Differentiate the following functions with respect to $x$ :

$\displaystyle \text{Question 1: }\frac{ x^2+1 }{ x+1 }$

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

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

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

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

$\\$

$\displaystyle \text{Question 2: }\frac{ 2x-1 }{ x^2+1 }$

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

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

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

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

$\displaystyle = \frac{2x^2 + 2 - 4 x^2 + 2x}{(x^2+1)^2}$

$\displaystyle = \frac{-2x^2 + 2x + 2 }{(x^2+1)^2}$

$\displaystyle = \frac{2 ( 1 +2 - x^2) }{(x^2+1)^2}$

$\\$

$\displaystyle \text{Question 3: }\frac{ x+e^x }{ 1+\log x }$

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

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

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

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

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

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

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

$\\$

$\displaystyle \text{Question 4: }\frac{e^x - \tan x }{\cot x - x^n }$

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

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

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

$\displaystyle \frac{d}{dx} \Big( \frac{e^x - \tan x }{\cot x - x^n } \Big) = \frac{(\cot x - x^n)(e^x - \sec^2 x) - (e^x - \tan x)(- \mathrm{cosec}^2 x - n x^{n-1})}{(\cot x - x^n)^2}$

$\displaystyle = \frac{(\cot x - x^n)(e^x - \sec^2 x) + (e^x - \tan x)( \mathrm{cosec}^2 x + n x^{n-1})}{(\cot x - x^n)^2}$

$\\$

$\displaystyle \text{Question 5: }\frac{ ax^2+bx+c }{ px^2+qx+r }$

$\displaystyle \text{Let } u = ax^2+bx+c ; \hspace{1.0cm} v = px^2+qx+r$

$\displaystyle \text{Then, } u' = 2ax+b ; \hspace{1.0cm} v' = 2px+q$

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

$\displaystyle \frac{d}{dx} \Big( \frac{ ax^2+bx+c }{ px^2+qx+r } \Big) = \frac{(px^2+qx+r )(2ax+b) - (ax^2+bx+c)(2px+q)}{(px^2+qx+r )^2}$

$\displaystyle = \frac{2apx^3+2aqx^2+2arx + bpx^2 + bqx + br - 2apx^3 - 2bpx^2 - 2pcx - aq x^2 - cq}{(px^2+qx+r )^2}$

$\displaystyle = \frac{(aq-bp)x^2+2(ar-xp)x + br - cq}{(px^2+qx+r )^2}$

$\\$

$\displaystyle \text{Question 6: }\frac{ x }{ 1+\tan x }$

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

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

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

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

$\\$

$\displaystyle \text{Question 7: }\frac{ 1 }{ ax^2+bx+c }$

$\displaystyle \text{Let } u = 1 ; \hspace{1.0cm} v = ax^2+bx+c$

$\displaystyle \text{Then, } u' = 0 ; \hspace{1.0cm} v' = 2ax+b$

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

$\displaystyle \frac{d}{dx} \Big( \frac{ 1 }{ ax^2+bx+c } \Big) = \frac{(ax^2+bx+c)(0) - (1)(2ax+b)}{(ax^2+bx+c)^2} = \frac{-(2ax+b)}{(ax^2+bx+c)^2}$

$\\$

$\displaystyle \text{Question 8: }\frac{ e^x }{ 1+x^2 }$

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

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

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

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

$\displaystyle = \frac{e^x + x^2 e^x - 2xe^x}{(1+x^2)^2}$

$\displaystyle = \frac{e^x ( 1 + x^2 - 2x)}{(1+x^2)^2}$

$\displaystyle = \frac{e^x (1-x)^2}{(1+x^2)^2}$

$\\$

$\displaystyle \text{Question 9: }\frac{ e^x+\sin x }{ 1+ \log x }$

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

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

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

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

$\displaystyle = \frac{x( 1 + \log x) ( e^x + \cos x ) - ( e^x +\sin x)}{( 1+ \log x)^2}$

$\\$

$\displaystyle \text{Question 10: }\frac{ x \tan x }{ \sec x + \tan x }$

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

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

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

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

$\displaystyle = \frac{ x \sec^3 x + x \sec^2 \tan x + \sec x \tan x + \tan^2 x - x \sec x\tan^2x - x \tan x \sec^2 x }{(\sec x + \tan x)^2}$

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

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

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

$\\$

$\displaystyle \text{Question 11: }\frac{x \sin x }{ 1 + \cos x }$

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

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

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

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

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

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

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

$\displaystyle = \frac{(1 + \cos x)[(x \cos x + \sin x) + x ( 1 - \cos x)] }{(1 + \cos x)^2}$

$\displaystyle = \frac{(x \cos x + \sin x) + x ( 1 - \cos x) }{(1 + \cos x)}$

$\displaystyle = \frac{(x + \sin x) }{(1 + \cos x)}$

$\\$

$\displaystyle \text{Question 12: }\frac{2^x \cot x }{ \sqrt{x} }$

$\displaystyle \text{Let } u = 2^x \cot x ; \hspace{1.0cm} v = \sqrt{x}$

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

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

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

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

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

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

$\\$

$\displaystyle \text{Question 13: }\frac{ \sin x - x \cos x }{ x \sin x + \cos x }$

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

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

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

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

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

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

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

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

$\\$

$\displaystyle \text{Question 14: }\frac{ x^2 - x + 1 }{ x^2 + x + 1 }$

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

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

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

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

$\displaystyle = \frac{ 2x^3 + 2x^2 + 2x - x^2 - x - 1 - 2x^3 + 2x^2 - 2x - x^2 + x - 1 }{(x^2 + x + 1)^2}$

$\displaystyle = \frac{ 2x^2 - 2 }{(x^2 + x + 1)^2}$

$\displaystyle = \frac{ 2(x^2 - 1) }{(x^2 + x + 1)^2}$

$\\$

$\displaystyle \text{Question 15: }\frac{\sqrt{a}+\sqrt{x} }{ \sqrt{a}-\sqrt{x}}$

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

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

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

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

$\displaystyle = \frac{ \sqrt{a}-\sqrt{x}+\sqrt{a}+\sqrt{x} }{2 \sqrt{x}(\sqrt{a}-\sqrt{x})^2}$

$\displaystyle = \frac{ \sqrt{a} }{ \sqrt{x}(\sqrt{a}-\sqrt{x})^2}$

$\\$

$\displaystyle \text{Question 16: }\frac{ a+\sin x }{ 1 + a \sin x }$

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

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

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

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

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

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

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

$\\$

$\displaystyle \text{Question 17: }\frac{ 10^{x} }{ \sin x }$

$\displaystyle \text{Let } u = 10^{x} ; \hspace{1.0cm} v = \sin x$

$\displaystyle \text{Then, } u' = 10^x \log 10 ; \hspace{1.0cm} v' = \cos x$

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

$\displaystyle \frac{d}{dx} \Big( \frac{ 10^{x} }{ \sin x } \Big) = \frac{(\sin x)(10^x \log 10) - (10^{x})(\cos x)}{(\sin x)^2}$

$\displaystyle = \frac{ \sin x \cdot 10^x \cdot \log 10 - 10^x \mathrm{cosec} x \cdot \cot x }{\sin^2 x}$

$\displaystyle = 10^x \cdot \log 10 \cdot \mathrm{cosec} x - 10^x \mathrm{cosec} x \cdot \cot x$

$\displaystyle = 10^x \cdot \mathrm{cosec} x ( \log 10 - \cot x)$

$\\$

$\displaystyle \text{Question 18: }\frac{ 1+3^x }{ 1 - 3^x }$

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

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

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

$\displaystyle \frac{d}{dx} \Big( \frac{ 1+3^x }{ 1 - 3^x } \Big) = \frac{(1 - 3^x)(3^x \log 3) - (1+3^x)(- 3^x \log 3)}{(1 - 3^x)^2}$

$\displaystyle = \frac{ 3^x \log 3 - 3^{2x} \log 3 + 3^x \log 3 + 3^{2x} \log 3 }{(1 - 3^x)^2}$

$\displaystyle = \frac{ 2 \cdot 3^x \cdot \log 3 }{(1 - 3^x)^2}$

$\\$

$\displaystyle \text{Question 19: }\frac{ 3^x }{x+ \tan x }$

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

$\displaystyle \text{Then, } u' =3^x \log 3 ; \hspace{1.0cm} v' = 1+ \sec^2 x$

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

$\displaystyle \frac{d}{dx} \Big( \frac{u}{v} \Big) = \frac{(x+ \tan x)(3^x \log 3) - (3^x)(1+ \sec^2 x)}{(x+ \tan x)^2}$

$\displaystyle = \frac{ 3^x [( x + \tan x ) \log 3 - ( 1 + \sec^2 x)] }{(x+ \tan x)^2}$

$\\$

$\displaystyle \text{Question 20: }\frac{ 1 + \log x }{ 1 - \log x }$

$\displaystyle \text{Let } u = 1 + \log x ; \hspace{1.0cm} v =1 - \log x$

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

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

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

$\displaystyle = \frac{ 1 - \log x + 1 + \log x }{x(1 - \log x)^2}$

$\displaystyle = \frac{ 2 }{x(1 - \log x)^2}$

$\\$

$\displaystyle \text{Question 21: } \frac{ 4x+ 5 \sin x }{ 3x + 7 \cos x }$

$\displaystyle \text{Let } u = 4x+ 5 \sin x ; \hspace{1.0cm} v = 3x + 7 \cos x$

$\displaystyle \text{Then, } u' = 4 + 5 \cos x ; \hspace{1.0cm} v' = 3 - 7 \sin x$

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

$\displaystyle \frac{d}{dx} \Big( \frac{ 4x+ 5 \sin x }{ 3x + 7 \cos x } \Big) = \frac{(3x + 7 \cos x)(4 + 5 \cos x) - (4x+ 5 \sin x)(3 - 7 \sin x)}{(3x + 7 \cos x)^2}$

$\displaystyle = \frac{ 12x + 15 x \cos x + 28 \cos x + 35 \cos^2 x - 12 x + 28 x \sin x - 15 \sin x + 35 \sin^2 x }{(3x + 7 \cos x)^2}$

$\displaystyle = \frac{ 15 x \cos x + 28 x \sin x +28 \cos x 15 \sin x + 35 ( \sin^2 x + \cos^2 x ) }{(3x + 7 \cos x)^2}$

$\displaystyle = \frac{ 15 x \cos x + 28 x \sin x +28 \cos x 15 \sin x + 35 }{(3x + 7 \cos x)^2}$

$\\$

$\displaystyle \text{Question 22: }\frac{ x }{1 + \tan x }$

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

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

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

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

$\\$

$\displaystyle \text{Question 23: }\frac{ a + b \sin x }{ c + d \cos x }$

$\displaystyle \text{Let } u = a + b \sin x ; \hspace{1.0cm} v = c + d \cos x$

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

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

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

$\displaystyle = \frac{ bc \cos s + bd \cos^2 x + ad \sin x + bd \sin^2 x }{(c + d \cos x)^2}$

$\displaystyle = \frac{ bc \cos x + ad \sin x + bd ( \sin^2 x + \cos^2 x) }{(c + d \cos x)^2}$

$\displaystyle = \frac{ bc \cos x + ad \sin x + bd }{(c + d \cos x)^2}$

$\\$

$\displaystyle \text{Question 24: }\frac{ px^2 + qx + r }{ ax+b }$

$\displaystyle \text{Let } u = px^2 + qx + r ; \hspace{1.0cm} v =ax+b$

$\displaystyle \text{Then, } u' = 2px+q ; \hspace{1.0cm} v' =a$

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

$\displaystyle \frac{d}{dx} \Big( \frac{ px^2 + qx + r }{ ax+b } \Big) = \frac{(ax+b )(2px+q) - (px^2 + qx + r)(a)}{(ax+b )^2}$

$\displaystyle = \frac{2apx^2 + aq x + 2bpx + bq - apx^2 - aqx - ar }{(ax+b )^2}$

$\displaystyle = \frac{ apx^2 + 2bpx + bq - ar }{(ax+b )^2}$

$\\$

$\displaystyle \text{Question 25: }\frac{ \sec x - 1 }{\sec x + 1 }$

$\displaystyle \text{Let } u = \sec x - 1 ; \hspace{1.0cm} v = \sec x + 1$

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

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

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

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

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

$\\$

$\displaystyle \text{Question 26: }\frac{ x^5 - \cos x }{ \sin x }$

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

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

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

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

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

$\displaystyle = \frac{ -x^5 \cos x + 5x^4 \sin x + 1 }{(\sin x )^2}$

$\\$

$\displaystyle \text{Question 27: }\frac{ x + \cos x }{ \tan x }$

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

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

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

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

$\\$

$\displaystyle \text{Question 28: }\frac{x^n }{ \sin x }$

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

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

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

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

$\\$

$\displaystyle \text{Question 29: }\frac{ ax+b }{ px^2 + qx + r }$

$\displaystyle \text{Let } u = ax+b ; \hspace{1.0cm} v =px^2 + qx + r$

$\displaystyle \text{Then, } u' = a ; \hspace{1.0cm} v' = 2px+q$

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

$\displaystyle \frac{d}{dx} \Big( \frac{ ax+b }{ px^2 + qx + r } \Big) = \frac{(px^2 + qx + r)(a) - (ax+b)(2px+q )}{(px^2 + qx + r)^2}$

$\displaystyle = \frac{ apx^2 + aqx + ar - 2apx^2 - 2bpx - aqx - bq }{(px^2 + qx + r)^2}$

$\displaystyle = \frac{ -apx^2 - 2bpx+ar - bq }{(px^2 + qx + r)^2}$

$\\$

$\displaystyle \text{Question 30: }\frac{1 }{ax^2+bx+c }$

$\displaystyle \text{Let } u = 1 ; \hspace{1.0cm} v = ax^2+bx+c$
$\displaystyle \text{Then, } u' = 0 ; \hspace{1.0cm} v' = 2ax+b$
$\displaystyle \text{Using the quotient rule: }$
$\displaystyle \frac{d}{dx} \Big( \frac{u}{v} \Big) = \frac{(ax^2+bx+c)(0) - (1)(2ax+b)}{(ax^2+bx+c)^2} = \frac{-(2ax+b)}{(ax^2+bx+c)^2}$