Y = 2x^3sinx - 3xcosx Find the derivative of the equation?

1 Answer
Mar 4, 2017

dy/dx=(2x^3-3)cosx+(6x^2+3x)sinxdydx=(2x33)cosx+(6x2+3x)sinx

Explanation:

Each term has to be differentiated using the color(blue)"product rule"product rule

"Given "f(x)=g(x).h(x)" then"Given f(x)=g(x).h(x) then

color(red)(bar(ul(|color(white)(2/2)color(black)(f'(x)=g(x)h'(x)+h(x)g'(x))color(white)(2/2)|)))larr" product rule"

color(magenta)"First term"

"for "f(x)= 2x^3sinx

"here "g(x)=2x^3rArrg'(x)=6x^2

h(x)=sinxrArrh'(x)=cosx

rArrf'(x)=2x^3(cosx)+6x^2(sinx)to(1)

color(magenta)"Second term"

f(x)=3xcosx

"here "g(x)=3xrArrg'(x)=3

h(x)=cosxrArrh'(x)=-sinx

rArrf'(x)=3x(-sinx)+3cosxto(2)

"Combining differentiated terms, that is " (1)-(2)

dy/dx=2x^3cosx+6x^2sinx+3xsinx-3cosx

.>rArrdy/dx=(2x^3-3)cosx+(6x^2+3x)sinx