Differentiate the following functions with respect to x :

\displaystyle \textbf{Question 1: }\frac{x^2+1}{x+1}
\displaystyle \text{Answer:}
\displaystyle \text{Let }u=x^2+1,\hspace{1cm}v=x+1.
\displaystyle \text{Then }u'=2x,\hspace{1cm}v'=1.
\displaystyle \text{Using the quotient rule,}
\displaystyle \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\,u'-u\,v'}{v^2}
\displaystyle \frac{d}{dx}\left(\frac{x^2+1}{x+1}\right)=\frac{(x+1)(2x)-(x^2+1)(1)}{(x+1)^2}
\displaystyle =\frac{2x^2+2x-x^2-1}{(x+1)^2}
\displaystyle =\frac{x^2+2x-1}{(x+1)^2}
\displaystyle \\

\displaystyle \textbf{Question 2: }\frac{2x-1}{x^2+1}
\displaystyle \text{Answer:}
\displaystyle \text{Let }u=2x-1,\hspace{1cm}v=x^2+1.
\displaystyle \text{Then }u'=2,\hspace{1cm}v'=2x.
\displaystyle \text{Using the quotient rule,}
\displaystyle \frac{d}{dx}\left(\frac{2x-1}{x^2+1}\right)=\frac{(x^2+1)(2)-(2x-1)(2x)}{(x^2+1)^2}
\displaystyle =\frac{2x^2+2-4x^2+2x}{(x^2+1)^2}
\displaystyle =\frac{-2x^2+2x+2}{(x^2+1)^2}
\displaystyle =\frac{2(-x^2+x+1)}{(x^2+1)^2}
\displaystyle \\

\displaystyle \textbf{Question 3: }\frac{x+e^x}{1+\log x}
\displaystyle \text{Answer:}
\displaystyle \text{Let }u=x+e^x,\hspace{1cm}v=1+\log x.
\displaystyle \text{Then }u'=1+e^x,\hspace{1cm}v'=\frac{1}{x}.
\displaystyle \text{Using the quotient rule,}
\displaystyle \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{vu'-uv'}{v^2}.
\displaystyle \frac{d}{dx}\left(\frac{x+e^x}{1+\log x}\right)
\displaystyle =\frac{(1+\log x)(1+e^x)-(x+e^x)\left(\frac{1}{x}\right)}{(1+\log x)^2}
\displaystyle =\frac{x(1+\log x)(1+e^x)-(x+e^x)}{x(1+\log x)^2}
\displaystyle =\frac{x+xe^x+x\log x+xe^x\log x-x-e^x}{x(1+\log x)^2}
\displaystyle =\frac{x\log x+xe^x\log x+xe^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 \\

\displaystyle \textbf{Question 4: }\frac{e^x-\tan x}{\cot x-x^n}
\displaystyle \text{Answer:}
\displaystyle \text{Let }u=e^x-\tan x,\hspace{1cm}v=\cot x-x^n.
\displaystyle \text{Then }u'=e^x-\sec^2x,\hspace{1cm}v'=-\mathrm{cosec}^2x-nx^{n-1}.
\displaystyle \text{Using the quotient rule,}
\displaystyle \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{vu'-uv'}{v^2}.
\displaystyle \frac{d}{dx}\left(\frac{e^x-\tan x}{\cot x-x^n}\right)
\displaystyle =\frac{(\cot x-x^n)(e^x-\sec^2x)-(e^x-\tan x)(-\mathrm{cosec}^2x-nx^{n-1})}{(\cot x-x^n)^2}
\displaystyle =\frac{(\cot x-x^n)(e^x-\sec^2x)+(e^x-\tan x)(\mathrm{cosec}^2x+nx^{n-1})}{(\cot x-x^n)^2}
\displaystyle \\

\displaystyle \textbf{Question 5: }\frac{ax^2+bx+c}{px^2+qx+r}
\displaystyle \text{Answer:}
\displaystyle \text{Let }u=ax^2+bx+c,\hspace{1cm}v=px^2+qx+r.
\displaystyle \text{Then }u'=2ax+b,\hspace{1cm}v'=2px+q.
\displaystyle \text{Using the quotient rule,}
\displaystyle \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{vu'-uv'}{v^2}.
\displaystyle \frac{d}{dx}\left(\frac{ax^2+bx+c}{px^2+qx+r}\right)
\displaystyle =\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}{(px^2+qx+r)^2}
\displaystyle \hspace{0.5cm}-\frac{2apx^3+aqx^2+2bpx^2+bqx+2cpx+cq}{(px^2+qx+r)^2}
\displaystyle =\frac{(aq-bp)x^2+2(ar-cp)x+br-cq}{(px^2+qx+r)^2}
\displaystyle \\

\displaystyle \textbf{Question 6: }\frac{x}{1+\tan x}
\displaystyle \text{Answer:}
\displaystyle \text{Let }u=x,\hspace{1cm}v=1+\tan x.
\displaystyle \text{Then }u'=1,\hspace{1cm}v'=\sec^2x.
\displaystyle \text{Using the quotient rule,}
\displaystyle \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{vu'-uv'}{v^2}.
\displaystyle \frac{d}{dx}\left(\frac{x}{1+\tan x}\right)
\displaystyle =\frac{(1+\tan x)(1)-x\sec^2x}{(1+\tan x)^2}
\displaystyle =\frac{1+\tan x-x\sec^2x}{(1+\tan x)^2}
\displaystyle \\

\displaystyle \textbf{Question 7: }\frac{1}{ax^2+bx+c}
\displaystyle \text{Answer:}
\displaystyle \text{Let }u=1,\hspace{1cm}v=ax^2+bx+c.
\displaystyle \text{Then }u'=0,\hspace{1cm}v'=2ax+b.
\displaystyle \text{Using the quotient rule,}
\displaystyle \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{vu'-uv'}{v^2}.
\displaystyle \frac{d}{dx}\left(\frac{1}{ax^2+bx+c}\right)
\displaystyle =\frac{(ax^2+bx+c)(0)-(1)(2ax+b)}{(ax^2+bx+c)^2}
\displaystyle =-\frac{2ax+b}{(ax^2+bx+c)^2}
\displaystyle \\

\displaystyle \textbf{Question 8: }\frac{e^x}{1+x^2}
\displaystyle \text{Answer:}
\displaystyle \text{Let }u=e^x,\hspace{1cm}v=1+x^2.
\displaystyle \text{Then }u'=e^x,\hspace{1cm}v'=2x.
\displaystyle \text{Using the quotient rule,}
\displaystyle \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{vu'-uv'}{v^2}.
\displaystyle \frac{d}{dx}\left(\frac{e^x}{1+x^2}\right)
\displaystyle =\frac{(1+x^2)e^x-e^x(2x)}{(1+x^2)^2}
\displaystyle =\frac{e^x+x^2e^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 \\

\displaystyle \textbf{Question 9: }\frac{e^x+\sin x}{1+\log x}
\displaystyle \text{Answer:}
\displaystyle \text{Let }u=e^x+\sin x,\hspace{1cm}v=1+\log x.
\displaystyle \text{Then }u'=e^x+\cos x,\hspace{1cm}v'=\frac{1}{x}.
\displaystyle \text{Using the quotient rule,}
\displaystyle \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{vu'-uv'}{v^2}.
\displaystyle \frac{d}{dx}\left(\frac{e^x+\sin x}{1+\log x}\right)
\displaystyle =\frac{(1+\log x)(e^x+\cos x)-(e^x+\sin x)\left(\frac{1}{x}\right)}{(1+\log x)^2}
\displaystyle =\frac{x(1+\log x)(e^x+\cos x)-(e^x+\sin x)}{x(1+\log x)^2}
\displaystyle \\

\displaystyle \textbf{Question 10: }\frac{x\tan x}{\sec x+\tan x}
\displaystyle \text{Answer:}
\displaystyle \text{Let }u=x\tan x,\hspace{1cm}v=\sec x+\tan x.
\displaystyle \text{Then }u'=x\sec^2x+\tan x,
\displaystyle v'=\sec x\tan x+\sec^2x.
\displaystyle \text{Using the quotient rule,}
\displaystyle \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{vu'-uv'}{v^2}.
\displaystyle \frac{d}{dx}\left(\frac{x\tan x}{\sec x+\tan x}\right)
\displaystyle =\frac{(\sec x+\tan x)(x\sec^2x+\tan x)-x\tan x(\sec x\tan x+\sec^2x)}{(\sec x+\tan x)^2}
\displaystyle =\frac{(\sec x+\tan x)(x\sec^2x+\tan x)-x\tan x\sec x(\sec x+\tan x)}{(\sec x+\tan x)^2}
\displaystyle =\frac{(\sec x+\tan x)\left(x\sec^2x+\tan x-x\sec x\tan x\right)}{(\sec x+\tan x)^2}
\displaystyle =\frac{x\sec^2x+\tan x-x\sec x\tan x}{\sec x+\tan x}
\displaystyle =\frac{x\sec x(\sec x-\tan x)+\tan x}{\sec x+\tan x}
\displaystyle \\

\displaystyle \textbf{Question 11: }\frac{x\sin x}{1+\cos x}
\displaystyle \text{Answer:}
\displaystyle \text{Let }u=x\sin x,\hspace{1cm}v=1+\cos x.
\displaystyle \text{Then }u'=x\cos x+\sin x,\hspace{1cm}v'=-\sin x.
\displaystyle \text{Using the quotient rule,}
\displaystyle \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{vu'-uv'}{v^2}.
\displaystyle \frac{d}{dx}\left(\frac{x\sin x}{1+\cos x}\right)
\displaystyle =\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^2x}{(1+\cos x)^2}
\displaystyle =\frac{(1+\cos x)(x\cos x+\sin x)+x(1-\cos^2x)}{(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)\left[x\cos x+\sin x+x(1-\cos x)\right]}{(1+\cos x)^2}
\displaystyle =\frac{x\cos x+\sin x+x-x\cos x}{1+\cos x}
\displaystyle =\frac{x+\sin x}{1+\cos x}
\displaystyle \\

\displaystyle \textbf{Question 12: }\frac{2^x\cot x}{\sqrt{x}}
\displaystyle \text{Answer:}
\displaystyle \text{Let }u=2^x\cot x,\hspace{1cm}v=\sqrt{x}.
\displaystyle \text{Then }u'=2^x\log 2\cot x-2^x\mathrm{cosec}^2x,
\displaystyle v'=\frac{1}{2\sqrt{x}}.
\displaystyle \text{Using the quotient rule,}
\displaystyle \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{vu'-uv'}{v^2}.
\displaystyle \frac{d}{dx}\left(\frac{2^x\cot x}{\sqrt{x}}\right)
\displaystyle =\frac{\sqrt{x}\left(2^x\log 2\cot x-2^x\mathrm{cosec}^2x\right)-2^x\cot x\left(\frac{1}{2\sqrt{x}}\right)}{x}
\displaystyle =\frac{x\left(2^x\log 2\cot x-2^x\mathrm{cosec}^2x\right)-2^{x-1}\cot x}{x\sqrt{x}}
\displaystyle =\frac{2^x\left(x\log 2\cot x-x\mathrm{cosec}^2x-\frac{1}{2}\cot x\right)}{x\sqrt{x}}
\displaystyle =\frac{2^x\left(-x\mathrm{cosec}^2x+x\cot x\log 2-\frac{1}{2}\cot x\right)}{x^{\frac{3}{2}}}
\displaystyle \\

\displaystyle \textbf{Question 13: }\frac{\sin x-x\cos x}{x\sin x+\cos x}
\displaystyle \text{Answer:}
\displaystyle \text{Let }u=\sin x-x\cos x,\hspace{1cm}v=x\sin x+\cos x.
\displaystyle \text{Then }u'=\cos x-\left(x(-\sin x)+\cos x\right)=x\sin x,
\displaystyle v'=x\cos x+\sin x-\sin x=x\cos x.
\displaystyle \text{Using the quotient rule,}
\displaystyle \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{vu'-uv'}{v^2}.
\displaystyle \frac{d}{dx}\left(\frac{\sin x-x\cos x}{x\sin x+\cos x}\right)
\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^2x+x\sin x\cos x-x\sin x\cos x+x^2\cos^2x}{(x\sin x+\cos x)^2}
\displaystyle =\frac{x^2(\sin^2x+\cos^2x)}{(x\sin x+\cos x)^2}
\displaystyle =\frac{x^2}{(x\sin x+\cos x)^2}
\displaystyle \\

\displaystyle \textbf{Question 14: }\frac{x^2-x+1}{x^2+x+1}
\displaystyle \text{Answer:}
\displaystyle \text{Let }u=x^2-x+1,\hspace{1cm}v=x^2+x+1.
\displaystyle \text{Then }u'=2x-1,\hspace{1cm}v'=2x+1.
\displaystyle \text{Using the quotient rule,}
\displaystyle \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{vu'-uv'}{v^2}.
\displaystyle \frac{d}{dx}\left(\frac{x^2-x+1}{x^2+x+1}\right)
\displaystyle =\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 \\

\displaystyle \textbf{Question 15: }\frac{\sqrt{a}+\sqrt{x}}{\sqrt{a}-\sqrt{x}}
\displaystyle \text{Answer:}
\displaystyle \text{Let }u=\sqrt{a}+\sqrt{x},\hspace{1cm}v=\sqrt{a}-\sqrt{x}.
\displaystyle \text{Then }u'=\frac{1}{2\sqrt{x}},\hspace{1cm}v'=-\frac{1}{2\sqrt{x}}.
\displaystyle \text{Using the quotient rule,}
\displaystyle \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{vu'-uv'}{v^2}.
\displaystyle \frac{d}{dx}\left(\frac{\sqrt{a}+\sqrt{x}}{\sqrt{a}-\sqrt{x}}\right)
\displaystyle =\frac{(\sqrt{a}-\sqrt{x})\left(\frac{1}{2\sqrt{x}}\right)-(\sqrt{a}+\sqrt{x})\left(-\frac{1}{2\sqrt{x}}\right)}{(\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{2\sqrt{a}}{2\sqrt{x}(\sqrt{a}-\sqrt{x})^2}
\displaystyle =\frac{\sqrt{a}}{\sqrt{x}(\sqrt{a}-\sqrt{x})^2}
\displaystyle \\

\displaystyle \textbf{Question 16: }\frac{a+\sin x}{1+a\sin x}
\displaystyle \text{Answer:}
\displaystyle \text{Let }u=a+\sin x,\hspace{1cm}v=1+a\sin x.
\displaystyle \text{Then }u'=\cos x,\hspace{1cm}v'=a\cos x.
\displaystyle \text{Using the quotient rule,}
\displaystyle \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{vu'-uv'}{v^2}.
\displaystyle \frac{d}{dx}\left(\frac{a+\sin x}{1+a\sin x}\right)
\displaystyle =\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{(1-a^2)\cos x}{(1+a\sin x)^2}
\displaystyle \\

\displaystyle \textbf{Question 17: }\frac{10^x}{\sin x}
\displaystyle \text{Answer:}
\displaystyle \text{Let }u=10^x,\hspace{1cm}v=\sin x.
\displaystyle \text{Then }u'=10^x\log10,\hspace{1cm}v'=\cos x.
\displaystyle \text{Using the quotient rule,}
\displaystyle \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{vu'-uv'}{v^2}.
\displaystyle \frac{d}{dx}\left(\frac{10^x}{\sin x}\right)
\displaystyle =\frac{(\sin x)(10^x\log10)-(10^x)(\cos x)}{\sin^2x}
\displaystyle =10^x\log10\cdot\mathrm{cosec}x-10^x\mathrm{cosec}x\cot x
\displaystyle =10^x\mathrm{cosec}x(\log10-\cot x)
\displaystyle \\

\displaystyle \textbf{Question 18: }\frac{1+3^x}{1-3^x}
\displaystyle \text{Answer:}
\displaystyle \text{Let }u=1+3^x,\hspace{1cm}v=1-3^x.
\displaystyle \text{Then }u'=3^x\log3,\hspace{1cm}v'=-3^x\log3.
\displaystyle \text{Using the quotient rule,}
\displaystyle \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{vu'-uv'}{v^2}.
\displaystyle \frac{d}{dx}\left(\frac{1+3^x}{1-3^x}\right)
\displaystyle =\frac{(1-3^x)(3^x\log3)-(1+3^x)(-3^x\log3)}{(1-3^x)^2}
\displaystyle =\frac{3^x\log3-3^{2x}\log3+3^x\log3+3^{2x}\log3}{(1-3^x)^2}
\displaystyle =\frac{2\cdot3^x\log3}{(1-3^x)^2}
\displaystyle \\

\displaystyle \textbf{Question 19: }\frac{3^x}{x+\tan x}
\displaystyle \text{Answer:}
\displaystyle \text{Let }u=3^x,\hspace{1cm}v=x+\tan x.
\displaystyle \text{Then }u'=3^x\log3,\hspace{1cm}v'=1+\sec^2x.
\displaystyle \text{Using the quotient rule,}
\displaystyle \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{vu'-uv'}{v^2}.
\displaystyle \frac{d}{dx}\left(\frac{3^x}{x+\tan x}\right)
\displaystyle =\frac{(x+\tan x)(3^x\log3)-(3^x)(1+\sec^2x)}{(x+\tan x)^2}
\displaystyle =\frac{3^x\left[(x+\tan x)\log3-(1+\sec^2x)\right]}{(x+\tan x)^2}
\displaystyle \\

\displaystyle \textbf{Question 20: }\frac{1+\log x}{1-\log x}
\displaystyle \text{Answer:}
\displaystyle \text{Let }u=1+\log x,\hspace{1cm}v=1-\log x.
\displaystyle \text{Then }u'=\frac{1}{x},\hspace{1cm}v'=-\frac{1}{x}.
\displaystyle \text{Using the quotient rule,}
\displaystyle \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{vu'-uv'}{v^2}.
\displaystyle \frac{d}{dx}\left(\frac{1+\log x}{1-\log x}\right)
\displaystyle =\frac{(1-\log x)\left(\frac{1}{x}\right)-(1+\log x)\left(-\frac{1}{x}\right)}{(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 \\

\displaystyle \textbf{Question 21: }\frac{4x+5\sin x}{3x+7\cos x}
\displaystyle \text{Answer:}
\displaystyle \text{Let }u=4x+5\sin x,\hspace{1cm}v=3x+7\cos x.
\displaystyle \text{Then }u'=4+5\cos x,\hspace{1cm}v'=3-7\sin x.
\displaystyle \text{Using the quotient rule,}
\displaystyle \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{vu'-uv'}{v^2}.
\displaystyle \frac{d}{dx}\left(\frac{4x+5\sin x}{3x+7\cos x}\right)
\displaystyle =\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+15x\cos x+28\cos x+35\cos^2x-12x+28x\sin x-15\sin x+35\sin^2x}{(3x+7\cos x)^2}
\displaystyle =\frac{15x\cos x+28x\sin x+28\cos x-15\sin x+35(\sin^2x+\cos^2x)}{(3x+7\cos x)^2}
\displaystyle =\frac{15x\cos x+28x\sin x+28\cos x-15\sin x+35}{(3x+7\cos x)^2}
\displaystyle \\

\displaystyle \textbf{Question 22: }\frac{x}{1+\tan x}
\displaystyle \text{Answer:}
\displaystyle \text{Let }u=x,\hspace{1cm}v=1+\tan x.
\displaystyle \text{Then }u'=1,\hspace{1cm}v'=\sec^2x.
\displaystyle \text{Using the quotient rule,}
\displaystyle \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{vu'-uv'}{v^2}.
\displaystyle \frac{d}{dx}\left(\frac{x}{1+\tan x}\right)
\displaystyle =\frac{(1+\tan x)(1)-x\sec^2x}{(1+\tan x)^2}
\displaystyle =\frac{1+\tan x-x\sec^2x}{(1+\tan x)^2}
\displaystyle \\

\displaystyle \textbf{Question 23: }\frac{a+b\sin x}{c+d\cos x}
\displaystyle \text{Answer:}
\displaystyle \text{Let }u=a+b\sin x,\hspace{1cm}v=c+d\cos x.
\displaystyle \text{Then }u'=b\cos x,\hspace{1cm}v'=-d\sin x.
\displaystyle \text{Using the quotient rule,}
\displaystyle \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{vu'-uv'}{v^2}.
\displaystyle \frac{d}{dx}\left(\frac{a+b\sin x}{c+d\cos x}\right)
\displaystyle =\frac{(c+d\cos x)(b\cos x)-(a+b\sin x)(-d\sin x)}{(c+d\cos x)^2}
\displaystyle =\frac{bc\cos x+bd\cos^2x+ad\sin x+bd\sin^2x}{(c+d\cos x)^2}
\displaystyle =\frac{bc\cos x+ad\sin x+bd(\sin^2x+\cos^2x)}{(c+d\cos x)^2}
\displaystyle =\frac{bc\cos x+ad\sin x+bd}{(c+d\cos x)^2}
\displaystyle \\

\displaystyle \textbf{Question 24: }\frac{px^2+qx+r}{ax+b}
\displaystyle \text{Answer:}
\displaystyle \text{Let }u=px^2+qx+r,\hspace{1cm}v=ax+b.
\displaystyle \text{Then }u'=2px+q,\hspace{1cm}v'=a.
\displaystyle \text{Using the quotient rule,}
\displaystyle \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{vu'-uv'}{v^2}.
\displaystyle \frac{d}{dx}\left(\frac{px^2+qx+r}{ax+b}\right)
\displaystyle =\frac{(ax+b)(2px+q)-a(px^2+qx+r)}{(ax+b)^2}
\displaystyle =\frac{2apx^2+aqx+2bpx+bq-apx^2-aqx-ar}{(ax+b)^2}
\displaystyle =\frac{apx^2+2bpx+bq-ar}{(ax+b)^2}
\displaystyle \\

\displaystyle \textbf{Question 25: }\frac{\sec x-1}{\sec x+1}
\displaystyle \text{Answer:}
\displaystyle \text{Let }u=\sec x-1,\hspace{1cm}v=\sec x+1.
\displaystyle \text{Then }u'=\sec x\tan x,\hspace{1cm}v'=\sec x\tan x.
\displaystyle \text{Using the quotient rule,}
\displaystyle \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{vu'-uv'}{v^2}.
\displaystyle \frac{d}{dx}\left(\frac{\sec x-1}{\sec x+1}\right)
\displaystyle =\frac{(\sec x+1)(\sec x\tan x)-(\sec x-1)(\sec x\tan x)}{(\sec x+1)^2}
\displaystyle =\frac{\sec^2x\tan x+\sec x\tan x-\sec^2x\tan x+\sec x\tan x}{(\sec x+1)^2}
\displaystyle =\frac{2\sec x\tan x}{(\sec x+1)^2}
\displaystyle \\

\displaystyle \textbf{Question 26: }\frac{x^5-\cos x}{\sin x}
\displaystyle \text{Answer:}
\displaystyle \text{Let }u=x^5-\cos x,\hspace{1cm}v=\sin x.
\displaystyle \text{Then }u'=5x^4+\sin x,\hspace{1cm}v'=\cos x.
\displaystyle \text{Using the quotient rule,}
\displaystyle \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{vu'-uv'}{v^2}.
\displaystyle \frac{d}{dx}\left(\frac{x^5-\cos x}{\sin x}\right)
\displaystyle =\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^2x+\cos^2x)}{(\sin x)^2}
\displaystyle =\frac{-x^5\cos x+5x^4\sin x+1}{(\sin x)^2}
\displaystyle \\

\displaystyle \textbf{Question 27: }\frac{x+\cos x}{\tan x}
\displaystyle \text{Answer:}
\displaystyle \text{Let }u=x+\cos x,\hspace{1cm}v=\tan x.
\displaystyle \text{Then }u'=1-\sin x,\hspace{1cm}v'=\sec^2x.
\displaystyle \text{Using the quotient rule,}
\displaystyle \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{vu'-uv'}{v^2}.
\displaystyle \frac{d}{dx}\left(\frac{x+\cos x}{\tan x}\right)=\frac{(\tan x)(1-\sin x)-(x+\cos x)(\sec^2x)}{(\tan x)^2}
\displaystyle \\

\displaystyle \textbf{Question 28: }\frac{x^n}{\sin x}
\displaystyle \text{Answer:}
\displaystyle \text{Let }u=x^n,\hspace{1cm}v=\sin x.
\displaystyle \text{Then }u'=nx^{n-1},\hspace{1cm}v'=\cos x.
\displaystyle \text{Using the quotient rule,}
\displaystyle \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{vu'-uv'}{v^2}.
\displaystyle \frac{d}{dx}\left(\frac{x^n}{\sin x}\right)=\frac{(\sin x)(nx^{n-1})-(x^n)(\cos x)}{(\sin x)^2}
\displaystyle =\frac{nx^{n-1}\sin x-x^n\cos x}{\sin^2x}
\displaystyle \\

\displaystyle \textbf{Question 29: }\frac{ax+b}{px^2+qx+r}
\displaystyle \text{Answer:}
\displaystyle \text{Let }u=ax+b,\hspace{1cm}v=px^2+qx+r.
\displaystyle \text{Then }u'=a,\hspace{1cm}v'=2px+q.
\displaystyle \text{Using the quotient rule,}
\displaystyle \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{vu'-uv'}{v^2}.
\displaystyle \frac{d}{dx}\left(\frac{ax+b}{px^2+qx+r}\right)
\displaystyle =\frac{(px^2+qx+r)(a)-(ax+b)(2px+q)}{(px^2+qx+r)^2}
\displaystyle =\frac{apx^2+aqx+ar-2apx^2-aqx-2bpx-bq}{(px^2+qx+r)^2}
\displaystyle =\frac{-apx^2-2bpx+ar-bq}{(px^2+qx+r)^2}
\displaystyle \\

\displaystyle \textbf{Question 30: }\frac{1}{ax^2+bx+c}
\displaystyle \text{Answer:}
\displaystyle \text{Let }u=1,\hspace{1cm}v=ax^2+bx+c.
\displaystyle \text{Then }u'=0,\hspace{1cm}v'=2ax+b.
\displaystyle \text{Using the quotient rule,}
\displaystyle \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{vu'-uv'}{v^2}.
\displaystyle \frac{d}{dx}\left(\frac{1}{ax^2+bx+c}\right)=\frac{(ax^2+bx+c)(0)-(1)(2ax+b)}{(ax^2+bx+c)^2}
\displaystyle =-\frac{2ax+b}{(ax^2+bx+c)^2}
\displaystyle \\

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