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DIFFERENTIATION

Differentiation

Exercise 11.1

Differentiate the following functions from first principles :

QUESTION 1

 1.~e^{-x}

Sol :

 f(x)=e^{-x}\\f(x+h)=e^{-(x+h)}\\\dfrac{d}{dx}f(x)=lim_{h\to0}~\dfrac{f(x+h)-f(x)}{h}\\=lim_{h\to0}~\dfrac{e^{-(x+h)}-e^{-x}}{h}\\=lim_{h\to0}~\dfrac{e^{-(x)}\times{e^{-h}}-e^{-x}}{h}\\=lim_{h\to0}~e^{-x}~\Bigg\{\dfrac{e^{-h}-1}{h}\Bigg\}\\=lim_{h\to0}~e^{-x}~\Bigg\{\dfrac{e^{-h}-1}{-h}\Bigg\}(-1)\\\bigg[Since,~{lim_{h\to0}}~{\dfrac{e^{h}-1}{h}}=1\bigg]\\=-e^{-x}

 

QUESTION 2

 2.~e^{3x}

Sol :

 f(x)=e^{3x}\\f(x+h)=e^{3(x+h)}\\\dfrac{d}{dx}f(x)=lim_{h\to0}~\dfrac{f(x+h)-f(x)}{h}\\=lim_{h\to0}~\dfrac{e^{3(x+h)}-e^{3x}}{h}\\=lim_{h\to0}~\dfrac{e^{3x}\times{e^{3h}}-e^{3x}}{h}\\=lim_{h\to0}~e^{3x}~\Bigg\{\dfrac{e^{3h}-1}{h}\Bigg\}\\=lim_{h\to0}~3e^{3x}~\Bigg\{\dfrac{e^{3h}-1}{3h}\Bigg\}\\\bigg[Since,~{lim_{h\to0}}~{\dfrac{e^{h}-1}{h}}=1\bigg]\\=3e^{3x}

QUESTION 3

 3.~e^{ax+b}

Sol :

 f(x)=e^{ax+b}\\f(x+h)=e^{a(x+h)+b}\\\dfrac{d}{dx}f(x)=lim_{h\to0}~\dfrac{f(x+h)-f(x)}{h}\\=lim_{h\to0}~\dfrac{e^{a(x+h)+b}-e^{ax+b}}{h}\\=lim_{h\to0}~\dfrac{e^{ax+b}\times{e^{ah}}-e^{ax+b}}{h}\\=lim_{h\to0}~e^{ax+b}~\Bigg\{\dfrac{e^{ah}-1}{h}\Bigg\}\\=lim_{h\to0}~e^{ax+b}~\Bigg\{\dfrac{e^{ah}-1}{ah}\Bigg\}\\\bigg[Since,~{lim_{h\to0}}~{\dfrac{e^{h}-1}{h}}=1\bigg]\\=e^{ax+b}

 

QUESTION 4

 4.~e^{cos~x}

Sol :

 f(x)=e^{cos~x}\\f(x+h)=e^{\sqrt{2x}}\\\dfrac{d}{dx}f(x)=lim_{h\to0}~\dfrac{f(x+h)-f(x)}{h}\\=lim_{h\to0}~\dfrac{e^{\sqrt{2(x+h)}}-e^{\sqrt{2x}}}{h}\\=lim_{h\to0}~e^{\sqrt{2x}}\Bigg\{\dfrac{e^{\sqrt{2(x+h)}-\sqrt{2x}}-1}{h}\Bigg\}\\=lim_{h\to0}~e^{cos~x}~\Bigg\{\dfrac{e^{cos~(x+h)-cos~x}-1}{{{cos~(x+h)-cos~x}}}\Bigg\}\times\dfrac{cos~(x+h)-cos~x}{h}\\\bigg[Since,~{lim_{h\to0}}~{\dfrac{e^{h}-1}{h}}=1\bigg]\\=lim_{h\to0}~e^{cos~x}~\Bigg\{\dfrac{cos~(x+h)-cos~x}{h}\Bigg\}\\\bigg[Since,~cos~A-cos~B=-2sin\dfrac{A+B}{2}sin\dfrac{A-B}{2}\bigg]\\=lim_{h\to0}~e^{cos~x}\times\Bigg(\dfrac{-2sin\dfrac{(x+h)+x}{2}sin\dfrac{(x+h)-x}{2}}{h}\Bigg)\\=e^{cos~x}lim_{h\to0}\times\dfrac{-2sin\bigg(\dfrac{2x+h}{2}\bigg)sin\bigg(\dfrac{h}{2}\bigg)}{h}\\=e^{cos~x}lim_{h\to0}\times\dfrac{-2sin\bigg(\dfrac{2x+h}{2}\bigg)}{2}\times\dfrac{sin\bigg(\dfrac{h}{2}\bigg)}{\bigg(\dfrac{h}{2}\bigg)}\\\bigg[Since,~lim_{h\to0}~\dfrac{sin~x}{x}=1\bigg]\\=e^{cos~x}lim_{h\to0}\times\dfrac{-2sin\bigg(\dfrac{2x+h}{2}\bigg)}{2}\times{1}\\=e^{cos~x}lim_{h\to0}\times{-sin\bigg(\dfrac{2x+h}{2}\bigg)}\\=e^{cos~x}\times{-sinx}\\=-sin~x~e^{cos~x}

 

QUESTION 5

 5.~e^{\sqrt{2x}}

Sol :

f(x)=e^{\sqrt{2x}}\\f(x+h)=e^{\sqrt{2(x+h)}}\\\dfrac{d}{dx}f(x)=lim_{h\to0}~\dfrac{f(x+h)-f(x)}{h}\\=lim_{h\to0}~\dfrac{e^{\sqrt{2(x+h)}}-e^{\sqrt{2x}}}{h}\\=lim_{h\to0}~e^{\sqrt{2x}}~\dfrac{e^{\sqrt{2(x+h)}-\sqrt{2x}}-1}{h}\\=lim_{h\to0}~e^{\sqrt{2x}}~\Bigg(\dfrac{e^{\sqrt{2(x+h)}-\sqrt{2x}}-1}{\sqrt{2(x+h)}-\sqrt{2x}}\Bigg)\Big(\dfrac{\sqrt{2(x+h)}-\sqrt{2x}}{h}\Big)\\\bigg[Since,~lim_{h\to0}~\dfrac{e^h-1}{h}=1\bigg]\\=lim_{h\to0}~\dfrac{\sqrt{2(x+h)}-\sqrt{2x}}{h}\\\big[Rationalising~numerator\big]\\=lim_{h\to0}~\dfrac{\sqrt{2(x+h)}-\sqrt{2x}}{h}\times\dfrac{\sqrt{2(x+h)}+\sqrt{2x}}{\sqrt{2(x+h)}+\sqrt{2x}}\\=lim_{h\to0}~\dfrac{2(x+h)-2x}{h\big(\sqrt{2(x+h)}+\sqrt{2x}\big)}\\=lim_{h\to0}~\dfrac{2(x+h)-2x}{h\big(\sqrt{2(x+h)}+\sqrt{2x}\big)}\\=lim_{h\to0}~\dfrac{2x+2h-2x}{h\big(\sqrt{2(x+h)}+\sqrt{2x}\big)}\\=lim_{h\to0}~\dfrac{2h}{h\big(\sqrt{2(x+h)}+\sqrt{2x}\big)}\\=lim_{h\to0}~\dfrac{2}{\big(\sqrt{2(x+h)}+\sqrt{2x}\big)}\\=\dfrac{e^{\sqrt{2x}}}{\sqrt{2x}}

 

Question 6

 6.~log~cos~x

Sol :

 f(x)=log~cos~x\\f(x+h)=log~cos~(x+h)\\\dfrac{d}{dx}f(x)=lim_{h\to0}~\dfrac{f(x+h)-f(x)}{h}\\=lim_{h\to0}~\dfrac{log~cos~(x+h)-log~cos~x}{h}\\=lim_{h\to0}~\Bigg\{\dfrac{log\dfrac{cos~x+h}{cos~x}}{h}\Bigg\}\\\Bigg[Since,~log~A-log~B=log\dfrac{A}{B}\Bigg]\\=lim_{h\to0}\dfrac{log\Bigg\{1+\dfrac{cos~x+h}{cos~x}-1\Bigg\}}{h}\\=lim_{h\to0}\dfrac{log\Bigg\{1+\dfrac{cos~(x+h)-cos~x}{cos~x}\Bigg\}}{h\times\dfrac{cos~(x+h)-cos~x}{cos~x}}\times{\dfrac{cos~(x+h)-cos~x}{cos~x}}\\\Bigg[Since,~lim_{x\to0}~\dfrac{log(1+x)}{x}=1\Bigg]\\=lim_{h\to0}{\dfrac{cos~(x+h)-cos~x}{cos~x\times{h}}}\\=lim_{h\to0}{\dfrac{-2sin\dfrac{x+h+x}{2}sin\dfrac{x+h-x}{2}}{cos~x\times{h}}}\\=-2~lim_{h\to0}{\dfrac{sin\dfrac{2x+h}{2}sin\dfrac{h}{2}}{2~cos~x\times\dfrac{h}{2}}}\\\Bigg[Since,lim_{h\to0}~\dfrac{sin~x}{x}=1\Bigg]\\=\dfrac{-2~sinx}{~2~cos~x}\\=-tan~x

Question 7

 (i).~e^{\sqrt{cot~x}}

Sol :

  f(x)=e^{\sqrt{cot~x}}\\f(x+h)=e^{\sqrt{cot(x+h)}}\\\dfrac{d}{dx}f(x)=lim_{h\to0}~\dfrac{f(x+h)-f(x)}{h}\\=lim_{h\to0}~\dfrac{e^{\sqrt{cot(x+h)}}-e^{\sqrt{cot~x}}}{h}\\=lim_{h\to0}~\dfrac{e^{cot~x}\big(e^{\sqrt{cot(x+h)}-\sqrt{cot~x}}-1\big)}{h}\\=e^{cot~x}~lim_{h\to0}~\Bigg(\dfrac{e^{\sqrt{cot(x+h)}-\sqrt{cot~x}}-1}{\sqrt{cot(x+h)}-\sqrt{cot~x}}\Bigg)\times\Bigg(\dfrac{\sqrt{cot(x+h)}+\sqrt{cot~x}}{h}\Bigg)\\\Big[Since,~lim_{x\to0}~\dfrac{e^x-1}{x}=1~and~rationalising~numerator~\Big]\\=e^{\sqrt{cot~x}}~lim_{h\to0}~\dfrac{\sqrt{cot(x+h)}-\sqrt{cot~x}}{h}\times\dfrac{\sqrt{cot(x+h)}+\sqrt{cot~x}}{\sqrt{cot(x+h)}+\sqrt{cot~x}}\\=e^{\sqrt{cot~x}}~lim_{h\to0}~\dfrac{cot(x+h)-cot~x}{h\times{\sqrt{cot(x+h)}}+\sqrt{cot~x}}\\\Big[Since,~cot(A-B)=\dfrac{cotAcotB+1}{cotB-cotA}\Big]\\=e^{\sqrt{cot~x}}~lim_{h\to0}~\dfrac{\dfrac{cot(x+h)cot~x+1}{cot(x-x-h)}}{h\bigg(\sqrt{cot(x+h)}+\sqrt{cot~x}\bigg)}\\=e^{\sqrt{cot~x}}~lim_{h\to0}~\dfrac{\dfrac{cot(x+h)cot~x+1}{cot(-h)}}{h\bigg(\sqrt{cot(x+h)}+\sqrt{cot~x}\bigg)}\\=e^{\sqrt{cot~x}}~lim_{h\to0}~\dfrac{cot(x+h)cot~x+1}{cot~(-h)\times{h\bigg(\sqrt{cot(x+h)+\sqrt{cot~x}}\bigg)}}\\=-e^{\sqrt{cot~x}}~lim_{h\to0}~\dfrac{cot(x+h)cot~x+1}{\dfrac{h}{tan~h}\bigg(\sqrt{cot(x+h)+\sqrt{cot~x}}\bigg)}\\\Big[Since, lim_{x\to0}~\dfrac{tan~x}{x}=1\Big]\\=-\dfrac{e^{\sqrt{cot~x}}\times(cot^2x+1)}{2\sqrt{cot~x}}\\\Big[cot^2x+1=cosec^2x\Big]\\=-\dfrac{e^{\sqrt{cot~x}}\times{cosec^2x}}{2\sqrt{cot~x}}

 

 (ii).~x^{2}e^{x}

Sol :

 f(x)=x^2{e^x}\\f(x+h)=(x+h)^2e^{(x+h)}\\\dfrac{d}{dx}f(x)=lim_{h\to0}~\dfrac{f(x+h)-f(x)}{h}\\=lim_{h\to0}~\dfrac{(x+h)^2e^{(x+h)}-x^2{e^x}}{h}\\\big[(a+b)^2=(x+h)^2\big]\\=lim_{h\to0}~\Bigg(\dfrac{x^2e^{(x+h)}-x^2{e^x}}{h}+\dfrac{h^2e^{(x+h)}}{h}+\dfrac{2xhe^{(x+h)}}{h}\Bigg)\\=lim_{h\to0}~\Bigg(\dfrac{x^2e^{(x+h)}-x^2{e^x}}{h}+he^{(x+h)}+2xe^{(x+h)}\Bigg)\\=lim_{h\to0}~\Bigg({x^2{e^x}}\times\dfrac{(e^{x}-1)}{h}+he^{(x+h)}+2xe^{(x+h)}\Bigg)\\\Bigg[Since,lim_{x\to0}\dfrac{e^x-1}{x}\Bigg]\\=lim_{h\to0}~\Bigg({x^2{e^x}}+he^{(x+h)}+2xe^{(x+h)}\Bigg)\\={x^2{e^x}}+2xe^x\\=e^x({x^2}+2x)

 

 (iii).~log~cosec~x

Sol :

 f(x)=log~cosec~x\\f(x+h)=log~cosec~(x+h)\\\dfrac{d}{dx}f(x)=lim_{h\to0}~\dfrac{f(x+h)-f(x)}{h}\\=lim_{h\to0}~\dfrac{log~cosec~(x+h)-log~cosec~x}{h}\\=lim_{h\to0}~\Bigg\{\dfrac{log\dfrac{cosec~x+h}{cosec~x}}{h}\Bigg\}\\\Bigg[Since,~log~A-log~B=log\dfrac{A}{B}\Bigg]\\=lim_{h\to0}\dfrac{log\Bigg\{1+\dfrac{sin~x}{sin~(x+h)}-1\Bigg\}}{h}\\=lim_{h\to0}\dfrac{log\Bigg\{1+\dfrac{sin~x-sin~(x+h)}{sin~(x+h)}\Bigg\}}{h\times\dfrac{sin~x-sin~(x+h)}{sin~(x+h)}}\times{\dfrac{sin~x-sin~(x+h)}{sin~(x+h)}}\\\Bigg[Since,~lim_{x\to0}~\dfrac{log(1+x)}{x}=1\Bigg]\\=lim_{h\to0}{\dfrac{sin~x-sin~(x+h)}{sin~(x+h)\times{h}}}\\=lim_{h\to0}{\dfrac{2cos\bigg(\dfrac{x+x+h}{2}\bigg)sin\bigg(\dfrac{x-x-h}{2}\bigg)}{sin~(x+h)\times{h}}}\\\Bigg[Since,Sin~A-Sin~B=2cos\dfrac{A+B}{2}sin\dfrac{A-B}{2}\Bigg]\\=lim_{h\to0}{\dfrac{2cos\bigg(\dfrac{2x+h}{2}\bigg)sin\bigg(\dfrac{-h}{2}\bigg)}{sin~(x+h)\times{h}}}\\=lim_{h\to0}{\dfrac{2cos\bigg(\dfrac{2x+h}{2}\bigg)}{sin~(x+h)}}\times\dfrac{sin\bigg(\dfrac{-h}{2}\bigg)}{-\dfrac{h}{2}\times{-2}}\\\bigg[Since,lim_{x\to0}\dfrac{Sinx}{x}=1\bigg]\\=lim_{h\to0}{\dfrac{cos\bigg(\dfrac{2x+h}{2}\bigg)}{-sin~(x+h)}}\\=-cot~x

 

 (iv).~sin^{-1}(2x+3)

Sol :

 f(x)=sin^{-1}~(2x+3)\\f(x+h)=sin^{-1}~2(x+h)+3\\f(x+h)=sin^{-1}~2x+2h+3\\\dfrac{d}{dx}f(x)=lim_{h\to0}~\dfrac{f(x+h)-f(x)}{h}\\=lim_{h\to0}~\dfrac{sin^{-1}~(2x+2h+3)-sin^{-1}~(2x+3)}{h}\\=lim_{h\to0}~\dfrac{sin^{-1}\big[~(2x+2h+3)(\sqrt{1-(2x+3)^2})-(2x+3)\sqrt{1-(2x+2h+3)^2}~\big]}{h}\\\bigg[Since,Sin^{-1}x-Sin^{-1}y=Sin^{-1}[x\sqrt{1-y^2}-y\sqrt{1-x^2}]\bigg]\\=lim_{h\to0}~\dfrac{sin^{-1}\big[~(2x+2h+3)(\sqrt{1-(2x+3)^2})-(2x+3)\sqrt{1-(2x+2h+3)^2}~\big]}{{\big[~(2x+2h+3)(\sqrt{1-(2x+3)^2})-(2x+3)\sqrt{1-(2x+2h+3)^2}~\big]}}\times\dfrac{\big[~(2x+2h+3)(\sqrt{(2x+3)^2})-(2x+3)\sqrt{1-(2x+2h+3)^2}~\big]}{h}\\\big[Since,lim_{x\to0}\dfrac{Sin^{-1}x}{x}=1\big]\\=lim_{h\to0}~\dfrac{(2x+2h+3)(\sqrt{1-(2x+3)^2})-(2x+3)\sqrt{1-(2x+2h+3)^2}}{h}\\\big[on~rationalising~numerator\big]\\=lim_{h\to0}~\dfrac{(2x+2h+3)^2{\{1-(2x+3)^2\}}-(2x+3)^2{\{1-(2x+2h+3)^2}\}}{h\times{(2x+2h+3)(\sqrt{(2x+3)^2})+(2x+3)\sqrt{1-(2x+2h+3)^2}}}\\=lim_{h\to0}~\dfrac{[~(2x+3)^2+4h^2+4h(2x+3)](1-(2x+3)^2)-(2x+3)^2[1-(2x+3)^2-4h^2-4h(2x+3)~]}{h\times{(2x+2h+3)(\sqrt{(2x+3)^2})+(2x+3)\sqrt{1-(2x+2h+3)^2}}}\\=lim_{h\to0}~\dfrac{[~(2x+3)^2+4h^2+4h(2x+3)-(2x+3)^4-4h^2(2x+3)^2-(2x+3)^2+(2x+3)^4+4h^2(2x+3)^2+4h(2x+3)^3~]}{h\times{(2x+2h+3)(\sqrt{(2x+3)^2})+(2x+3)\sqrt{1-(2x+2h+3)^2}}}\\=lim_{h\to0}~\dfrac{4h[h+(2x+3)]}{h\times{(2x+2h+3)(\sqrt{(2x+3)^2})+(2x+3)\sqrt{1-(2x+2h+3)^2}}}\\=lim_{h\to0}~\dfrac{4(2x+3)}{h\times{(2x+2h+3)(\sqrt{(2x+3)^2})+(2x+3)\sqrt{1-(2x+2h+3)^2}}}\\=lim_{h\to0}\dfrac{4(2x+3)}{2(2x+3)\sqrt{1-(2x+3)^2}}\\=\dfrac{2}{\sqrt{1-(2x+3)^2}}

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