Differentiate the following functions with respect to x :

\displaystyle \textbf{Question 1: }x^3\sin x
\displaystyle \text{Answer:}
\displaystyle \text{Let }u=x^3,\hspace{1cm}v=\sin x.
\displaystyle \text{Then }u'=3x^2,\hspace{1cm}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)
\displaystyle =x^2(x\cos x+3\sin x)
\displaystyle \\

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

\displaystyle \textbf{Question 3: }x^2e^x\log x
\displaystyle \text{Answer:}
\displaystyle \text{Let }u=x^2,\hspace{1cm}v=e^x,\hspace{1cm}w=\log x.
\displaystyle \text{Then }u'=2x,\hspace{1cm}v'=e^x,\hspace{1cm}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^2e^x\log x)=2xe^x\log x+x^2e^x\log x+x^2e^x\cdot\frac{1}{x}
\displaystyle =2xe^x\log x+x^2e^x\log x+xe^x
\displaystyle =xe^x(2\log x+x\log x+1)
\displaystyle \\

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

\displaystyle \textbf{Question 5: }x^n\log_a x
\displaystyle \text{Answer:}
\displaystyle \text{Let }u=x^n,\hspace{1cm}v=\log_a x=\frac{\log_e x}{\log_e a}.
\displaystyle \text{Then }u'=nx^{n-1},\hspace{1cm}v'=\frac{1}{x\log_e 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\cdot\frac{1}{x\log_e a}+\log_a x\,(nx^{n-1})
\displaystyle =x^{n-1}\cdot\frac{1}{\log_e a}+n x^{n-1}\log_a x
\displaystyle =x^{n-1}\left(\frac{1}{\log_e a}+n\log_a x\right)
\displaystyle \\

\displaystyle \textbf{Question 6: }(x^3+x^2+1)\sin x
\displaystyle \text{Answer:}
\displaystyle \text{Let }u=x^3+x^2+1,\hspace{1cm}v=\sin x.
\displaystyle \text{Then }u'=3x^2+2x,\hspace{1cm}v'=\cos x.
\displaystyle \text{Using the product rule,}
\displaystyle \frac{d}{dx}(uv)=uv'+vu'.
\displaystyle \frac{d}{dx}\big((x^3+x^2+1)\sin x\big)=(x^3+x^2+1)\cos x+(3x^2+2x)\sin x
\displaystyle \\

\displaystyle \textbf{Question 7: }\sin x\cos x
\displaystyle \text{Answer:}
\displaystyle \text{Let }u=\sin x,\hspace{1cm}v=\cos x.
\displaystyle \text{Then }u'=\cos x,\hspace{1cm}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 x
\displaystyle =-\sin^2x+\cos^2x=\cos2x
\displaystyle \\

\displaystyle \textbf{Question 8: }\frac{2^x\cot x}{\sqrt{x}}
\displaystyle \text{Answer:}
\displaystyle \frac{2^x\cot x}{\sqrt{x}}=2^x\cdot\cot x\cdot x^{-\frac12}
\displaystyle \text{Let }u=2^x,\hspace{0.8cm}v=\cot x,\hspace{0.8cm}w=x^{-\frac12}.
\displaystyle \text{Then }u'=2^x\log_e2,\hspace{0.8cm}v'=-\mathrm{cosec}^2x,\hspace{0.8cm}w'=-\frac12x^{-\frac32}.
\displaystyle \text{Using the product rule,}
\displaystyle \frac{d}{dx}(uvw)=u'vw+uv'w+uvw'.
\displaystyle \frac{d}{dx}\left(\frac{2^x\cot x}{\sqrt{x}}\right)=2^x\log_e2\cdot\cot x\cdot x^{-\frac12}+2^x(-\mathrm{cosec}^2x)x^{-\frac12}+2^x\cot x\left(-\frac12x^{-\frac32}\right)
\displaystyle =2^x\log_e2\cdot\cot x\cdot\frac{1}{\sqrt{x}}-2^x\mathrm{cosec}^2x\cdot\frac{1}{\sqrt{x}}-\frac{2^x\cot x}{2x\sqrt{x}}
\displaystyle =\frac{2^x}{\sqrt{x}}\left(\log_e2\cdot\cot x-\mathrm{cosec}^2x-\frac{\cot x}{2x}\right)
\displaystyle \\

\displaystyle \textbf{Question 9: }x^2\sin x\log x
\displaystyle \text{Answer:}
\displaystyle \text{Let }u=x^2,\hspace{0.8cm}v=\sin x,\hspace{0.8cm}w=\log x.
\displaystyle \text{Then }u'=2x,\hspace{0.8cm}v'=\cos x,\hspace{0.8cm}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\sin x\log x+x^2\cos x\log x+x^2\sin x\cdot\frac{1}{x}
\displaystyle =2x\sin x\log x+x^2\cos x\log x+x\sin x
\displaystyle \\

\displaystyle \textbf{Question 10: }x^5e^x+x^6\log x
\displaystyle \text{Answer:}
\displaystyle \text{Let }u_1=x^5,\hspace{1cm}v_1=e^x.
\displaystyle \text{Let }u_2=x^6,\hspace{1cm}v_2=\log x.
\displaystyle \text{Then }u_1'=5x^4,\hspace{1cm}v_1'=e^x.
\displaystyle \text{Then }u_2'=6x^5,\hspace{1cm}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^5e^x+x^6\log x)=x^5e^x+5x^4e^x+x^6\cdot\frac{1}{x}+6x^5\log x
\displaystyle =x^5e^x+5x^4e^x+x^5+6x^5\log x
\displaystyle =x^4\left(xe^x+5e^x+x+6x\log x\right)
\displaystyle \\

\displaystyle \textbf{Question 11: }(x\sin x+\cos x)(x\cos x-\sin x)
\displaystyle \text{Answer:}
\displaystyle \text{Let }u=x\sin x+\cos x,\hspace{1cm}v=x\cos x-\sin x.
\displaystyle \text{Then }u'=x\cos x+\sin x-\sin x=x\cos x,
\displaystyle 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^2x-x\sin x\cos x+x^2\cos^2x-x\sin x\cos x
\displaystyle =x^2(\cos^2x-\sin^2x)-2x\sin x\cos x
\displaystyle =x^2\cos2x-x\sin2x
\displaystyle =x(x\cos2x-\sin2x)
\displaystyle \\

\displaystyle \textbf{Question 12: }(x\sin x+\cos x)(e^x+x^2\log x)
\displaystyle \text{Answer:}
\displaystyle \text{Let }u=x\sin x+\cos x,\hspace{1cm}v=e^x+x^2\log x.
\displaystyle \text{Then }u'=x\cos x+\sin x-\sin x=x\cos x,
\displaystyle v'=e^x+x+2x\log x.
\displaystyle \text{Using the product rule,}
\displaystyle \frac{d}{dx}(uv)=uv'+vu'.
\displaystyle \frac{d}{dx}\Big((x\sin x+\cos x)(e^x+x^2\log x)\Big)
\displaystyle =(x\sin x+\cos x)(e^x+x+2x\log x)+x\cos x(e^x+x^2\log x)
\displaystyle \\

\displaystyle \textbf{Question 13: }(1-2\tan x)(5+4\sin x)
\displaystyle \text{Answer:}
\displaystyle \text{Let }u=1-2\tan x,\hspace{1cm}v=5+4\sin x.
\displaystyle \text{Then }u'=-2\sec^2x,\hspace{1cm}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)
\displaystyle =(1-2\tan x)(4\cos x)+(5+4\sin x)(-2\sec^2x)
\displaystyle =4\cos x-8\sin x-10\sec^2x-8\sec x\tan x
\displaystyle =4\left(\cos x-2\sin x-\frac{5}{2}\sec^2x-2\sec x\tan x\right)
\displaystyle \\

\displaystyle \textbf{Question 14: }(1+x^2)\cos x
\displaystyle \text{Answer:}
\displaystyle \text{Let }u=1+x^2,\hspace{1cm}v=\cos x.
\displaystyle \text{Then }u'=2x,\hspace{1cm}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
\displaystyle =-\sin x-x^2\sin x+2x\cos x
\displaystyle \\

\displaystyle \textbf{Question 15: }\sin^2x
\displaystyle \text{Answer:}
\displaystyle \frac{d}{dx}(\sin^2x)=\frac{d}{dx}(\sin x\cdot\sin x)
\displaystyle =\sin x\cos x+\cos x\sin x
\displaystyle =2\sin x\cos x=\sin2x
\displaystyle \\

\displaystyle \textbf{Question 16: }\log_{x^2}x
\displaystyle \text{Answer:}
\displaystyle \text{For }x>0,
\displaystyle \frac{d}{dx}\left(\log_{x^2}x\right)=\frac{d}{dx}\left(\frac{\log x}{\log x^2}\right)
\displaystyle =\frac{d}{dx}\left(\frac{\log x}{2\log x}\right)
\displaystyle =\frac{d}{dx}\left(\frac12\right)=0
\displaystyle \\

\displaystyle \textbf{Question 17: }e^x\log\sqrt{x}\tan x
\displaystyle \text{Answer:}
\displaystyle \text{Let }u=e^x,\hspace{0.8cm}v=\log\sqrt{x},\hspace{0.8cm}w=\tan x.
\displaystyle \text{Then }u'=e^x,\hspace{0.8cm}v'=\frac{1}{2x},\hspace{0.8cm}w'=\sec^2x.
\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\log\sqrt{x}\tan x+e^x\cdot\frac{1}{2x}\tan x+e^x\log\sqrt{x}\sec^2x
\displaystyle =e^x\left(\log x^{\frac12}\tan x+\frac{\tan x}{2x}+\log x^{\frac12}\sec^2x\right)
\displaystyle =\frac{e^x}{2}\left(\log x\cdot\tan x+\frac{\tan x}{x}+\log x\cdot\sec^2x\right)
\displaystyle \\

\displaystyle \textbf{Question 18: }x^3e^x\cos x
\displaystyle \text{Answer:}
\displaystyle \text{Let }u=x^3,\hspace{1cm}v=e^x,\hspace{1cm}w=\cos x.
\displaystyle \text{Then }u'=3x^2,\hspace{1cm}v'=e^x,\hspace{1cm}w'=-\sin x.
\displaystyle \text{Using the product rule,}
\displaystyle \frac{d}{dx}(uvw)=u'vw+uv'w+uvw'.
\displaystyle \frac{d}{dx}(x^3e^x\cos x)=3x^2e^x\cos x+x^3e^x\cos x-x^3e^x\sin x
\displaystyle =x^2e^x(3\cos x+x\cos x-x\sin x)
\displaystyle \\

\displaystyle \textbf{Question 19: }\frac{x^2\cos\frac{\pi}{4}}{\sin x}
\displaystyle \text{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{1cm}v=\cos\frac{\pi}{4},\hspace{1cm}w=\mathrm{cosec}\,x.
\displaystyle \text{Then }u'=2x,\hspace{1cm}v'=0,\hspace{1cm}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}\left(x^2\cos\frac{\pi}{4}\,\mathrm{cosec}\,x\right)
\displaystyle =2x\cos\frac{\pi}{4}\,\mathrm{cosec}\,x-x^2\cos\frac{\pi}{4}\,\mathrm{cosec}\,x\cot x
\displaystyle =\cos\frac{\pi}{4}\left(2x\,\mathrm{cosec}\,x-x^2\mathrm{cosec}\,x\cot x\right)
\displaystyle =\cos\frac{\pi}{4}\left(\frac{2x}{\sin x}-\frac{x^2\cot x}{\sin x}\right)
\displaystyle \\

\displaystyle \textbf{Question 20: }x^4(5\sin x-3\cos x)
\displaystyle \text{Answer:}
\displaystyle \text{Let }u=x^4,\hspace{1cm}v=5\sin x-3\cos x.
\displaystyle \text{Then }u'=4x^3,\hspace{1cm}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 =5x^4\cos x+3x^4\sin x+20x^3\sin x-12x^3\cos x
\displaystyle \\

\displaystyle \textbf{Question 21: }(2x^2-3)\sin x
\displaystyle \text{Answer:}
\displaystyle \text{Let }u=2x^2-3,\hspace{1cm}v=\sin x.
\displaystyle \text{Then }u'=4x,\hspace{1cm}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 \\

\displaystyle \textbf{Question 22: }x^5(3-6x^{-9})
\displaystyle \text{Answer:}
\displaystyle \text{Let }u=x^5,\hspace{1cm}v=3-6x^{-9}.
\displaystyle \text{Then }u'=5x^4,\hspace{1cm}v'=54x^{-10}.
\displaystyle \text{Using the product rule,}
\displaystyle \frac{d}{dx}(uv)=uv'+vu'.
\displaystyle \frac{d}{dx}\Big(x^5(3-6x^{-9})\Big)=x^5(54x^{-10})+5x^4(3-6x^{-9})
\displaystyle =54x^{-5}+15x^4-30x^{-5}
\displaystyle =15x^4+24x^{-5}
\displaystyle \\

\displaystyle \textbf{Question 23: }x^{-4}(3-4x^{-5})
\displaystyle \text{Answer:}
\displaystyle \text{Let }u=x^{-4},\hspace{1cm}v=3-4x^{-5}.
\displaystyle \text{Then }u'=-4x^{-5},\hspace{1cm}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}\cdot20x^{-6}-4x^{-5}(3-4x^{-5})
\displaystyle =20x^{-10}-12x^{-5}+16x^{-10}
\displaystyle =-12x^{-5}+36x^{-10}
\displaystyle \\

\displaystyle \textbf{Question 24: }x^{-3}(5+3x)
\displaystyle \text{Answer:}
\displaystyle \text{Let }u=x^{-3},\hspace{1cm}v=5+3x.
\displaystyle \text{Then }u'=-3x^{-4},\hspace{1cm}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}-15x^{-4}-9x^{-3}
\displaystyle =-15x^{-4}-6x^{-3}
\displaystyle \\

\displaystyle \textbf{Question 25: }\frac{ax+b}{cx+d}
\displaystyle \text{Answer:}
\displaystyle \text{We know } \frac{d}{dx}\left(\frac{f(x)}{g(x)}\right)=\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{1cm}g(x)=cx+d.
\displaystyle \text{Then }f'(x)=a,\hspace{1cm}g'(x)=c.
\displaystyle \frac{d}{dx}\left(\frac{ax+b}{cx+d}\right)=\frac{(cx+d)a-(ax+b)c}{(cx+d)^2}
\displaystyle =\frac{ad-bc}{(cx+d)^2}
\displaystyle \\

\displaystyle \textbf{Question 26: }(ax+b)^n(cx+d)^m
\displaystyle \text{Answer:}
\displaystyle \text{Let }u=(ax+b)^n,\hspace{1cm}v=(cx+d)^m.
\displaystyle \text{Then }u'=an(ax+b)^{n-1},\hspace{1cm}v'=cm(cx+d)^{m-1}.
\displaystyle \text{Using the product rule,}
\displaystyle \frac{d}{dx}(uv)=uv'+vu'.
\displaystyle \frac{d}{dx}\Big((ax+b)^n(cx+d)^m\Big)=(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}\left\{cm(ax+b)+an(cx+d)\right\}
\displaystyle \\

\displaystyle \textbf{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.}
\displaystyle \text{Answer:}
\displaystyle \textbf{Product Rule (1st Method)}
\displaystyle \text{Let }u=1+2\tan x,\hspace{1cm}v=5+4\cos x.
\displaystyle \text{Then }u'=2\sec^2x,\hspace{1cm}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)
\displaystyle =(1+2\tan x)(-4\sin x)+2\sec^2x(5+4\cos x)
\displaystyle =-4\sin x-8\tan x\sin x+10\sec^2x+8\sec x
\displaystyle =-4\sin x+10\sec^2x+\left(\frac{8}{\cos x}-\frac{8\sin^2x}{\cos x}\right)
\displaystyle =-4\sin x+10\sec^2x+8\left(\frac{1-\sin^2x}{\cos x}\right)
\displaystyle =-4\sin x+10\sec^2x+8\left(\frac{\cos^2x}{\cos x}\right)
\displaystyle =-4\sin x+10\sec^2x+8\cos x
\displaystyle \textbf{2nd Method}
\displaystyle (1+2\tan x)(5+4\cos x)=5+4\cos x+10\tan x+8\tan x\cos x
\displaystyle =5+4\cos x+10\tan x+8\sin x
\displaystyle \therefore \frac{d}{dx}\Big((1+2\tan x)(5+4\cos x)\Big)
\displaystyle =0-4\sin x+10\sec^2x+8\cos x
\displaystyle =-4\sin x+10\sec^2x+8\cos x
\displaystyle \text{Using both the methods, we get the same result.}
\displaystyle \\

\displaystyle \text{(i) }(3x^2+2)^2
\displaystyle \textbf{Product Rule (1st Method)}
\displaystyle \text{Let }u=3x^2+2,\hspace{1cm}v=3x^2+2.
\displaystyle \text{Then }u'=6x,\hspace{1cm}v'=6x.
\displaystyle \text{Using the product rule,}
\displaystyle \frac{d}{dx}(uv)=uv'+vu'.
\displaystyle \frac{d}{dx}\Big((3x^2+2)^2\Big)=(3x^2+2)(6x)+6x(3x^2+2)
\displaystyle =12x(3x^2+2)=36x^3+24x
\displaystyle \textbf{2nd Method}
\displaystyle \frac{d}{dx}\Big((3x^2+2)^2\Big)=\frac{d}{dx}(9x^4+12x^2+4)
\displaystyle =36x^3+24x
\displaystyle \text{Using both the methods we get the same result.}
\displaystyle \\

\displaystyle \text{(ii) }(x+2)(x+3)
\displaystyle \textbf{Product Rule (1st Method)}
\displaystyle \text{Let }u=x+2,\hspace{1cm}v=x+3.
\displaystyle \text{Then }u'=1,\hspace{1cm}v'=1.
\displaystyle \text{Using the product rule,}
\displaystyle \frac{d}{dx}(uv)=uv'+vu'.
\displaystyle \frac{d}{dx}\Big((x+2)(x+3)\Big)=(x+2)+(x+3)=2x+5
\displaystyle \textbf{2nd Method}
\displaystyle \frac{d}{dx}(x^2+5x+6)=2x+5
\displaystyle \text{Using both the methods we get the same result.}
\displaystyle \\

\displaystyle \text{(iii) }(3\sec x-4\mathrm{cosec}x)(-2\sin x+5\cos x)
\displaystyle \textbf{Product Rule (1st Method)}
\displaystyle \text{Then }u'=3\sec x\tan x+4\mathrm{cosec}x\cot x,\hspace{1cm}v'=-2\cos x-5\sin x.
\displaystyle \frac{d}{dx}\Big((3\sec x-4\mathrm{cosec}x)(-2\sin x+5\cos x)\Big)
\displaystyle =-6-15\tan x+8\cot x+20-6\tan^2x-8\cot x+15\tan x+20\cot^2x
\displaystyle =14-6(\sec^2x-1)+20(\mathrm{cosec}^2x-1)
\displaystyle =-6\sec^2x+20\mathrm{cosec}^2x
\displaystyle \textbf{2nd Method}
\displaystyle (3\sec x-4\mathrm{cosec}x)(-2\sin x+5\cos x)
\displaystyle =-6\tan x+15+8-20\cot x
\displaystyle =23-6\tan x-20\cot x
\displaystyle \therefore \frac{d}{dx}= -6\sec^2x+20\mathrm{cosec}^2x
\displaystyle \text{Using both the methods we get the same result.}
\displaystyle \\

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