Question

Find the differential of y=x³-2x²+5

Answer 1

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

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

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

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

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

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

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

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

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