12th Class Math New Book Federal Board Exercise 2.6
12th Class Math New Book Federal Board Exercise 2.6
Lets check the derivative of trigonometric functions first.
$$\frac d{dx}(Sin\;x)\;=\;Cos\;x\\\frac d{dx}(Cos\;x)\;=\;-Sin\;x\\\frac d{dx}(Tan\;x)\;=\;Sec^2\;x\\\frac d{dx}(Cot\;x)\;=\;-Cosec^2\;x\\\frac d{dx}(Ssc\;x)\;=\;Sec\;x\;Tan\;x\\\frac d{dx}(Cosec\;x)\;=-Cosecx\;Cotx$$
Find the derivative of the given functions.
Q1 :-
$$y=x^2-C\mathrm{os}x$$
$$\frac d{dx}y=\frac d{dx}(x^2-C\mathrm{os}x)\\\frac{dy}{dx}=\frac d{dx}x^2-\frac d{dx}Cosx$$
$$=2x^{2-1}-(-Sinx)\\=2x+Sinx$$
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Q2 :-
$$y=4x^3+x+Sinx$$
$$\frac d{dx}y=\frac d{dx}(4x^3+x+Sinx)\\\frac{dy}{dx}=\frac d{dx}4x^3+\frac d{dx}x+\frac d{dx}Sinx$$
$$=4(3x^{3-1})+1+Cosx\\=12x^2+1+Cosx$$
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Q3 :-
$$y=3\;Cosx-5\;Cotx$$
$$\frac d{dx}y=\frac d{dx}(3\;Cosx-5\;Cotx)\\\frac{dy}{dx}=\frac d{dx}(3\;Cosx)-\frac d{dx}(5\;Cotx)$$
$$=3\frac d{dx}Cosx-5\frac d{dx}Cotx\\=3(-Sinx)-5(-Cosec^2x)\\=-3\;Sinx+5\;Cosec^2x$$
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Q4 :-
$$y=\;Sinx\;Cosx$$
$$\frac d{dx}y=\frac d{dx}\;Sinx\;Cosx$$
$$\frac{dy}{dx}=Sinx\frac d{dx}Cosx\\+\;Cosx\frac d{dx}Sinx$$
$$=Sinx\;(-Sinx)+Cosx(Cosx)\\=-Sin^2x+Cos^2x\\=Cos^2x-Sin^2x\\=Cos\;2x$$
Another Method:-
$$y=\;Sinx\;Cosx$$
$$y=\frac{2\;Sinx\;Cosx}2$$
$$\frac d{dx}y=\frac d{dx}\frac{1\;}2Sin\;2x$$
$$=\frac{1\;}2\frac d{dx}Sin\;2x\\=\frac12Cos\;2x\frac d{dx}2x$$
$$=\frac12Cos\;2x\;(2)\\=Cos\;2x$$
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Q5 :-
$$y=(x^2+Sinx)\;Secx$$
$$\frac d{dx}y=\frac d{dx}(x^2+Sinx)\;Secx$$
$$\frac{dy}{dx}=(x^2+Sinx)\;\frac d{dx}\;Secx\\+\;Secx\frac d{dx}(x^2+Sinx)\;$$
$$\frac{dy}{dx}=(x^2+Sinx)\;(Secx\;\tan x)\\+Secx(\frac d{dx}x^2+\frac d{dx}Sinx)$$
$$\frac{dy}{dx}=(x^2+Sinx)\;(Secx\;\tan x)\\+Secx\;(2x+Cosx)\\\\$$
$$\frac{dy}{dx}=x^2Secx\;\tan x+Sinx\;Secx\;\tan x\\+2x\;Secx\;+Secx.Cosx\\\\$$
$$=x^2\frac1{Cosx}.\frac{Sinx}{Cosx}+Sinx.\frac1{Cosx}.\frac{Sinx}{Cosx}\\+2x.\frac1{Cosx}+Secx.Cosx\\\\$$
$$=x^2\frac{Sinx}{Cos^2x}+\frac{Sin^2x}{Cos^2x}\\+\frac{2x}{Cosx}+\frac1{Cosx}.Cosx\\\\$$
$$=x^2\frac{Sinx}{Cos^2x}+\frac{Sin^2x}{Cos^2x}\\+\frac{2x}{Cosx}+1\\\\$$
Another Method:-
$$y=(x^2+Sinx)\;Secx$$
$$y=x^2\;Secx+Sinx.Secx\\=x^2Secx\;+Sinx.\frac1{Cosx}\\=x^2Secx+\tan x$$
$$\frac d{dx}y=\frac d{dx}(x^2Secx+\tan x)\\\frac{dy}{dx}=\frac d{dx}(x^2Secx)\frac d{dx}(\tan x)\\\\$$
$$=x^2\frac d{dx}Secx+Secx\frac d{dx}x^2+Sec^2x\\\\$$
$$\frac{dy}{dx}=x^2(Secx\;\tan x)+Secx\;(2x)+Sec^2x\\=x^2Sec\;\tan x+2x\;Secx+Sec^2x\\\\$$
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Q6 :-
$$y=\frac{5-Cosx}{5+Sinx}\\\\$$
$$\frac{dy}{dx}=\frac d{dx}(\frac{5-Cosx}{5+Sinx})\\\\$$
$$\style{font-family:’Times New Roman’}{\style{font-size:8px}{\\=\frac{(5+Sinx){\displaystyle\frac d{dx}}(5-Cosx)-(5-Cosx){\displaystyle\frac d{dx}}(5+Sinx)}{{(5+Sinx)}^2}\\}}$$
$$=\frac{(5+Sinx)\;Sinx-(5-Cosx)\;Cosx}{{(5+Sinx)}^2}\\\\$$
$$=\frac{5+Sinx+Sin^2x-5Cosx+\;Cos^2x}{{(5+Sinx)}^2}\\=\frac{5(\;Sinx-Cosx\;)+1}{{(5+Sinx)}^2}\\\\$$
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Q7 :-
$$y=\frac{Secx}{1+\tan x}$$
$$\style{font-family:’Times New Roman’}{\style{font-size:12px}{\frac{dy}{dx}=\frac d{dx}(\frac{Secx}{1+\tan x})}}$$
$$\style{font-family:’Times New Roman’}{\style{font-size:8px}{=\frac{(1+\tan x){\displaystyle\frac d{dx}}Secx-Secx{\displaystyle\frac d{dx}}(1+\tan x)}{{(1+\tan x)}^2}}}$$
$$=\frac{(1+\tan x)\;(Secx\;\tan x)-Secx(Sec^2x)}{{(1+\tan x)}^2}$$
$$\frac{dy}{dx}=\frac{Secx(\tan x\;+\tan^2x-Sec^2x)}{{(1+\tan x)}^2}$$
$$\frac{dy}{dx}=\frac{Secx(\tan x\;+\tan^2x-(1+\tan^2x)}{{(1+\tan x)}^2}$$
$$=\frac{Secx(\tan x\;+\tan^2x-1-\tan^2x)}{{(1+\tan x)}^2}\\=\frac{Secx\;(\tan x-1)}{{(1+\tan x)}^2}$$
Another Method:-
$$y=\frac{Secx}{1+\tan x}$$
$$y=\frac1{\displaystyle\frac{Cosx}{1+{\displaystyle\frac{Sinx}{Cosx}}}}$$
$$=\frac1{\displaystyle\frac{Cosx}{\displaystyle\frac{Cosx+Sinx}{Cosx}}}$$
$$=\frac1{Cosx}\times\frac{Cosx}{Cosx+Sinx}\\=\frac1{Cosx+Sinx}$$
$$y={(Cosx+Sinx)}^{-1}$$
$$\frac d{dx}y=\frac d{dx}{(Cosx+Sinx)}^{-1}$$
$$\frac{dy}{dx}=-1\;(Cosx+Sinx)\frac d{dx}{(Cosx+Sinx)}$$
$$=-1\;{(Cosx+Sinx)}^{-2}\;(-Sinx+Cosx)$$
$$=\frac{-1\;}{{(Cosx+Sinx)}^2}.\;(-Sinx+Cosx)$$
$$=\frac{Sinx+Cosx\;}{{(Cosx+Sinx)}^2}$$
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Q8 :-
$$y=\frac{Sinx}{x^2+Sinx}$$
$$\frac{dy}{dx}=\frac d{dx}(\frac{Sinx}{x^2+Sinx})$$
$$\style{font-family:’Times New Roman’}{\style{font-size:8px}{=\frac{(x^2+Sinx){\displaystyle\frac d{dx}}Sinx-Sinx{\displaystyle\frac d{dx}}(x^2+Sinx)}{{(x^2+Sinx)}^2}}}$$
$$=\frac{(x^2+Sinx)Cosx-Sinx\;(2x+Cosx)}{{(x^2+Sinx)}^2}$$
$$=\frac{x^2Cosx+Sinx\;Cosx-2x\;Sinx-Sinx\;Cosx}{{(x^2+Sinx)}^2}$$
$$=\frac{x\;(x\;Cosx-2\;Sinx)}{{(x^2+Sinx)}^2}$$
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Q9 :-
$$y=\frac{Cotx}{x+1}$$
$$\frac d{dx}y=\frac d{dx}\frac{Cotx}{x+1}$$
$$\frac{dy}{dx}=\frac{(x+1){\displaystyle\frac d{dx}}Cotx-Cotx{\displaystyle\frac d{dx}}(x+1)}{{(x+1)}^2}$$
$$\frac{dy}{dx}=\frac{(x+1)\;(-Cosec^2x)-Cotx\;(1)}{{(x+1)}^2}$$
$$\frac{dy}{dx}=\frac{-x\;Cosec^2x-Cosec^2x-Cotx\;}{{(x+1)}^2}$$
$$=\frac{-x\;Cosec^2x-(1+Cot^2x)-Cotx}{{(x+1)}^2}$$
$$=\frac{-x\;Cosec^2x-1-Cot^2x-Cotx}{{(x+1)}^2}$$
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Q10 :-
$$y=(1+Cosx)\;(x-Sinx)$$
$$\frac d{dx}y=\frac d{dx}(1+Cosx)\;(x-Sinx)$$
$$\frac{dy}{dx}=(1+Cosx)\;\frac d{dx}(x-Sinx)\\+(x-Sinx)\frac d{dx}(1+Cosx)$$
$$=(1+Cosx)(1-Cosx)\;\\+(x-Sinx)\;(-Sinx)$$
$$=1-Cos^2x+Sin^2x-x\;Sinx\\$$
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Lets check the derivative of inverse trigonometric functions
$$\frac d{dx}Sin^{-1}x\;\;=\frac1{\sqrt{1-x^2}}\\\\\frac d{dx}Cos^{-1}x=\frac{-1}{\sqrt{1-x^2}}\\\\\frac d{dx}\tan^{-1}x\;=\frac1{1+x^2}\\\\\frac d{dx}Cot^{-1}x\;\;=\frac{-1}{1+x^2}\\\\\frac d{dx}Sec^{-1}x\;=\frac1{x\sqrt{x^2-1}}\\\\\frac d{dx}Cosec^{-1}x\;=\frac{-1}{x\sqrt{x^2-1}}$$
Q11 :-
$$y=Sin^{-1\;}(5x-1)$$
$$\frac d{dx}y=\frac d{dx}Sin^{-1\;}(5x-1)$$
$$\frac{dy}{dx}=\frac1{\sqrt{1-{(5x-1)}^2}}\frac d{dx}(5x-1)$$
$$=\frac1{\sqrt{1-(25x^2+1-10x)}}$$
$$=\frac5{\sqrt{-25x^2+10x}}$$
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Q12 :-
$$y=4\;Cot^{-1\;}\frac x2$$
$$\frac d{dx}y=\frac d{dx}4\;Cot^{-1\;}(\frac x2)$$
$$\frac{dy}{dx}=4\frac d{dx}\;Cot^{-1\;}(\frac x2)$$
$$\frac{dy}{dx}=4\frac{-1}{1+{({\displaystyle\frac x2})}^2}\frac d{dx}\;(\frac x2)$$
$$=\frac{-4}{1+{\displaystyle\frac{x^2}4}}.\frac12\frac d{dx}x$$
$$=\frac{-4}{\displaystyle\frac{4+x^2}4}.\frac12(1)$$
$$=\frac{-8}{4+x^2}$$
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Q13 :-
$$y=\frac{Sin^{-1}x}{Sinx}$$
$$\frac d{dx}y=\frac d{dx}\frac{Sin^{-1}x}{Sinx}$$
$$\frac{dy}{dx}=\frac{Sinx{\displaystyle\frac d{dx}}Sin^{-1}x-Sin^{-1}x{\displaystyle\frac d{dx}}Sinx}{{(Sinx)}^2}\\\\$$
$$=\frac{Sinx{\displaystyle\frac1{\sqrt{1-x^2}}-}Sin^{-1}x.Cosx}{{(Sinx)}^2}\\\\$$
$$=\frac{Sinx-\sqrt{1-x^2}Cosx.Sin^{-1}x}{\displaystyle\frac{\sqrt{1-x^2}}{Sin^2x}}$$
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Q14 :-
$$y=\frac{Sec^{-1}x}x$$
$$\frac d{dx}y=\frac d{dx}\frac{Sec^{-1}x}x$$
$$\frac{dy}{dx}=\frac{x{\displaystyle\frac d{dx}}Sec^{-1}x-Sec^{-1}x{\displaystyle\frac d{dx}}x}{x^2}$$
$$=\frac{x{\displaystyle\frac1{x\sqrt{x^2-1}}}-Sec^{-1}x.1}{x^2}$$
$$=\frac{1-\sqrt{x^2-1}-Sec^{-1}x}{x^2\sqrt{x^2-1}}$$
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Q15 :-
$$y=x\;Sin^{-1}x+Cos^{-1}x\\y=x(Sin^{-1}x+Cos^{-1}x)$$
$$\frac d{dx}y=x\frac d{dx}(Sin^{-1}x+Cos^{-1}x)\\+(Sin^{-1}x+Cos^{-1}x)\frac d{dx}x$$
$$\frac{dy}{dx}=x(\frac1{\sqrt{1-x^2}}+\frac{-1}{\sqrt{1-x^2}})\\+(Sin^{-1}x+Cos^{-1}x)\;1$$
$$=x(0)+Sin^{-1}x+Cos^{-1}x\\=Sin^{-1}x+Cos^{-1}x$$
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Q16 :-
$$y=\frac1{\tan^{-1}\;x^2}$$
$$\frac d{dx}y=\frac d{dx}\frac1{\tan^{-1}\;x^2}$$
$$\frac{dy}{dx}=\frac{\tan^{-1}\;x^2{\displaystyle\frac d{dx}}1-1{\displaystyle\frac d{dx}}\tan^{-1}\;x^2}{{(\tan^{-1}\;x^2)}^2}$$
$$=\frac{\tan^{-1}\;x^2(0)-1{\displaystyle\frac1{1+{(x^2)}^2}}.{\displaystyle\frac d{dx}}\;x^2}{{(\tan^{-1}\;x^2)}^2}$$
$$=\frac{{\displaystyle\frac1{1+x^4}}.2x}{{(\tan^{-1}x^2)}^2}\\=\frac{-2x}{(1+x^4){(\tan^{-1}x^2)}^2}$$
Another Method:-
$$y=\frac1{\tan^{-1}x^2}$$
$$\frac d{dx}y=\frac d{dx}{(\tan^{-1}x^2)}^{-1}\\\\$$
$$\frac{dy}{dx}=-1\;{(\tan^{-1}x^2)}^{-1-1}\frac d{dx}\tan^-x^2$$
$$=-1\;{(\tan^{-1}x^2)}^{-2}\frac1{1+{(x^2)}^2}\frac d{dx}x^2$$
$$=\frac{-1}{{(\tan^{-1}x^2)}^2}.\frac1{1+x^4}.2x$$
$$=\frac{-2x}{(1+x^4){(\tan^{-1}x^2)}^2}$$
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