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Find the Y upon DX for the following functions why is equal to bracket X -1 bracket X -2 upon route x

Find The Y Upon DX For The Following Functions Why Is Equal To Bracket X 1 Bracket X 2 Upon Route X class=

Sagot :

Rewrite the equation as

[tex]y = (x-1) (x-2) x^{-1/2}[/tex]

Then by the product rule, the derivative is

[tex]\dfrac{dy}{dx} = (x-2) x^{-1/2} + (x-1) x^{-1/2} - \dfrac12 (x-1) (x-2) x^{-3/2}[/tex]

and we can factorize this as

[tex]\dfrac{dy}{dx} = \dfrac12 x^{-3/2} \left(2 (x-2) x^{3/2-1/2} + 2 (x-1) x^{3/2-1/2} - (x-1) (x-2)\right)[/tex]

[tex]\dfrac{dy}{dx} = \dfrac12 x^{-3/2} \left(2 (x-2) x + 2 (x-1) x - (x-1) (x-2)\right)[/tex]

[tex]\dfrac{dy}{dx} = \dfrac12 x^{-3/2} (3x^2 - 3x - 2)[/tex]

[tex]\dfrac{dy}{dx} = \dfrac{3x^2 - 3x - 2}{2x^{3/2}}[/tex]

and optionally expanded once more (if only to match the provided "Ans") to

[tex]\dfrac{dy}{dx} = \dfrac32 x^{2-3/2} - \dfrac32 x^{1-3/2} - x^{-3/2}[/tex]

[tex]\dfrac{dy}{dx} = \dfrac32 x^{1/2} - \dfrac32 x^{-1/2} - x^{-3/2}[/tex]

[tex]\dfrac{dy}{dx} = \dfrac32 \sqrt x - \dfrac3{2\sqrt x} - \dfrac1{\sqrt{x^3}}[/tex]

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