Question

Find derivative using product rule y=(3-x²) (x³-x+1)

Answer 1

Explanation:

=d[(3-x^2) (x^3-x+1)]/dX

=(x^3-x+1) d(3-x^2)/dx + (3-x^2)d(x^3-x+1)/dx

=(x^3 - x +1) ( -2x) + (3-x^2) ( 3x^2 -1)

=-2x^4 +2x^2-2x + 9x^2 -3 - 3x^4 +x^2

= -5x^4+12x^2 -2x -3

Answer 2

Answer:

$\bigstar{\bold{\dfrac{d}{dx} (y)=-5x^{4}+12x^{2}-2x-3}}$

Explanation:

$\Large{\underline{\underline{\bf{Given:}}}}$

• y = (3 - x²) × (x³ - x  +1)

$\Large{\underline{\underline{\bf{To\:Find:}}}}$

$\sf{\dfrac{d}{dx}(y)}$

$\Large{\underline{\underline{\bf{Identities\:used:}}}}$

$\sf{\dfrac{d}{dx}(x^{n} )=nx^{n-1} }$

$\sf{\dfrac{d}{dx} (x)=1}$

$\sf{\dfrac{d}{dy} (k)=0}$

$\Large{\underline{\underline{\bf{Solution:}}}}$

➟ Here we have to find the derivative of the given function

➟ $\sf{\dfrac{d}{dx}(y)=\dfrac{d}{dx}\bigg((3-x^{2} )\times (x^{3} -x+1)\bigg)}$

➟  The product rule given by,

    $\sf{\dfrac{d}{dx} (uv)=u'v + v'u}$

➟ Using the rule,

    $\sf{\dfrac{d}{dx}(y)=\bigg(\dfrac{d}{dx}(3-x^{2})\times (x^{3}-x+1)+\dfrac{d}{dx}(x^{3}-x+1)\times (3-x^{2} )\bigg)}$

➟ Simplifying and using the identities,

    $\sf{\dfrac{d}{dx}(y)=-2x(x^{3} -x + 1)+3x^{2}-1(3-x^{2} )}$

   $\sf{\dfrac{d}{dx}(y)=-2x^{4} +2x^{2} -2x +9x^{2} -3x^{4}-3+x^{2} }$

➟ Cancelling the like terms,

    $\sf{\dfrac{d}{dx} (y)=-5x^{4}+12x^{2}-2x-3}$

➟ Hence the derivative of the given function is -5x⁴ + 12x² - 2x - 3

    $\boxed{\bold{\dfrac{d}{dx} (y)=-5x^{4}+12x^{2}-2x-3}}$

$\Large{\underline{\underline{\bf{More\:Identities:}}}}$

$\sf{\dfrac{d}{dy} (\sqrt{x} )=\dfrac{1}{2\sqrt{x} } }$

$\sf{\dfrac{d}{dy} (\dfrac{1}{x^{n} } )=\dfrac{-n}{x^{n+1} } }$