12th Class Math New Book Federal Board Exercise 2.4
12th Class Math New Book Federal Board Exercise 2.4. Class 12 Math New Book Federal Board Chapter 2 Notes
Q1 :- Find derivative of the functions.
a) :- $$y\;=\;x^9$$
Rule:-
$$\style{font-size:20px}{\frac d{dx}x^n\;=\;n\;x^{(n-1)}}$$
Differentiating both Sides w.r.t x
$$\frac d{dx}\;y=\frac d{dx}\;x^9\\\frac{dy}{dx}\;=\;9x^{9-1}$$
$$\frac{dy}{dx}\;=\;9x^8$$
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b) :- $$\;f\;(x)=4x^\frac13$$
Differentiating both sides w.r.t x
$$\;\;\frac d{dx}f\;(x)=\;\frac d{dx}4x^\frac13\\f’\;(x)\;=4\frac d{dx}\;x^\frac13$$
$$4(\;\frac13\;x^{\frac13-1}\;)\\\frac43\;=\;x^\frac{-2}3$$
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c) :-
f (x) = 9
Differentiating both w.r.t x
$$\;\;\frac d{dx}f\;(x)=\;\frac d{dx}\;9\\f’\;(x)\;=\;0$$
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d) :- $$f\;(x)\;=6x^{3\;}+\;3x^2-10$$
Rule:-
$$\frac d{dx}(f+g)\;=\frac d{dx}f\;+\;\frac d{dx}g$$
Differentiating both sides w.r.t x
$$\;\frac d{dx}f\;(x)\;=\frac d{dx}(6x^{3\;}+\;3x^2-10)$$
$$f’\;(x)\;=\frac d{dx}6x^{3\;}+\frac d{dx}\;3x^2-\frac d{dx}10)\\=\;6\frac d{dx}x^3+3\frac d{dx}x^2-0$$
$$=\;6\;(\;3x^{3-1})+\;3(2x^{2-1})\\=\;18\;x^2\;+\;6x$$
Q2 :- $$Determine\;f’\;(x)\;$$
a) :- $$\;f(x)\;=\;\sqrt5$$
Differentiating both sides w.r.t x
$$\frac d{dx}f(x)\;=\frac d{dx}\;\sqrt5\\f’\;(x)\;=\;0$$
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b):- $$f(x)\;=\sqrt5\;x$$
Differentiating both sides w.r.t x
$$\frac d{dx}f(x)\;=\sqrt5\frac d{dx}x\\f’\;(x)\;=\sqrt{5\;}\frac d{dx}x$$
$$=\sqrt{5\;}\;(\;1x^{1-1})\;\\=\;\sqrt{5\;}(\;1(1)\;)\\=\;\sqrt5$$
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c) :- $$f\;(x)\;=5\;\sqrt x$$
Differentiating both sides w.r.t x
$$\frac d{dx}f\;(x)\;=\frac d{dx}(5\;\sqrt x)\\f’f(x)\;=\;5\frac d{dx}\sqrt x$$
$$=\;5(\frac d{dx}x^\frac12\;)\\=\;5(\frac12x^{\frac12-1}\;)$$
$$=\;5(\frac12x^\frac{-1}2\;)\\=\;5(\frac1{2x^{1/2}})$$
$$=\;5(\frac1{2\sqrt x})$$
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d) :- $$f\;(x)\;=\;\sqrt{5x}\\f\;(x)\;=\;\sqrt5\sqrt x$$
Differentiating both sides w.r.t x
$$\frac d{dx}f\;(x)\;=\;\frac d{dx}(\sqrt5\sqrt x)\\f’\;(x)\;=\;\sqrt5\;\frac d{dx}\;\sqrt x$$
$$=\;\sqrt5\;(\frac d{dx}\;x^{\frac12\;})\\=\;\sqrt5\;(\frac12\;x^{\frac12-1\;})$$
$$f'(x)=\;\sqrt5\;(\frac12\;x^{\frac{-1}2\;})\\=5\;(\frac1{2x^{1/2}})$$
$$=5\;(\frac1{2\sqrt x})\\=\;\frac{\sqrt5}{2\sqrt x}$$
Q3 :- $$Determine\;f’\;(x)\:$$
a) :- $$\;f\;(x)\:=x^2\;(\;x^3+5\;)$$
Differentiating both sides w.r.t x
$$\frac d{dx}\;f\;(x)\:=\frac d{dx}x^2\;(\;x^3+5\;)\\f’\;(x)=\frac d{dx}(x^5+5x^2)$$
$$=\frac d{dx}x^5\;\frac d{dx}5x^2\\=5x^{5-1}+5\frac d{dx}x^2$$
$$=\;5x^4+5(2x^{2-1})\\=\;5x^4+\;10x$$
Second method
a) :-
$$\;f\;(x)\:=x^2\;(\;x^3+5\;)$$
Product Rule:-
$$\frac d{dx}\;uv=u\;\frac d{dx}v\;+\;v\;\frac d{dx}u$$
Differentiating both sides w.r.t x
$$\frac d{dx}\;f\;(x)\:=\frac d{dx}\{\;x^2\;(\;x^3+5\;)\;\}$$
$$f’\;(x)\;=\;x^2\frac d{dx}\;(x^3+5)+\;(x^3+5)\;\frac d{dx}x^2$$
$$=\;x^2(\frac d{dx}x^3\;+\frac d{dx}5)+\;(x^3+5)\;2x^{2-1}$$
$$=\;x^2(3x^{3-1}\;+0)+\;(x^3+5)\;2x\\=\;x^2(3x^2)+(x^3+5)2x$$
$$=\;3x^4+\;2x^4+\;10x\\=\;5x^4+\;10x$$
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b) :-
f (x) = ( x+9 ) ( x-9 )
Differentiating both sides w.r.t x
$$\frac d{dx}f\;(x)\;=\frac d{dx}(x+9(\;(x-9)\\f’\;(x)\;=\frac d{dx}(x^2-81)$$
$$=\frac d{dx}x^2-\frac d{dx}81\\=\;2x-0\\=\;2x$$
b) :-
2nd Method:-
f (x) = ( x+9 ) ( x-9 )
Product rule:–
$$\frac d{dx}(uv)\\=\;u\frac d{dx}v+v\frac d{dx}u$$
Differentiating both sides w.r.t x
$$f’\;(x)\;=(x+9)\frac d{dx}(x-9)\\+\;(x-9)\frac d{dx}(x+9)$$
$$=(x+9)(\frac d{dx}x-\frac d{dx}9)\\+\;(x-9)(\frac d{dx}x+\frac d{dx}9)$$
$$=\;(x+9)(1-0)\;+\;(x-9)(1+0)\\=\;(x+9)(1)\;+(x-9)(1)\;$$
$$=x+9\;+\;x-9\\=\;2x$$
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c) :- $$f\;(x)\;={(x^2+x^3)}^3$$
Differentiating both sides w.r.t x
$$\frac d{dx}f\;(x)\;=\frac d{dx}{(x^2+x^3)}^3\\$$
$$f’\;(x)\;=\frac d{dx}(\;{(x^2)}^3+{(x^3)}^3\;+3{(x^2)}^2(x^3)\;+3(x^2){(x^3)}^2\;)$$
$$=\frac d{dx}(\;x^6+x^9+3x^7+3x^8)$$
$$=\frac d{dx}\;x^6+\frac d{dx}x^9+\frac d{dx}3x^7+\frac d{dx}3x^8\\=\;6x^{6-1}+9x^{9-1}+3\frac d{dx}x^7+3\frac d{dx}x^8\\$$
$$=6x^5+9x^8+21x^6+24x^7\\\\$$
c) :- $$f\;(x)\;={(x^2+x^3)}^3$$
Differentiating both sides w.r.t x
$$\frac d{dx}f\;(x)\;=\frac d{dx}{(x^2+x^3)}^3\\$$
$$f’\;(x)\;=3{(x^2+x^3)}^{3-1}\frac d{dx}(x^2+x^3)\\=\;3{(x^2+x^3)}^2(\frac d{dx}x^2+\frac d{dx}x^3)\\$$
$$=\;3{(x^2+x^3)}^2(2x^{2-1}+3x^{3-1})\\=\;3{(x^2+x^3)}^2(2x+3x^2)\\$$
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d) :- $$f\;(x)=-3x^{-8}+2\sqrt x\\$$
Differentiating both sides w.r.t x
$$\frac d{dx}f\;(x)=\frac d{dx}(-3x^{-8}+2\sqrt x)\\f’\;(x)\;=\frac d{dx}(-3x^{-8})+\frac d{dx}(2\sqrt x)$$
$$=\;-3\frac d{dx}x^{-8}+2\frac d{dx}\sqrt x\\=\;-3(-8x^{-8-1})+2\frac1{2\sqrt x}\\$$
$$=\;24x^{-9}+\frac1{\sqrt x}\\$$
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e) :- $$f\;(x)\;=\;ax^3+bx^3+cx+d$$
Differentiating both sides w.r.t x
$$\frac d{dx}f\;(x)\;=\frac d{dx}(\;ax^3+bx^3+cx+d\;)\\$$
$$f’\;(x)\;=\frac d{dx}\;ax^3+\frac d{dx}bx^3+\frac d{dx}cx+\frac d{dx}d\;)\\=\;a\frac d{dx}x^3+b\frac d{dx}x^2+c\frac d{dx}x+0$$
$$=\;a(3x^{3-1})+(2x^{2-1})\;+c\;(1)\\=\;3ax^2+2bx\;+c$$
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f) :- $$f\;(x)\;=x^{24}+2x^\frac12+3x^8+9x^4\\$$
Differentiating both sides w.r.t x
$$\frac d{dx}f\;(x)\;=\frac d{dx}(x^{24}+2x^\frac12+3x^8+9x^4)\\$$
$$f’\;(x)\;=\frac d{dx}x^{24}+\frac d{dx}2x^\frac12+\frac d{dx}3x^8+\frac d{dx}9x^4\\=24x^{24-1}+2\frac d{dx}x^\frac12+3\frac d{dx}x^8+9\frac d{dx}x^4$$
$$=\;24x^{23}+2\frac1{2\sqrt x}+3(8x^{8-1})+9(4x^{4-1})\\=\;24x^{23}+\frac1{\sqrt x}+24x^7+36x^3$$
Q4 :- $$Find\;\frac{dy}{dx}$$
a) :- $$y\;=\frac{x+2x^{\displaystyle\frac32}}{\sqrt x}\\y\;=\frac x{\sqrt x}+\frac{2x^{\displaystyle\frac32}}{\sqrt x}$$
$$y\;=\sqrt x+2x$$
Differentiating both sides w.r.t x
$$\frac d{dx}y\;=\frac d{dx}\sqrt x+2x\\\frac{dy}{dx}=\frac d{dx}\sqrt x+\frac d{dx}2x$$
$$=\;\frac1{2\sqrt x}+2\frac d{dx}x\\=\frac1{2\sqrt x}+2(1)\\=\frac1{2\sqrt x}+2$$
2nd Method:-
$$y=\frac{x+2x^{3/2}}{\sqrt x}$$
Quotient Rule:-
$$\frac d{dx}\frac uv=\frac{v{\displaystyle\frac d{dx}}u-u{\displaystyle\frac d{dx}}v}{v^2}$$
Differentiating both sides w.r.t x
$$\frac d{dx}y=\frac d{dx}\frac{x+2x^{3/2}}{\sqrt x}$$
$$\frac{dy}{dx}=\frac{\sqrt x{\displaystyle\frac d{dx}}(x+2x^{\displaystyle\frac32})-(x+2x^{\displaystyle\frac32}){\displaystyle\frac d{dx}}\sqrt x}{{(\sqrt x)}^2}$$
$$=\frac{\sqrt x({\displaystyle\frac d{dx}}x+{\displaystyle\frac d{dx}}2x^{\displaystyle\frac32})-(x+2x^{\displaystyle\frac32}){\displaystyle\frac1{2\sqrt x}}}x$$
$$=\frac{\sqrt x(1+2{\displaystyle\frac d{dx}}x^{\displaystyle\frac32})-{\displaystyle\frac{(x+2x^{\displaystyle\frac32})}{2\sqrt x}}}{2\sqrt x}\\=\frac{\sqrt x(1+2.{\displaystyle\frac32}x^{{\displaystyle\frac32}-1})-({\displaystyle\frac x{2\sqrt x}}+{\displaystyle\frac{2x^{3/2}}{2\sqrt x}})}x$$
$$\frac{dy}{dx}=\frac{\sqrt x(1+3\;x^{\displaystyle\frac12})-({\displaystyle\frac12}\sqrt x+x)}x$$
$$=\frac{\sqrt x+3x-{\displaystyle\frac12}\sqrt x-x}x\\=\frac{{\displaystyle\frac12}\sqrt x+2x}x$$
$$=(\frac12\sqrt x+2x)\frac1x\\=(\frac12\sqrt x)\frac1x+2x(\frac1x)$$
$$=\frac1{2\sqrt x}+2$$
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b) :- $$y=(x^3-5)(2x+3)\\y=2x^4+3x^3-10x-15$$
Differentiating both sides w.r.t x
$$\frac d{dx}y=\frac d{dx}(2x^4+3x^3-10x-15)$$
$$\frac{dy}{dx}=\frac d{dx}2x^4+\frac d{dx}3x^3-\frac d{dx}10x-\frac d{dx}15$$
$$=\;2(4x^{4-1})+(3(3x^{3-1})-10(1)\;-0\\=\;8x^3+9x^2-10$$
2nd Method:-
$$y=(x^3+5)(2x+3)$$
Differentiating both sides w.r.t x
$$\frac d{dx}y=\frac d{dx}(x^3+5)(2x+3)$$
$$\frac{dy}{dx}=(x^3+5)\frac d{dx}(2x+3)+(2x+3)\frac d{dx}(x^3-5)\\$$
$$\frac{dy}{dx}=(x^3+5)(\frac d{dx}2x+\frac d{dx}3)+(2x+3)(\frac d{dx}x^3-\frac d{dx}5)\\$$
$$=(x^3+5)(2(1)+0)+(2x+3)(3x^2-0)\\=(x^3-5)(2)+(2x+3)(3x^2)\\\\$$
$$=2x^3-10+6x^3+9x^2\\=8x^3+9x^2-10\\$$
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c) :- $$y=(4x^2-3)(7x^2+x)\\y=28x^4+4x^3-21x^2-3x\\$$
Differentiating both sides w.r.t x
$$\frac d{dx}y=\frac d{dx}(28x^4+4x^3-21x^2-3x)\\$$
$$\frac{dy}{dx}=\frac d{dx}28x^4+\frac d{dx}4x^3-\frac d{dx}21x^2-\frac d{dx}3x\\$$
$$=28(4x^3)+4(3x^2)-21(2x)-3(1)\\=112x^3+12x^2-42x-3\\$$
2nd Method:-
$$y=(4x^2-3)(7x^2+x)\\$$
$$\frac d{dx}y=\frac d{dx}(4x^2-3)(7x^2+x)\\$$
$$\frac{dy}{dx}=(4x^2-3)\frac d{dx}(7x^2+x)+(7x^2+x)\frac d{dx}(4x^2-3)\\$$
$$=(4x^2-3)(\frac d{dx}7x^2+\frac d{dx}x)+(7x^2+x)(\frac d{dx}4x^2-\frac d{dx}3)\\$$
$$=(4x^2-3)(7(2x)+1)+(7x^2+x)(4(2x)-0)\\=(4x^2-3)(14x+1)+(7x^2+x)(8x)\\$$
$$=56x^3+4x^2-42x-3+56x^3+8x^2\\=112x^3+12x^2-42x-3\\$$
Q5 :- Find slope of the tangent at x=1
a) :-$$f\;(x)\;=x^2+3x\\\frac d{dx}f\;(x)=\frac d{dx}(x^2+3x)\\$$
$$f’\;(x)=\frac d{dx}x^2+\frac d{dx}3x\\=2x+3(1)\\$$
$$f’\;(x)=2x+3\\f’\;(1)=2(1)+3=5\\$$
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b) :- $$f\;(x)=x^4-x^2\\\frac d{dx}f\;(x)=\frac d{dx}(x^4-x^2)$$
$$f’\;(x)=\frac d{dx}x^4-\frac d{dx}x^2\\f’\;(x)=4x^3-2x$$
$$f’\;(1)=4{(1)}^3-2(1)\\=\;4-2\;=\;2$$
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