12th Class Math New Book Federal Board Exercise 2.5

12th Class Math New Book Federal Board Exercise 2.5 solution.

Find the derivative.

Power rule:–

$$\frac d{dx}(x^n)=\;n\;x^{n-1}\\$$

Product rule:-

$$\frac d{dx}\;(u\;v)=\;u\;\frac d{dx}v\;+\;v\;\frac d{dx}u\\$$

Quotient rule:-

$$\frac d{dx}\;(\frac uv)=\;\frac{\;v\;\frac d{dx}u\;-\;u\;\frac d{dx}v}{v^2}\\$$

Q1 :-

$$y=\frac1x$$

Differentiating both sides   w.r.t   x

$$\frac d{dx}y=\frac d{dx}(\frac1{x^2})$$

We will apply Quotient law

$$\frac{dy}{dx}=\frac{x{\displaystyle\frac d{dx}}1-1{\displaystyle\frac d{dx}}x}{x^2}$$

$$=\;\frac{x(0)-1(1)}{x^2}\\=\;\frac{0-1}{x^2}=\frac{-1}{x^2}$$

Another Method:-

$$y=\frac1x$$

Differentiating both sides   w.r.t   x

$$\frac d{dx}y=\frac d{dx}(\frac1x)\\\frac{dy}{dx}=\frac d{dx}x^{-1}$$

We will apply power rule

$$\frac{dy}{dx}=\;-1x^{-1-1}\\=\;-1x^{-2}\\=\;\frac{-1}{x^2}$$

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Q2  :-

$$y=(x^2-7)(x^2+4x+2)$$

Differentiating both sides   w.r.t   x

$$\frac d{dx}y=\frac d{dx}(x^2-7)(x^2+4x+2)$$

We will apply product rule

$$\frac{dy}{dx}=(x^2-7)\frac d{dx}(x^2+4x+2)\\+(x^2+4x+2)\frac d{dx}(x^2-7)$$

$$=(x^2-7)\;(2x+4+0)\\+(x^4+4x+2)\;(2x-0)\\=(x^2-7)(2x+4)+(x^4+4x+2)(2x)$$

$$=2x^3+4x^2-14x-28+2x^3+8x^2+4x\\=4x^2+12x^2-10-28$$

Another Method:-

$$y=(x^2-7)(x^2+4x+2)\\=x^4+4x^3+2x^2-17x^2-28x-14\\=x^4+4x^3-5x^2-28x-14$$

Differentiating both sides   w.r.t   x

$$\frac d{dx}y=\frac d{dx}(x^4+4x^3-5x^2-28x-14)$$

$$\frac{dy}{dx}=4x^3+4(3x^2\_-5(2x)-28(1)-0\\=\;4x^3+12x^2-10x-28$$

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Q3 :-

$$y=(7x+1)(x^4-x^3-9)$$

Differentiating both sides   w.r.t   x

$$\frac d{dx}y=\frac d{dx}(7x+1)(x^4-x^3-9)$$

$$\frac{dy}{dx}=(7x+1)\frac d{dx}\;(x^4-x^3-9)\\+(x^4-x^3-9)\frac d{dx}\;(7x-1)$$

$$=(7x+1)(4x^3-3x^2-0)\\+(x^4-x^3-9)(7-0)\\=28x^4-21x^3-63x\\+4x^3-3x^2-9+7x^4-7x^3-63\\=35x^4-24x^3-3x^2-126x-9\\$$

Another Method:-

$$y=(7x+1)(x^4-x^3-9)\\y=7x^5-7x^4-63x^2+x^4-x^3-9x\\=\;7x^5-6x^4-x^3-63x^2-9x\\$$

Differentiating both sides   w.r.t   x

$$\frac d{dx}y=\frac d{dx}(7x^5-6x^4-x^3-63x^2-9x)\\$$

$$\frac{dy}{dx}=7(5x^4)-6(4x^3)-3x^2-63(2x)-9(1)\\=\;35x^4-24x^3-3x^2-126x-9$$

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Q4 :-

$$y=\frac{3x+4}{x^2+1}$$

Differentiating both sides   w.r.t   x

$$\frac{dy}{dx}=\frac d{dx}\;(\frac{3x^2+5}{3x-1})\\$$

$$\style{font-size:16px}{=\frac{(x^2+1){\displaystyle\frac d{dx}}(3x+4)-(3x+4){\displaystyle\frac d{dx}}(x^2+1)}{{(x^2+1)}^2}}$$

$$=\frac{(x^2+1)(3(1)+0)-(3x+4)(2x+0)}{{(x^2+1)}^2}\\=\frac{3x^2+3-6x^2-8x}{{(x^2+1)}^2}\\$$

$$=\frac{-3x^2-8x+3}{{(x^2+1)}^2}\\$$

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Q5 :-

$$y=\frac{x-2}{x^4+x+1}$$

Differentiating both sides   w.r.t   x

$$\frac{dy}{dx}=\frac d{dx}(\frac{x-2}{x^4+x+1})\\\\\\$$

$$\style{font-size:12px}{=\frac{(x^4+x+1){\displaystyle\frac d{dx}}(x-2)-(x-2){\displaystyle\frac d{dx}}(x^4+x+1)}{(x^4+x+1)^2}}$$

$$=\frac{(x^4+x+1)(1-0)-(x-2)(4x^3+1+0)}{(x^4+x+1)^2}\\$$

$$=\frac{x^4+x+1-(4x^4+x-8x^3-2)}{(x^4+x+1)^2}\\$$

$$=\frac{x^4+x+1-4x^4-x+8x^3+2}{(x^4+x+1)^2}\\\frac{dy}{dx}=\frac{-3x^4+8x^2+3}{(x^4+x+1)^2}\\$$

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Q6 :-

$$y=\frac{3x^2+5}{3x-1}\\$$

Differentiating both sides   w.r.t   x

$$\frac{dy}{dx}=\frac d{dx}\;(\frac{3x^2+5}{3x-1})\\$$

$$=\frac{(3x-1){\displaystyle\frac d{dx}}(3x^2+5)-(3x^2+5){\displaystyle\frac d{dx}}(3x-1)}{{(3x-1)}^2}\\$$

$$\frac{dy}{dx}=\frac{(3x-1)(6x+0)-(3x^2+5)(3-0)}{{(3x-1)}^2}\\=\frac{(18x^2-6x))-9x^2-15}{{(3x-1)}^2}\\$$

$$=\frac{9x^2-6x-15}{{(3x-1)}^2}\\=\frac{3\;(3x-2x-5)}{{(3x-1)}^2}\\$$

$$\frac{dy}{dx}=\frac{3\;(3x^2-5x+3x-5)}{{(3x-1)}^2}\\=\frac{3\;(x(3x-5)(x+1)}{{(3x-1)}^2}\\$$

$$=\frac{3\;(3x-5)(x+1)}{{(3x-1)}^2}\\$$

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Q7 :-

$$y=(\frac1x+\frac1{x^2})(3x^2+27)\\y=3x^2+\frac{27}x+3x+\frac{27}{x^2}\\$$

Differentiating both sides   w.r.t   x

$$\frac d{dx}y=\frac d{dx}(3x^2+\frac{27}x+3x+\frac{27}{x^2})\\$$

$$\frac{dy}{dx}=3\frac d{dx}(x^2)+27\frac d{dx}\;(\frac1x)\\+3\frac d{dx}(x)+27\frac d{dx}\;(\frac1{x^2})$$

$$=3(2x)+27\frac d{dx}(x^{-1})+3(1)+27\frac d{dx}(x^{-2-1})\\$$

$$=6x+27(-1x^{-1-1})+3+27(-2x^{-2-1})\\=6x-27x^{-2}+3-54x^{-3}\\=6x-\frac{27}{x^2}+3-\frac{54}{x^3}$$

Another Method:- 

$$y=(\frac1x+\frac1{x^2})(3x^2+27)\\$$

Differentiating both sides   w.r.t   x

$$\frac d{dx}y=\frac d{dx}(\frac1x+\frac1{x^2})(3x^2+27)\\$$

$$\frac{dy}{dx}=(\frac1x+\frac1{x^2})\frac d{dx}(3x^3+27)\\+(3x^3+27)\frac d{dx}(\frac1x+\frac1{x^2})\\\\$$

$$=(\frac1x+\frac1{x^2})(3\frac d{dx}x^3+\frac d{dx}27)\\+(3x^3+27)(\frac d{dx}x^{-1}+\frac d{dx}x^{-2})\\\\$$

$$=(\frac1x+\frac1{x^2})\;(9x^2+0)\\+(3x^3+27)\;(x^{-2}+2x^{-3})\\\\$$

$$=9x+9-3x-6-27x^{-2}-54x^{-3}\\=6x-\frac{27}{x^2}-\frac{54}{x^3}+3$$

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Q8 :-

$$y=\frac{2-3x}{7-x}$$

Differentiating both sides   w.r.t   x

$$\frac d{dx}y=\frac d{dx}\frac{2-3x}{7-x}\\\frac{dy}{dx}=\frac{(7-x)(-3)-(2-3x)(0-1)}{{(7-x)}^2}$$

$$=\frac{(7-x)(-3)-(2-3x)(-1)}{{(7-x)}^2}\\=\frac{-21+3x+2-3x}{{(7-x)}^2}\\=\frac{-19}{{(7-x)}^2}$$

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Q9 :-

$$y=\frac{x^2-10x+2}{x^3-x}$$

Differentiating both sides   w.r.t   x

$$\style{font-size:12px}{\frac{dy}{dx}=\frac d{dx}(\frac{x^2-10x+2}{x^3-x})}$$

$$\style{font-size:8px}{=\frac{(x^3-x){\displaystyle\frac d{dx}}(x^2-10x+2)-(x^2-10x+2){\displaystyle\frac d{dx}}(x^3-x)}{{(x^3-x)}^2}}$$

$$\style{font-size:10px}{=\frac{(x^3-x)(2x-10)-(x^2-10x+2)(3x^2-1)}{{(x^3-x)}^2}}$$

$$=\frac{2x^4-10x^3-2x^2+10x-(3x^4-x^2-30x^3+10x+6x^2-2}{{(x^3-x)}^2}$$

$$=\frac{-x^4+20x^3-7x^2+2}{{(x^3-x)}^2}$$

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Q10 :-

$$y=\frac{x^4+2x^3-1}{x^2}$$

Differentiating both sides   w.r.t   x

$$\frac{dy}{dx}=\frac d{dx}(\frac{x^4+2x^3-1}{x^2})$$

$$=\frac{x^2{\displaystyle\frac d{dx}}(x^4+2x^3-1)-(x^4+2x^3-1){\displaystyle\frac d{dx}}x^2}{{(x^2)}^2}$$

$$=\frac{x^2(4x^3+2(3x^2)-0)-(x^4+2x^3-1).2x}{x^4}$$

$$=\frac{x^2(4x^3+6x^2)-2x(x^4+2x^3-1)}{x^4}\\=\frac{x\left\{x(4x^3+6x^2)-2(x^4+2x^3-1)\right\}}{x^4}$$

$$\frac{dy}{dx}=\frac{x\;(4x^4+6x^3-2x^4-4x^3+2)}{x^4}\\=\frac{4x^4-2x^4+6x^3-4x^3+2}{x^3}$$

$$=\frac{2x^4+2x^3+2}{x^3}\\=\frac{2\;(x^4+x^3+1)}{x^3}$$

$$=2\;(x+1+\frac1{x^3})\\=\;2x+2+\frac2{x^3}$$

Another Method:- 

$$y=\frac{x^4+2x^3-1}{x^2}\\y=x^2+2x-\frac1{x^2}$$

$$\frac d{dx}y=\frac d{dx}(x^2+2x-\frac1{x^2})\\\frac{dy}{dx}=\frac d{dx}x^2+\frac d{dx}2x-\frac d{dx}x^{-2}$$

$$=\;2x+2(1)\;-(-2x^{-3})\\=2x+2+\frac2{x^3}$$

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Q11 :-

$$y=\frac{10}{{(x^3-10)}^9}$$

Differentiating both sides   w.r.t   x

$$\frac d{dx}y=\frac d{dx}\frac{10}{{(x^3-10)}^9}$$

$$\frac{dy}{dx}=\frac{{(x^3-10)}^9{\displaystyle\frac d{dx}}(10){\displaystyle\frac d{dx}}{(x^3-10)}^9}{\left\{{(x^3-10)}^9\right\}^2}$$

$$=\frac{{(x^3-10)}^9(0)-10.9{(x^3-10)}^{9-1}{\displaystyle\frac d{dx}}{(x^3-10)}}{{(x^3-10)}^{18}}$$

$$=\frac{0-9{0\;(x^3-10)}^8{(3x^2-0)}}{{(x^3-10)}^{18}}$$

$$=\frac{-27{0\;x^2(x^3-10)}^8}{{(x^3-10)}^{18}}\\=\frac{-270x^2}{{(x^3-10)}^{10}}$$

Another Method:-

$$y=\frac{10}{{(x^3-10)}^9}\\y=10\;{(x^3-10)}^{-9}$$

Differentiating both sides   w.r.t   x

$$\frac d{dx}y=\frac d{dx}10\;{(x^3-10)}^{-9}\\\frac{dy}{dx}=10\frac d{dx}{(x^3-10)}^{-9}$$

$$\frac{dy}{dx}=10\;(-9\;{(x^3-10)}^{-9-1}\frac d{dx}(x^3-10)\;)$$

$$=-90\;{(x^3-10)}^{-10}(3x^2-0)\\=-270\;x^2{(x^3-10)}^{-10}$$

$$=\frac{-270x^2}{{(x^3-10)}^{10}}$$

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Q12 :-

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

Differentiating both sides   w.r.t   x

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

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

$$=\frac{(3x-2){\displaystyle.2}{(x^2+1)}^{2-1}{\displaystyle\frac d{dx}}{(x^2+1)-(x^2+1)}^2(3-0)}{{(3x-2)}^2}$$

$$=\frac{(3x-2){\displaystyle\;2}{(x^2+1)}{(2x+0)-(x^2+1)}^2(3)}{{(3x-2)}^2}$$

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

$$=\frac{{(x-1)}^2.2{(x+1)}^{2-1}{\displaystyle\frac d{dx}}(x+1).2{(x-1)}^{2-1}{\displaystyle\frac d{dx}}(x-1)}{{(x-1)}^4}$$

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

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

$$=\frac{2(x+1)\;\;(x-1-x-1)}{{(x-1)}^3}\;\\=\frac{-4(x+1)}{{(x-1)}^3}$$

$$\;\frac{dy}{dx}=\frac{4x(3x-2)(x^2+1)-3{(x^2+1)}^2}{{(3x-2)}^2}$$

$$\;\frac{dy}{dx}=\frac{(x^2+1)(12x^2-8x-3x^2-3)}{{(3x-2)}^2}$$

$$\;\frac{dy}{dx}=\frac{(x^2+1)(9x^2-8x-3)}{{(3x-2)}^2}$$

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Q13 :-

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

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

Differentiating both sides   w.r.t   x

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

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

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

$$\frac{dy}{dx}=2\;(\frac{x+1}{x-1})\frac{x-1-x-1}{{(x-1)}^2}\\=2(\frac{x+1}{x-1})(\frac{-2}{{(x-1)}^2})$$

$$=\frac{-4(x+1)}{{(x-1)}^3}$$

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Find slope of tangent at the indicated point.

Q14 :-

$$y=\frac{4x-1}x,\;x=-1\\y=\frac{4x}x-\frac1x\\y=4-x^{-1}$$

Differentiating both sides   w.r.t   x

$$\frac d{dx}y=\frac d{dx}(4-x^{-1})\\\\\\$$

$$\frac{dy}{dx}=0-(-1x^{-1-1})\\=0\;+\;x^{-2}\\\\\\$$

$$\frac{dy}{dx}=\frac1{x^2}\\\frac{dy}{dx}=\frac1{{(-1)}^2}\\=\frac11=1$$

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Q15 :-

$$y=\frac{54}{x^2+1},\;x=2\\y=54{(x^2+1)}^{-1}$$

Differentiating both sides   w.r.t   x

$$\frac d{dx}y=\frac d{dx}54{(x^2+1)}^{-1}$$

$$\frac{dy}{dx}=54\;(-1{(x^2+1)}^{-1-1}\frac d{dx}(x^2+1)\;)$$

$$=-54\;{(x^2+1)}^{-2}(2x+0)\\=-108x\;{(x^2+1)}^{-2}\;\;$$

$$\frac{dy}{dx}=\frac{-108x}{{(x^2+1)}^2}\\\frac{dy}{dx}=\frac{-108x}{(\;{{(2)}^2+1)}^2}$$

$$=\frac{-108(2)}{5\times5}\\=\frac{-216}{25}$$

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Q16 :-

$$y=\frac{2x+5}{x+2},\;x=1$$

$$\frac{dy}{dx}=\frac d{dx}(\frac{2x+5}{x+2})$$

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

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

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

$$\frac{dy}{dx}=\frac{-1}9\\$$

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Q17 :-

$$y=(2\sqrt x+1)(x^3-6),\;x=0\\y=(2x^\frac72-12x^\frac12+x^3-6)\\$$

$$\frac d{dx}y=\frac d{dx}(2x^\frac72-12x^\frac12+x^3-6)\\$$

$$=2.\frac72x^{\frac72-1}-12.\frac12x^{\frac12-1}\\+3x^{3-1}-0\\$$

$$=7x^\frac52-6x^\frac{-1}2+3x^2\\\frac{dy}{dx}=7x^\frac52-\frac6{\sqrt x}+3x^2\\$$

$$\frac{dy}{dx}=7{(0)}^\frac52-\frac6{\sqrt x}+3{(0)}^2\;\frac{dy}{dx}=0-\frac60+0\\\frac{dy}{dx}=-\infty\\$$

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