Differentiate the following functions with respect to $x$ :

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

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

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

$\displaystyle \text{Using the product rule: }$

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

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

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

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

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

$\displaystyle \text{Using the product rule: }$

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

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

$\\$

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

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

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

$\displaystyle \text{Using the product rule: }$

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

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

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

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

$\\$

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

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

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

$\displaystyle \text{Using the product rule: }$

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

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

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

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

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

$\displaystyle \text{Using the product rule: }$

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

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

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

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

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

$\displaystyle \text{Using the product rule: }$

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

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

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

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

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

$\displaystyle \text{Using the product rule: }$

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

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

$\\$

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

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

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

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

$\displaystyle \text{Using the product rule: }$

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

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

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

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

$\\$

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

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

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

$\displaystyle \text{Using the product rule: }$

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

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

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

$\\$

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

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

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

$\displaystyle \text{Using the product rule: }$

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

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

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

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

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

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

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

$\displaystyle \text{Using the product rule: }$

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

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

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

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

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

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

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

$\\$

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

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

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

$\displaystyle \text{Using the product rule: }$

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

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

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

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

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

$\\$

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

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

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

$\displaystyle \text{Using the product rule: }$

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

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

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

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

$\\$

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

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

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

$\displaystyle \text{Using the product rule: }$

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

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

$\\$

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

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

$\\$

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

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

$\\$

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

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

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

$\displaystyle \text{Using the product rule: }$

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

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

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

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

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

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

$\\$

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

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

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

$\displaystyle \text{Using the product rule: }$

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

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

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

$\\$

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

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

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

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

$\displaystyle \text{Using the product rule: }$

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

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

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

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

$\\$

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

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

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

$\displaystyle \text{Using the product rule: }$

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

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

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

$\\$

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

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

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

$\displaystyle \text{Using the product rule: }$

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

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

$\\$

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

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

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

$\displaystyle \text{Using the product rule: }$

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

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

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

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

$\\$

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

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

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

$\displaystyle \text{Using the product rule: }$

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

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

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

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

$\\$

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

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

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

$\displaystyle \text{Using the product rule: }$

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

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

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

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

$\\$

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

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

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

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

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

$\\$

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

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

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

$\displaystyle \text{Using the product rule: }$

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

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

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

$\\$

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

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.

$\\$

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

$\displaystyle \text{(i) } (3x^2+2)^2$

$\displaystyle \text{(i) } (x+2)(x+3)$

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

$\displaystyle \text{(i) } (3x^2+2)^2$

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.

$\\$

$\displaystyle \text{(i) } (x+2)(x+3)$

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.

$\\$

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

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.