Differentiate Exercise 1.1 (1)

 Differentiate Exercise 1.1

(1)Differentiate w.r.t x

$(i)(x^3-2x-1)^5$

Solution- 

`\frac{dy}{dx}=5(x^3-2x-1)^4\frac{d}{dx}(x^3-2x-1)`

`=5(x^3-2x-1)^4\times\frac{d}{dx}(x^3-2x-1)`

`=5(x^3-2x-1)^4\times3x^2-2`

`=5(3x^2-2)(x^3-2x-1)^4`

${(ii)\left(2x^{\frac{3}{2}}-3x^{\frac{4}{3}}-5\right)^{\frac{5}{2}}}$

Solution-

`=\frac{5}{2}\left(2x^{\frac{3}{2}}-3x^{\frac{4}{3}}-5\right)^{\frac{3}{2}}\frac{d}{dx}\left(2x^{\frac{3}{2}}-3x^{\frac{4}{3}}-5\right)`

`=\frac{5}{2}\left(2x^{\frac{3}{2}}-3x^{\frac{4}{3}}-5\right)^{\frac{3}{2}}\left(\frac{3}{2}2x^{\frac{1}{2}}-3\frac{4}{3}3x^{\frac{1}{3}}\right)`

$=\frac{5}{2}\left(2x^{\frac{3}{2}}-3x^{\frac{4}{3}}-5\right)^{\frac{3}{2}}\left(3\sqrt{x}-12\sqrt[3]{x}\right)$

$(iii)\sqrt{x^2+x-7}$

Solution-

$=\frac{1}{\sqrt{x^2+x-7}}\times\frac{d}{dx}(x^2+x-7)$

$=\frac{1}{2\sqrt{x^2+x-7}}\times(2x+1)$

$=\frac{2x+1}{2\sqrt{x^2+4x-7}}$

$(vi)\sqrt{x^2+\sqrt{x^2+1}}$

Solution-

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

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

$=\frac{1}{2\sqrt{x^2+\sqrt{x^2+1}}}\left(2x+\frac{2x}{2\sqrt{x^2+1}}\right)$

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

$=\frac{1}{\sqrt{x^2+\sqrt{x^2+1}}}x\left(\frac{2\sqrt{x^2+1}+1}{\sqrt{x^2+1}}\right)$

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

$(v)\frac{3}{5\sqrt[3]{(2x^2-7x-5)^5}}$

Solution-

$=\frac{3}{5}\frac{1}{(2x^2-7x-5)^{\frac{3}{2}\times 5}}$

$=\frac{3}{5}\frac{1}{(2x^2-7x-5)^{\frac{15}{2}}}$

$=\frac{3}{5}(2x^2-7x-5)^{\frac{-15}{2}}$

Now, differentiate with respect to x,

$\frac{dy}{dx}=\frac{3}{5}\frac{d}{dx}(2x^2-7x-5)^{\frac{-15}{2}}$

$=\frac{3\times-15}{5\times2}(2x^2-7x-5)^{\frac{-17}{2}}\frac{d}{dx}(2x^2-7x-5)$

$=\frac{9}{2}\frac{-1}{(2x^2-7x-5)^{\frac{17}{2}}}(4x-7)$

$(vi)\frac{3}{5\sqrt[3]{(2x^2-7x-5)^5}}$

Solution-

$=\frac{3}{5}\frac{1}{(2x^2-7x-5)^{\frac{3}{2}\times 5}}$

$=\frac{3}{5}\frac{1}{(2x^2-7x-5)^{\frac{15}{2}}}$

$=\frac{3}{5}(2x^2-7x-5)^{\frac{-15}{2}}$

Now, differentiate with respect to x,

$\frac{dy}{dx}=\frac{3}{5}\frac{d}{dx}(2x^2-7x-5)^{\frac{-15}{2}}$

$=\frac{3\times-15}{5\times2}(2x^2-7x-5)^{\frac{-17}{2}}\frac{d}{dx}(2x^2-7x-5)$

$=\frac{9}{2}\frac{-1}{(2x^2-7x-5)^{\frac{17}{2}}}(4x-7)$

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Differentiate Exercise 1.1