Answer to Question #155195 in Calculus for Phyroe

Question #155195

Find the differential of y=x³-2x²+5


1
Expert's answer
2021-01-21T11:24:15-0500

To  find  :  Differential  of  y=x32x2+5.Prerequisite:      1)  ddx(c)=0,  where  c  is  constant.To\;find\;:\;Differential\;of\;y=x^3-2x^2+5.\\ Pre-requisite:\;\;\;\\ 1)\;\dfrac{d}{dx}(c)=0,\;where\;c\;is\;constant.\\


2)  ddx(cf(x))=  cddx(f(x))2)\;\dfrac{d}{dx}(cf(x))=\;c\dfrac{d}{dx}(f(x))


3)  ddx(xn)=nxn1,    where  nZ.3)\;\dfrac{d}{dx}(x^n)=nx^{n-1},\;\;where\;n \in Z.


ddx(y)=ddx(x32x2+5)\therefore \dfrac{d}{dx}(y)=\dfrac{d}{dx}(x^3-2x^2+5)


                            =ddx(x3)ddx(2x2)+ddx(5)\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\dfrac{d}{dx}(x^3)-\dfrac{d}{dx}(2x^2)+\dfrac{d}{dx}(5)


                            =3(x31)2(2)(x21)+0\;\;\;\;\;\;\;\;\;\;\;\;\;\;=3(x^{3-1})-2(2)(x^{2-1})+0


                            =3x24x\;\;\;\;\;\;\;\;\;\;\;\;\;\;=3x^{2}-4x


dydx=3x24x\therefore \dfrac{dy}{dx}=3x^2-4x


The  differential  has  the  following  form:dy=(3x24x)dx\therefore The \; differential\; has\;the\; following\; form : dy=(3x^2-4x)dx




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