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

1 Answer
Jim G.
Mar 4, 2017

#dy/dx=(2x^3-3)cosx+(6x^2+3x)sinx#

Explanation:

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

#"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#

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