cos x \log x + x \sin x 

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\displaystyle \text{Question 10: }  x^5 e^x + x^6 \log x

Answer:

\displaystyle \text{Let } u_1 = x^5 ; \hspace{1.0cm}  v_1 = e^x           \displaystyle \text{Let } u_2 = x^6 ; \hspace{1.0cm}  v_2 = \log x 

\displaystyle \text{Then, } u_1' = 5x^4  ; \hspace{1.0cm}  v_1' = e^x           \displaystyle \text{Then, } u_2' = 6x^5  ; \hspace{1.0cm}  v_2' = \frac{1}{x}

\displaystyle \text{Using the product rule: }

\displaystyle \frac{d}{dx} ( u_1v_1+ u_2v_2 ) = u_1v_1'+v_1u_1' + u_2v_2'+v_2u_2'

\displaystyle \frac{d}{dx} (x^5 e^x + x^6 \log x) =  x^5 \cdot e^x + 5x^4 \cdot  e^x + x^6 \cdot  \frac{1}{x} + 6x^5 \log x

\displaystyle = x^5 e^x + 5 x^4 e^x + x^5 + 6x^5 \log x

\displaystyle = x^4 ( xe^x + 5e^x + x + 6x \log x)

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\displaystyle \text{Question 11: } (x \sin x + \cos x) ( x \cos x - \sin x) 

Answer:

\displaystyle \text{Let } u = x \sin x + \cos x ; \hspace{1.0cm}  v = x \cos x - \sin x

\displaystyle \text{Then, } u' = x \cos x + \sin x - \sin x = x \cos x  ; \hspace{1.0cm}  \\ \\ v' = -x \sin x + \cos x - \cos x = - x \sin x 

\displaystyle \text{Using the product rule: }

\displaystyle \frac{d}{dx} ( uv) = uv'+vu'

\displaystyle \frac{d}{dx} \Big( (x \sin x + \cos x) ( x \cos x - \sin x) \Big)

\displaystyle =  (x \sin x + \cos x) ( - x \sin x) + ( x \cos x - \sin x) ( x \cos x )

\displaystyle = - x^2 \sin^2 x - x \cos x \sin x + x^2 \cos^2 x - x \cos x \sin x   

\displaystyle = x^2 ( \cos^2 x - \sin^2 x) - x (2\cos x \sin x)

\displaystyle = x^2 \cos 2x - x \sin 2x 

\displaystyle = x ( x \cos 2x - \sin 2x)

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\displaystyle \text{Question 12: } (x \sin x + \cos x) ( e^x + x^2 \log x)  

Answer:

\displaystyle \text{Let } u = x \sin x + \cos x  ; \hspace{1.0cm}  v = e^x + x^2 \log x

\displaystyle \text{Then, } u' = x \cos x + \sin x - \sin x = x \cos x   ; \hspace{1.0cm}  v' = e^x + x + 2 x \log x

\displaystyle \text{Using the product rule: }

\displaystyle \frac{d}{dx} ( uv) = uv'+vu'

\displaystyle \frac{d}{dx} ((x \sin x + \cos x) ( e^x + x^2 \log x))

\displaystyle =  (x \sin x + \cos x) ( e^x + x + 2 x \log x) + (x \cos x) ( e^x + x^2 \log x)

\displaystyle = x \cdot \sin x  \cdot e^x + x \cdot \sin x \cdot x + x \cdot \sin x 2 x \cdot \log x + \cos x \cdot e^x + \cos x \cdot x + \cos x \cdot 2 x \cdot \log x + x \cdot \cos x \cdot e^x + x \cdot \cos x \cdot x^2 \cdot \log x  

\displaystyle = x \cos x (e^x + x^2  \log x) + (x \sin x + \cos x )( e^x + x + 2x \log x)

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\displaystyle \text{Question 13: } (1 - 2 \tan x)( 5 + 4 \sin x) 

Answer:

\displaystyle \text{Let } u = 1 - 2 \tan x ; \hspace{1.0cm}  v = 5 + 4 \sin x

\displaystyle \text{Then, } u' = -2 \sec^2 x  ; \hspace{1.0cm}  v' = 4 \cos x

\displaystyle \text{Using the product rule: }

\displaystyle \frac{d}{dx} ( uv) = uv'+vu'

\displaystyle \frac{d}{dx} \Big( (1 - 2 \tan x)( 5 + 4 \sin x) \Big) =  (1 - 2 \tan x) ( 4 \cos x) + ( 5 + 4 \sin x) (-2 \sec^2 x)

\displaystyle = 4 \cos x  - 8 \sin x - 10 \sec^2 x - 8 \sec x \tan x

\displaystyle = 4 (  \cos x  - 2 \sin x - \frac{5}{2} \sec^2 x - 2 \sec x \tan x )

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\displaystyle \text{Question 14: }  (1+x^2) \cos x

Answer:

\displaystyle \text{Let } u = 1+x^2 ; \hspace{1.0cm}  v = \cos x

\displaystyle \text{Then, } u' = 2x ; \hspace{1.0cm}  v' = - \sin x

\displaystyle \text{Using the product rule: }

\displaystyle \frac{d}{dx} ( uv) = uv'+vu'

\displaystyle \frac{d}{dx} \Big( (1+x^2) \cos x \Big) = (1+x^2)(-\sin x) + 2x \cos x  = - \sin x - x^2 \sin x + 2 x \cos x 

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\displaystyle \text{Question 15: } \sin^2 x 

Answer:

\displaystyle \frac{d}{dx} (\sin^2 x) = \frac{d}{dx} ( \sin x \sin x ) = \sin x \cos x + \cos x \sin x =  2 \sin x \cos x = \sin 2x 

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\displaystyle \text{Question 16: } \log_{x^2} x 

Answer:

\displaystyle \frac{d}{dx} \Big( \log_{x^2} x \Big) = \frac{d}{dx} \Big( \frac{\log x}{\log x^2} \Big) = \frac{d}{dx} \Big( \frac{\log x}{2 \log x} \Big) = \frac{d}{dx} \Big( \frac{1}{2} \Big) = 0

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\displaystyle \text{Question 17: }  e^x \log \sqrt{x} \tan x

Answer:

\displaystyle \text{Let } u = e^x ; \hspace{1.0cm}  v =\log \sqrt{x} ; \hspace{1.0cm}  w =\tan x

\displaystyle \text{Then, } u' = e^x  ; \hspace{1.0cm}  v' =\frac{1}{2x} ; \hspace{1.0cm}  w' = \sec^2 x

\displaystyle \text{Using the product rule: }

\displaystyle \frac{d}{dx} ( uvw) = u'vw+ uv'w + uvw'

\displaystyle \frac{d}{dx} (e^x \log \sqrt{x} \tan x) = e^x \cdot \log \sqrt{x} \cdot \tan x + e^x \cdot \frac{1}{2x} \cdot \tan x + e^x \cdot \log \sqrt{x} \cdot \sec^2 x

\displaystyle = e^x( \log \sqrt{x} \cdot \tan x + \frac{1}{2x} \cdot \tan x + \log \sqrt{x} \cdot \sec^2 x ) 

\displaystyle = e^x( \log x^{\frac{1}{2}} \cdot \tan x + \frac{1}{2x} \cdot \tan x + \log x^{\frac{1}{2}} \cdot \sec^2 x ) 

\displaystyle = e^x( \frac{1}{2} \cdot \log x \cdot \tan x + \frac{1}{2x} \cdot \tan x + \frac{1}{2} \cdot \log x \cdot \sec^2 x ) 

\displaystyle = \frac{e^x}{2} (  \log x \cdot \tan x + \frac{\tan x}{x} +  \log x \cdot \sec^2 x ) 

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\displaystyle \text{Question 18: }  x^3 e^x \cos x

Answer:

\displaystyle \text{Let } u = x^3 ; \hspace{1.0cm}  v =e^x ; \hspace{1.0cm}  w = \cos x

\displaystyle \text{Then, } u' = 3x^2 ; \hspace{1.0cm}  v' = e^x ; \hspace{1.0cm}  w' = -\sin x

\displaystyle \text{Using the product rule: }

\displaystyle \frac{d}{dx} ( uvw) = u'vw+ uv'w + uvw'

\displaystyle \frac{d}{dx} (x^3 e^x \cos x) =  3x^2 \cdot e^x \cdot \cos x + x^3 \cdot e^x \cdot \cos x + x^3 \cdot e^x \cdot (-\sin x)

\displaystyle =  x^2 \cdot e^x ( 3 \cos x + x \cos x - x \sin x

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\displaystyle \text{Question 19: } \frac{x^2 \cos \frac{\pi}{4} }{\sin x} 

Answer:

\displaystyle  \frac{x^2 \cos \frac{\pi}{4} }{\sin x}  = x^2 \cos \frac{\pi}{4} \mathrm{cosec} x

\displaystyle \text{Let } u = x^2 ; \hspace{1.0cm}  v = \cos \frac{\pi}{4} ; \hspace{1.0cm}  w =\mathrm{cosec} x

\displaystyle \text{Then, } u' = 2x ; \hspace{1.0cm}  v' = 0 ; \hspace{1.0cm}  w' = - \mathrm{cosec} x \cot x

\displaystyle \text{Using the product rule: }

\displaystyle \frac{d}{dx} ( uvw) = u'vw+ uv'w + uvw'

\displaystyle \frac{d}{dx} (x^2 \cos \frac{\pi}{4} \mathrm{cosec} x) =  2x \cdot \cos \frac{\pi}{4} \cdot \mathrm{cosec} x + x^2 \cdot (0) \cdot \mathrm{cosec} x + x^2 \cdot \cos \frac{\pi}{4} (- \mathrm{cosec} x \cot x )

\displaystyle = \cos \frac{\pi}{4} ( 2x \ \mathrm{cosec} x - x^2 \mathrm{cosec} x \cdot \cot x)

\displaystyle = \cos \frac{\pi}{4} \Big(   \frac{2x}{\sin x} - x^2 \frac{\cot x}{\sin x} \Big)

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\displaystyle \text{Question 20: }  x^4 ( 5 \sin x - 3 \cos x)

Answer:

\displaystyle \text{Let } u = x^4 ; \hspace{1.0cm}  v = 5 \sin x - 3 \cos x

\displaystyle \text{Then, } u' = 4x^3 ; \hspace{1.0cm}  v' = 5 \cos x + 3 \sin x 

\displaystyle \text{Using the product rule: }

\displaystyle \frac{d}{dx} ( uv) = uv'+vu'

\displaystyle \frac{d}{dx} \Big( x^4 ( 5 \sin x - 3 \cos x) \Big) =  x^4 ( 5 \cos x + 3 \sin x) + 4x^3 ( 5 \sin x - 3 \cos x) 

\displaystyle = 5 x^4 \cos x + 3 x^4 \sin x + 20 x^3 \sin x - 12 x^3 \cos x

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\displaystyle \text{Question 21: } (2x^2 - 3) \sin x 

Answer:

\displaystyle \text{Let } u = 2x^2 - 3 ; \hspace{1.0cm}  v = \sin x

\displaystyle \text{Then, } u' = 4x ; \hspace{1.0cm}  v' = \cos x 

\displaystyle \text{Using the product rule: }

\displaystyle \frac{d}{dx} ( uv) = uv'+vu'

\displaystyle \frac{d}{dx} \Big( (2x^2 - 3) \sin x \Big) =  (2x^2 - 3)\cos x + 4x \sin x

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\displaystyle \text{Question 22: }  x^5(3 - 6 x^{-9})

Answer:

\displaystyle \text{Let } u = x^5 ; \hspace{1.0cm}  v = 3 - 6 x^{-9}

\displaystyle \text{Then, } u' = 5x^4 ; \hspace{1.0cm}  v' = 54x^{-10}

\displaystyle \text{Using the product rule: }

\displaystyle \frac{d}{dx} ( uv) = uv'+vu'

\displaystyle \frac{d}{dx} \Big( x^5(3 - 6 x^{-9})  \Big) =  x^5(54x^{-10}) + 5x^4(3 - 6 x^{-9}) 

\displaystyle = 54x^{-5} + 15 x^4 - 30 x^{-5}

\displaystyle = 15x^4 + 24x^{-5}

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\displaystyle \text{Question 23: } x^{-4} (3 - 4x^{-5}) 

Answer:

\displaystyle \text{Let } u = x^{-4} ; \hspace{1.0cm}  v = 3 - 4x^{-5}

\displaystyle \text{Then, } u' = -4x^{-5} ; \hspace{1.0cm}  v' = 20x^{-6}

\displaystyle \text{Using the product rule: }

\displaystyle \frac{d}{dx} ( uv) = uv'+vu'

\displaystyle \frac{d}{dx} \Big( x^{-4} (3 - 4x^{-5}) \Big)  = x^{-4} \cdot 20x^{-6} -4x^{-5} (3 - 4x^{-5})

\displaystyle = 20 x^{-10} - 12 x^{-5} + 16 x^{-10}

\displaystyle = -12 x^{-5} + 36 x^{-10}

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\displaystyle \text{Question 24: } x^{-3} (5+3x) 

Answer:

\displaystyle \text{Let } u = x^{-3} ; \hspace{1.0cm}  v = 5+3x

\displaystyle \text{Then, } u' =-3x^{-4}  ; \hspace{1.0cm}  v' = 3

\displaystyle \text{Using the product rule: }

\displaystyle \frac{d}{dx} ( uv) = uv'+vu'

\displaystyle \frac{d}{dx} \Big( x^{-3} (5+3x) \Big) =  x^{-3} (3) -3x^{-4} (5+3x)

\displaystyle = 3x^{-3} - 15 x^{-4} - 9x^{-3}

\displaystyle = -15x^{-4} - 6 x^{-3}

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\displaystyle \text{Question 25: } \frac{ax+b}{cx+d} 

Answer:

\displaystyle \text{We know:   } \frac{d}{dx} \Bigg\{   \frac{f(x)}{g(x)} \Bigg\} = \frac{g(x) \frac{d}{dx} \{ f(x) \} -  f(x) \frac{d}{dx} \{ g(x) \}}{[g(x)] ^2}

\displaystyle \text{Let } f(x) = ax+b ; \hspace{1.0cm}  g(x) = cx+d

\displaystyle \text{Then, } f'(x) =a  ; \hspace{1.0cm}  g'(x) = c

\displaystyle \frac{d}{dx} \Bigg\{  \frac{ax+b}{cx+d} \Bigg\} = \frac{(cx+d) a -  (ax+b) c}{[cx+d] ^2} = \frac{ad-bc}{(cx+d)^2} 

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\displaystyle \text{Question 26: }  (ax+b)^n (cx+d)^m

Answer:

\displaystyle \text{Let } u = (ax+b)^n ; \hspace{1.0cm}  v =(cx+d)^m

\displaystyle \text{Then, } u' = an(ax+b)^{n-1} ; \hspace{1.0cm}  v' = cm (cx+d)^{m-1}

\displaystyle \text{Using the product rule: }

\displaystyle \frac{d}{dx} ( uv) = uv'+vu'

\displaystyle \frac{d}{dx} ((ax+b)^n (cx+d)^m) =  (ax+b)^n cm (cx+d)^{m-1} + an(ax+b)^{n-1} (cx+d)^m   

\displaystyle = (ax+b)^{n-1} (cx+d)^{m-1} \{ cm(ax+b) + an( cx+d) \}

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\displaystyle \text{Question 27: } \text{Differentiate in two ways, using product rule and otherwise, the  } \\ \\ \text{ function} (1 + 2 \tan x) (5 + 4 \cos x) . \text{ Verify that the answers are the same.  } 

Answer:

Product Rule ( 1st Method)

\displaystyle \text{Let } u = 1 + 2 \tan x ; \hspace{1.0cm}  v = 5 + 4 \cos x

\displaystyle \text{Then, } u' = 2 \sec^2  ; \hspace{1.0cm}  v' = - 4 \sin x

\displaystyle \text{Using the product rule: }

\displaystyle \frac{d}{dx} ( uv) = uv'+vu'

\displaystyle \frac{d}{dx} \Big( (1 + 2 \tan x) (5 + 4 \cos x) \Big) = (1 + 2 \tan x) (- 4 \sin x) + (2 \sec^2) (5 + 4 \cos x) 

\displaystyle = -4 \sin x - 8 \tan x \sin x + 10 \sec^2 x + 8 \sec x

\displaystyle = - 4 \sin x + 10 \sec^2 x + \Bigg(  \frac{8}{\cos x} - \frac{8 \sin^2 x}{\cos x} \Bigg) 

\displaystyle = - 4 \sin x + 10 \sec^2 x + 8\Bigg( \frac{1- \sin^2 x}{\cos x}  \Bigg) 

\displaystyle = - 4 \sin x + 10 \sec^2 x + 8\Bigg( \frac{\cos^2 x}{\cos x}  \Bigg) 

\displaystyle = - 4 \sin x + 10 \sec^2 x + 8 \cos x     

2nd Method

\displaystyle \frac{d}{dx} \Big( (1 + 2 \tan x) (5 + 4 \cos x) \Big) = 5 + 10 \tan x + 4 \cos x + 8 \sin x

\displaystyle = 0+ 10 \sec^2 x - 4 \sin x  + 8 \cos x

\displaystyle = - 4 \sin x + 10 \sec^2 x + 8 \cos x

Using both the methods we get the same results.

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\displaystyle \text{Question 28: } \text{Differentiate each of the following functions by the product rule } \\ \\ \text{ and the other method and verify that answer from both the method is same. }

\displaystyle \text{(i) } (3x^2+2)^2 

\displaystyle \text{(i) }  (x+2)(x+3)

\displaystyle \text{(i) } (3 \sec x - 4 \mathrm{cosec} x)(-2 \sin x + 5 \cos x) 

Answer:

\displaystyle \text{(i) } (3x^2+2)^2 

Product Rule ( 1st Method)

\displaystyle \text{Let } u = 3x^2+2 ; \hspace{1.0cm}  v =3x^2+2

\displaystyle \text{Then, } u' = 6x ; \hspace{1.0cm}  v' = 6x

\displaystyle \text{Using the product rule: }

\displaystyle \frac{d}{dx} ( uv) = uv'+vu'

\displaystyle \frac{d}{dx} ((3x^2+2)^2) = (3x^2+2) \cdot 6x + 6x \cdot (3x^2+2) = 12x((3x^2+2) = 36x^2 + 24x

2nd Method

\displaystyle \frac{d}{dx} \Big( ((3x^2+2)^2 \Big) = \frac{d}{dx} \Big( 9x^4+12x^2+4 \Big) = 36x^3+ 24x + 0= 6x^3+ 24x

Using both the methods we get the same results.

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\displaystyle \text{(i) }  (x+2)(x+3)

Product Rule ( 1st Method)

\displaystyle \text{Let } u = x+2 ; \hspace{1.0cm}  v = x+3

\displaystyle \text{Then, } u' = 1 ; \hspace{1.0cm}  v' = 1

\displaystyle \text{Using the product rule: }

\displaystyle \frac{d}{dx} ( uv) = uv'+vu'

\displaystyle \frac{d}{dx} ( (x+2)(x+3)) =  (x+2) \cdot 1 + 1 \cdot (x+3)  = 2x + 5

2nd Method

\displaystyle \frac{d}{dx} \Big( (x+2)(x+3) \Big) = \frac{d}{dx} \Big( x^2 + 5x + 6 \Big)  = 2x+5 + 0=2x+5

Using both the methods we get the same results.

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\displaystyle \text{(i) } (3 \sec x - 4 \mathrm{cosec} x)(-2 \sin x + 5 \cos x) 

Product Rule ( 1st Method)

\displaystyle \text{Let } u = 3 \sec x - 4 \mathrm{cosec} x ; \hspace{1.0cm}  v = -2 \sin x + 5 \cos x

\displaystyle \text{Then, } u' = 3 \sec x \cdot \tan x  + 4 \mathrm{cosec} x \cdot \cot x; \hspace{1.0cm}  v' = - 2 \cos x - 5 \sin x

\displaystyle \text{Using the product rule: }

\displaystyle \frac{d}{dx} ( uv) = uv'+vu'

\displaystyle \frac{d}{dx} \Big( (3 \sec x - 4 \mathrm{cosec} x)(-2 \sin x + 5 \cos x) \Big) = 

\displaystyle = (3 \sec x - 4 \mathrm{cosec} x)(- 2 \cos x - 5 \sin x)  + (3 \sec x \cdot \tan x  + 4 \mathrm{cosec} x \cdot \cot x)(-2 \sin x + 5 \cos x)  

\displaystyle = - 6 - 15 \tan x +8 \cot x + 20 - 6 \tan^2 x -8  \cot x + 15 \tan x + 20 \cot^2 x  

\displaystyle = - 6   + 20 - 6 \tan^2 x   + 20 \cot^2 x 

\displaystyle = - 6   + 20 - 6( \sec^2 x - 1)    + 20 ( \mathrm{cosec}^2 x -1) x  

\displaystyle = 14  - 6 \sec^2 x + 6 + 20 \mathrm{cosec}^2 x -  20 

\displaystyle = - 6 \sec^2 x + 20 \mathrm{cosec}^2 x 

2nd Method

\displaystyle \frac{d}{dx} \Big( (3 \sec x - 4 \mathrm{cosec} x)(-2 \sin x + 5 \cos x) \Big)

\displaystyle = \frac{d}{dx} \Big(  - 6 \sec x \cdot \sin x + 15 \sec x \cdot \cos x + 8 \mathrm{cosec} x \cdot \sin x - 20 \mathrm{cosec} x \cdot \cos x   \Big)

\displaystyle = \frac{d}{dx} \Big(  - 6 \tan x + 15 + 8 - 20 \cot x \Big)

\displaystyle = - \sec^2 x + 20 \ \mathrm{cosec}^2 x

Using both the methods we get the same results.

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